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Research Article Evaluation of Density Matrix and Helmholtz Free Energy for Harmonic Oscillator Asymmetric Potential via Feynmans Approach Piyarut Moonsri 1 and Artit Hutem 2 1 Chemistry Division, Faculty of Science and Technology, Phetchabun Rajabhat University, Phetchabun 67000, ailand 2 Physics Division, Faculty of Science and Technology, Phetchabun Rajabhat University, Phetchabun 67000, ailand Correspondence should be addressed to Artit Hutem; [email protected] Received 18 August 2014; Revised 8 October 2014; Accepted 9 October 2014; Published 2 November 2014 Academic Editor: Miquel Sol` a Copyright © 2014 P. Moonsri and A. Hutem. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We apply a Feynmans technique for calculation of a canonical density matrix of a single particle under harmonic oscillator asymmetric potential and solving the Bloch equation of the statistical mechanics system. e density matrix ( P ) and kinetic energy per unit length ( ) can be directly evaluated from the solving solutions. From the evaluation, it was found that both of the density matrix and kinetic energy per unit length depended on the parameter of the value of asymmetric potential (), the value of axes-shiſt potential (), and temperature (T). Comparison of the Helmholtz free energy was derived by the Feynmans technique and the path-integral method. e results illustrated are slightly different. 1. Introduction We consider a quantum mechanical ensemble of identical systems in situations where the description is incomplete. When the description of the system is incomplete, the state of the system is described by means of what is called a density matrix. In the statistical point of view, a state of quantum ensemble is described by a density operator (or statistical operator) P ; the corresponding matrix is called a density matrix. A knowledge of this matrix enables us to calculate the mean value of physical quantity of the system and also the probabilities of various values of such a quantity. March and Murray used the Bloch equation to calculate the canonical density matrix in a single particle framework which may be related directly to the generalized canonical density matrix, containing the Fermi-Dirac function [1]. Howard et al. used the Bloch equation evaluate the electron density P and kinetic energy () for the one dimensional sech 2 () potential. Howard et al. used tool employed is the Slater sum, which satisfies a partial differential equation [2]. For calculation of the canonical density matrix in terms of the Slater sum and kinetic energy per unit length in terms of the Slater sum see March and Nieto [3]. To study problems of the canonical ensemble, we focus on the calculation the canonical density matrix and the Helmholtz free energy of a single particle under asymmetric harmonic oscillator potential. e scheme of the paper is as follows. In Section 2, there is detailing with the density matrix. In Section 3, we show the calculation of the canonical density matrix (or charge bond- order matrix) and the Helmholtz free energy by Feynmans approach. In Section 4, we representation of the evaluation of the density matrix and the Helmholtz free energy by the path-integral method (()) method. In Section 5 we make a presentation of numerical results for the Helmholtz free energy of a single particle under asymmetric harmonic oscillator potential. e conclusion and discussion are given in Section 6. 2. Statistical Density Matrix A density matrix is a matrix that describes a quantum system in a mixed state, a statistical ensemble of several quantum states. is should be contrasted with a single state Hindawi Publishing Corporation Advances in Physical Chemistry Volume 2014, Article ID 912054, 9 pages http://dx.doi.org/10.1155/2014/912054

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Page 1: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Research ArticleEvaluation of Density Matrix and Helmholtz FreeEnergy for Harmonic Oscillator Asymmetric Potential viaFeynmans Approach

Piyarut Moonsri1 and Artit Hutem2

1 Chemistry Division Faculty of Science and Technology Phetchabun Rajabhat University Phetchabun 67000 Thailand2 Physics Division Faculty of Science and Technology Phetchabun Rajabhat University Phetchabun 67000 Thailand

Correspondence should be addressed to Artit Hutem magoohootemyahoocom

Received 18 August 2014 Revised 8 October 2014 Accepted 9 October 2014 Published 2 November 2014

Academic Editor Miquel Sola

Copyright copy 2014 P Moonsri and A Hutem This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We apply a Feynmans technique for calculation of a canonical density matrix of a single particle under harmonic oscillatorasymmetric potential and solving the Bloch equation of the statistical mechanics system The density matrix (P

119906) and kinetic

energy per unit length (120591119871) can be directly evaluated from the solving solutions From the evaluation it was found that both of the

density matrix and kinetic energy per unit length depended on the parameter of the value of asymmetric potential (120582) the value ofaxes-shift potential (119892) and temperature (T) Comparison of the Helmholtz free energy was derived by the Feynmans techniqueand the path-integral method The results illustrated are slightly different

1 Introduction

We consider a quantum mechanical ensemble of identicalsystems in situations where the description is incompleteWhen the description of the system is incomplete the state ofthe system is described by means of what is called a densitymatrix In the statistical point of view a state of quantumensemble is described by a density operator (or statisticaloperator) P

119906 the corresponding matrix is called a density

matrix A knowledge of this matrix enables us to calculatethe mean value of physical quantity of the system and alsothe probabilities of various values of such a quantity

March and Murray used the Bloch equation to calculatethe canonical density matrix in a single particle frameworkwhich may be related directly to the generalized canonicaldensity matrix containing the Fermi-Dirac function [1]Howard et al used the Bloch equation evaluate the electrondensity P

119906and kinetic energy 120591

119871(119909) for the one dimensional

sech2(119909) potential Howard et al used tool employed is theSlater sum which satisfies a partial differential equation [2]

For calculation of the canonical density matrix in termsof the Slater sum and kinetic energy per unit length in terms

of the Slater sum see March and Nieto [3] To study problemsof the canonical ensemble we focus on the calculation thecanonical density matrix and the Helmholtz free energyof a single particle under asymmetric harmonic oscillatorpotential

The scheme of the paper is as follows In Section 2 thereis detailing with the density matrix In Section 3 we show thecalculation of the canonical density matrix (or charge bond-order matrix) and the Helmholtz free energy by Feynmansapproach In Section 4 we representation of the evaluationof the density matrix and the Helmholtz free energy bythe path-integral method (120581(119910)) method In Section 5 wemake a presentation of numerical results for the Helmholtzfree energy of a single particle under asymmetric harmonicoscillator potential The conclusion and discussion are givenin Section 6

2 Statistical Density Matrix

A density matrix is a matrix that describes a quantumsystem in a mixed state a statistical ensemble of severalquantum states This should be contrasted with a single state

Hindawi Publishing CorporationAdvances in Physical ChemistryVolume 2014 Article ID 912054 9 pageshttpdxdoiorg1011552014912054

2 Advances in Physical Chemistry

vector that describes a quantum system in a pure state Thedensity matrix is the quantum-mechanical analogue to aphase-space probability measure (probability distribution ofposition and momentum) in classical statistical mechanicsExplicitly suppose a quantum system may be found in state10038161003816100381610038161205951⟩ with probability 119875

1or it may be found in state |120595

2⟩ with

probability1198752or it may be found in state |120595

3⟩with probability

1198753 and so on The density matrix for this system is

P = sum119899

119875119899

1003816100381610038161003816120595119899⟩ ⟨1205951198991003816100381610038161003816 (1)

where (|120595119899⟩) need not be orthogonal and sum

119899119875119899= 1 By

choosing an orthonormal basis (|119906119895⟩) one may resolve the

density operator into the density matrix whose elements are

P119894119895= sum119899

119875119899⟨119906119895| 120595119899⟩ ⟨120595

119899| 119906119894⟩ (2)

For an operator A (which describes an observable A of thesystem) the expectation value ⟨A⟩ is given by

⟨A⟩ = sum119899

119875119899⟨120595119899

10038161003816100381610038161003816A10038161003816100381610038161003816120595119899⟩ = sum

119894

⟨119906119894

10038161003816100381610038161003816PA

10038161003816100381610038161003816119906119894⟩ = tr (PA)

(3)

In the canonical ensemble the probability that the system isin state |119899⟩ is given by

119875119899=

119890minus120573E119899

sum119899119890120573E119899

(4)

and the thermal average of an arbitrary operator A is

⟨A⟩ = sum119899

⟨11989910038161003816100381610038161003816A10038161003816100381610038161003816119899⟩(

119890minus120573E119899

sum119899119890120573E119899

)

=1

Q119873(120573)

sum119899

⟨11989910038161003816100381610038161003816A10038161003816100381610038161003816119899⟩ 119890

minus120573E119899

(5)

The operator

P =1

Q119873(120573)

sum119899

|119899⟩ ⟨119899| 119890minus120573E119899 (6)

is commonly referred to as density operator and its matrixrepresentation as the density matrix Here Q

119873(120573) is the

partition function

3 A Linear Harmonic Oscillator AsymmetricPotential by Feynmans Approach

Next we consider the case of a linear harmonics oscillator ofasymmetric potential whose Hamiltonian is given by [4]

H119878=2

2119898+1

21198981205962

1199092

+ radic119898120596

ℎ120596

2120582(119909 minus 119892radic

119898120596) (7)

we note that this is a time-independent Hamiltonian ofharmonic oscillator asymmetric axes-shift potential system

Substituting in the time-independent Hamiltonian equation(7) into the Bloch equation [3] we can rewrite the differentialequation completely in terms of P

119906(120585 120585 119891) as

minus120597P119906

120597120573= minus

ℎ2

2119898

1205972P119906

1205971199092+1

21198981205962

1199092

P119906

+ radic119898120596

ℎ120596

2120582(119909 minus 119892radic

119898120596) P

119906

(8)

We define some new dimensionless variable The positionvariable 119909 = 119909 is replaced with the dimensionless variable120585

120585 = radic119898120596

ℎ119909 119909

2

=ℎ

1198981205961205852

120573 =2119891

ℎ120596 (9)

Substituting in for 119909 in terms of 120585 we can rewrite thedifferential equation completely in terms of 120585 and P

119906as

minus120597P119906

120597119891= minus

1205972P119906

1205971205852+ 1205852

P119906+ 120582120585P

119906minus 120582119892P

119906 (10)

Knowing the Gaussian dependence of the density matrix P119906

in case of the free particle we try the following form

P119906(120585 120585 119891) asymp 119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

120582119892P119906= 120582119892119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

minus120597P119906

120597119891= [120585

2119889119886 (119891)

119889119891+ 120585

119889119887 (119891)

119889119891+119889119888 (119891)

119889119891]

times 119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

minus1205972P

119906

1205971205852= (2119886 (119891) minus 4119886

2

(119891) 1205852

minus 4119886 (119891) 119887 (119891) 120585 minus 1198872

(119891))

times 119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

1205852

P119906= 1205852

119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

120582120585P119906= 120582120585119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

(11)

Substituting expansion equation (11) into (10) leads us to theequation

119889119886 (119891)

1198891198911205852

+119889119887 (119891)

119889119891120585 +

119889119888 (119891)

119889119891

= (1 minus 41198862

(119891)) 1205852

+ (120582 minus 4119886 (119891) 119887 (119891)) 120585

+ (2119886 (119891) minus 120582119892 minus 1198872

(119891))

(12)

Advances in Physical Chemistry 3

Equating the coefficients for 1205852 120585 and 1205850 = 1 on both sides ofthe last equation (12) gives

119889119886 (119891)

119889119891= (1 minus 4119886

2

(119891)) (13)

119889119887 (119891)

119889119891= (120582 minus 4119886 (119891) 119887 (119891)) (14)

119889119888 (119891)

119889119891= (2119886 (119891) minus 120582119892 minus 119887

2

(119891)) (15)

respectively Computing (13) is integrated to give

119886 (119891) =1

2coth (2119891) (16)

Equation (14) is the linear first-order ordinary differentialequation to give [5 6]

120582 =119889119887 (119891)

119889119891+ 2 coth (2119891) 119887 (119891)

119887 (119891) = 119890minusint 2 coth(2119891)119889119891

(int120582119890int 2 coth(2119891)119889119891

119889119891 + 119860)

119887 (119891) =120582

2coth (2119891) + 119860csch (2119891)

(17)

Equation (15) is integrated with respect to 119891 as119889119888 (119891)

119889119891= (coth (2119891) minus 120582119892 minus (120582

2coth (2119891) +119860csch (2119891))

2

)

119888 (119891) = int coth (2119891) 119889119891 minus int120582119892119889119891

minus int(120582

2coth (2119891) + 119860csch (2119891))

2

119889119891

119888 (119891) =1

8(minus2119891120582 (4119892 + 120582) + (4119860

2

+ 1205822

) coth (2119891)

+ 4 ln (sinh (2119891))

+2119860120582csch (119891) sech (119891) minus ln119861) (18)

where 119861 120582 and 119892 are constants Substituting 119886(119891) 119887(119891) and119888(119891) of (16) (17) and (18) respectively into (11) we mayrewrite density matrix asP119906

=119861

radicsinh (2119891)

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

)) coth (2119891)

+(120585119860 +119860120582

2) csch (2119891) minus

119891120582

4(4119892 + 120582) ])

(19)

Thus the Taylor series expansion of sinh(2119891) coth(2119891) about119891 rarr 0 is given by

lim119891rarr0

sinh (2119891) asymp 2119891 lim119891rarr0

coth (2119891) asymp 1

2119891 (20)

we obtain the density matrix final form

P119906asymp

119861

radic2119891

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

))1

2119891

+(120585119860 +119860120582

2)1

2119891minus120582119891

4(4119892 + 120582)])

(21)

The initial condition is

P119906= 120575 (119909 minus ) for 119891 = 0 (22)

or

P119906= radic

119898120596

ℎ120575 (120585 minus 120585) for 119891 = 0 (23)

For low 119891 = (ℎ1205962) 120573 = ℎ1205962119896119861119879 (high temperature) the

particle should act almost like a free particle as its probablekinetic energy is so high Therefore we expect that for low119891 the density matrix for a harmonic oscillator will be givenapproximately by [7]

P119906119891119903119890

(120585 120585 119891) asymp radic119898120596

4120587ℎ119891119890minus(120585minus

120585)2

4119891

(24)

Comparing the value of 119860 119861 in (21) with (24) identifiesthe solution as

119861

radic2119891= radic

119898120596

4120587ℎ119891997888rarr 119861 = radic

119898120596

2120587ℎ

1205852

minus 2120585 120585 + 1205852

= 1205852

+ (120582 + 2119860) 120585

+ (1198602

+1205822

4+ 119860120582 minus 4119892120582119891

2

minus 1205822

1198912

)

119860 = minus( 120585 +120582

2)

(25)

Substituting (25) into (19) and setting 120585 = 120585 produce

P119906(120585 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus[(120585+(1205822))

2

((cosh(2119891)minus1) sinh(2119891))minus(1205821198914)(120582+4119892)]

(26)

Using the identity of trigonometric tanh(1199092) =

(cosh(119909) minus 1) sinh(119909) and setting 120585 = radic119898120596ℎ119909 into

4 Advances in Physical Chemistry

(26) we can rewrite the canonical density matrix or chargebond-order matrix [8] as

P119906(119909 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus(1198981205964ℎ)[(2119909+radicℎ119898120596120582)

2 tanh(119891)minus(120582119891ℎ119898120596)(120582+4119892)]

(27)

The obtained results are neatly summarized in Figure 1 Itis the probability density for finding the system at 119909 we noteas a Gaussian distribution in 119909 Let us define four new thevariables 120574 120578 120583 and 120601 as function of 119891 by

120574 (119891) = radic119898120596

2120587ℎ sinh (2119891) 120578 (119891) =

119898120596

4ℎtanh (119891)

120583 (119891) =120582119891

4(120582 + 4119892) 120601 = radic

119898120596120582

(28)

we can rewrite the density matrix completely in terms of thenew variable (120574 120578 120583 120601) as

P119906(119909 119891) asymp 120574119890

minus(120578(2119909+120601)2

minus120583)

(29)

The expectation value of 1199092 in the density matrix P119906(119909 119891)

can be determined from the definition of the expectationvalues as

⟨1199092

⟩ =intinfin

minusinfin

1199092P119906(119909 119891) 119889119909

intinfin

minusinfin

P119906(119909 119891) 119889119909

=120574radic120587119890

120583

1612057832

2radic120578

120574radic120587119890120583(1 + 2120578120601

2

)

⟨1199092

⟩ =ℎ

2119898120596coth(

ℎ120596120573

2) +

41198981205961205822

(30)

The average of the potential energy (119864119904) is thus

⟨119864119904⟩ =

1

21198981205962

⟨1199092

⟨119864119904⟩ =

ℎ120596

4coth(

ℎ120596120573

2) +

ℎ120596

81205822

(31)

Thekinetic energy per unit length 120591119871(119909 119891) is defined from

the canonical density matrix P119906(119909 119891) as [3]

120591119871(119909 119891) = minus

ℎ2

2119898

1205972

1205971199092P119906(119909 119891) (32)

By substituting (29) into (32) we can write the kinetic energyper unit length completely as follows

120591119871(119909 119879) =

ℎ120596

2tanh( ℎ120596

2119896119861119879)radic

119898120596

2120587ℎ sinh (ℎ120596119896119861119879)

times 119890(120582ℎ1205968119896

119861119879)(120582+4119892)

times [

[

2 minus119898120596

ℎtanh( ℎ120596

2119896119861119879)(2119909 + radic

119898120596120582)

2

]

]

times 119890minus(1198981205964ℎ) tanh(ℎ1205962119896

119861119879)(2119909+radicℎ119898120596120582)

2

(33)

The obtained results from (33) are summarized in Figure 2Now the Helmholtz free energy of the harmonic oscilla-

tor asymmetric potential system (119865New) is given by

119865New = minus1

120573lnQ

119873(120573) (34)

or

Q119873(120573) = 119890

minus120573119865New (35)

The density matrix of the system in the integrationrepresentation will be given by

119890minus120573119865new = int

infin

minusinfin

P119906(119909 120573) 119889119909 (36)

Substituting (29) into (36) we obtain for the partitionfunction of the particle in harmonic oscillator asymmetricpotential system

119890minus120573119865new = int

infin

minusinfin

120574119890minus(120578(2119909+120601)

2

minus120583)

119889119909

=120574

2radic120587

120578

=119890(1205821198914)(120582+4119892)

2 sinh (119891)

(37)

Thus we have the following results

119865new =1

120573ln (2 sinh (119891)) minus

120582119891

4120573(120582 + 4119892) (38)

This is the Helmholtz free energy for a one-dimensionalharmonic oscillator asymmetric potential (119865new) as alreadyderived When we write the Helmholtz free energy in newvariable is 119866new = 2119865newℎ120596 119891 = ℎ1205961205732 into (38) we obtainthat

119866new =1

119891ln (2 sinh (119891)) minus 120582

4(120582 + 4119892) (39)

One may summarize the numerical results of Helmholtzfree energy for harmonic oscillator asymmetric potential by

Advances in Physical Chemistry 5

x

05

10

15

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

120588u(x120573)

(a) Density Matrix 120582 = 1 119892 = 1

05

10

15

20

x

155 100minus5minus10minus15

1205821 = 01

1205822 = 20

1205823 = 35

120588u(x120573)

(b) Density Matrix 119892 = 1 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

g1 = 01

g2 = 25

g3 = 55

120588u(x120573)

(c) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

x

155 100minus5minus10minus15

g1 = minus01

g2 = minus45

g3 = minus85

120588u(x120573)

(d) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

1205821 = minus01

1205822 = minus40

1205823 = minus75

120588u(x120573)

(e) Density Matrix 119892 = 1 119879 = 15

Figure 1 Plot of the canonical density matrix of a single particle for harmonic oscillator asymmetric potential system

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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International Journal of

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Journal of

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Advances in

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CatalystsJournal of

Page 2: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

2 Advances in Physical Chemistry

vector that describes a quantum system in a pure state Thedensity matrix is the quantum-mechanical analogue to aphase-space probability measure (probability distribution ofposition and momentum) in classical statistical mechanicsExplicitly suppose a quantum system may be found in state10038161003816100381610038161205951⟩ with probability 119875

1or it may be found in state |120595

2⟩ with

probability1198752or it may be found in state |120595

3⟩with probability

1198753 and so on The density matrix for this system is

P = sum119899

119875119899

1003816100381610038161003816120595119899⟩ ⟨1205951198991003816100381610038161003816 (1)

where (|120595119899⟩) need not be orthogonal and sum

119899119875119899= 1 By

choosing an orthonormal basis (|119906119895⟩) one may resolve the

density operator into the density matrix whose elements are

P119894119895= sum119899

119875119899⟨119906119895| 120595119899⟩ ⟨120595

119899| 119906119894⟩ (2)

For an operator A (which describes an observable A of thesystem) the expectation value ⟨A⟩ is given by

⟨A⟩ = sum119899

119875119899⟨120595119899

10038161003816100381610038161003816A10038161003816100381610038161003816120595119899⟩ = sum

119894

⟨119906119894

10038161003816100381610038161003816PA

10038161003816100381610038161003816119906119894⟩ = tr (PA)

(3)

In the canonical ensemble the probability that the system isin state |119899⟩ is given by

119875119899=

119890minus120573E119899

sum119899119890120573E119899

(4)

and the thermal average of an arbitrary operator A is

⟨A⟩ = sum119899

⟨11989910038161003816100381610038161003816A10038161003816100381610038161003816119899⟩(

119890minus120573E119899

sum119899119890120573E119899

)

=1

Q119873(120573)

sum119899

⟨11989910038161003816100381610038161003816A10038161003816100381610038161003816119899⟩ 119890

minus120573E119899

(5)

The operator

P =1

Q119873(120573)

sum119899

|119899⟩ ⟨119899| 119890minus120573E119899 (6)

is commonly referred to as density operator and its matrixrepresentation as the density matrix Here Q

119873(120573) is the

partition function

3 A Linear Harmonic Oscillator AsymmetricPotential by Feynmans Approach

Next we consider the case of a linear harmonics oscillator ofasymmetric potential whose Hamiltonian is given by [4]

H119878=2

2119898+1

21198981205962

1199092

+ radic119898120596

ℎ120596

2120582(119909 minus 119892radic

119898120596) (7)

we note that this is a time-independent Hamiltonian ofharmonic oscillator asymmetric axes-shift potential system

Substituting in the time-independent Hamiltonian equation(7) into the Bloch equation [3] we can rewrite the differentialequation completely in terms of P

119906(120585 120585 119891) as

minus120597P119906

120597120573= minus

ℎ2

2119898

1205972P119906

1205971199092+1

21198981205962

1199092

P119906

+ radic119898120596

ℎ120596

2120582(119909 minus 119892radic

119898120596) P

119906

(8)

We define some new dimensionless variable The positionvariable 119909 = 119909 is replaced with the dimensionless variable120585

120585 = radic119898120596

ℎ119909 119909

2

=ℎ

1198981205961205852

120573 =2119891

ℎ120596 (9)

Substituting in for 119909 in terms of 120585 we can rewrite thedifferential equation completely in terms of 120585 and P

119906as

minus120597P119906

120597119891= minus

1205972P119906

1205971205852+ 1205852

P119906+ 120582120585P

119906minus 120582119892P

119906 (10)

Knowing the Gaussian dependence of the density matrix P119906

in case of the free particle we try the following form

P119906(120585 120585 119891) asymp 119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

120582119892P119906= 120582119892119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

minus120597P119906

120597119891= [120585

2119889119886 (119891)

119889119891+ 120585

119889119887 (119891)

119889119891+119889119888 (119891)

119889119891]

times 119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

minus1205972P

119906

1205971205852= (2119886 (119891) minus 4119886

2

(119891) 1205852

minus 4119886 (119891) 119887 (119891) 120585 minus 1198872

(119891))

times 119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

1205852

P119906= 1205852

119890minus(119886(119891)120585

2

+119887(119891)120585+119888(119891))

120582120585P119906= 120582120585119890

minus(119886(119891)1205852

+119887(119891)120585+119888(119891))

(11)

Substituting expansion equation (11) into (10) leads us to theequation

119889119886 (119891)

1198891198911205852

+119889119887 (119891)

119889119891120585 +

119889119888 (119891)

119889119891

= (1 minus 41198862

(119891)) 1205852

+ (120582 minus 4119886 (119891) 119887 (119891)) 120585

+ (2119886 (119891) minus 120582119892 minus 1198872

(119891))

(12)

Advances in Physical Chemistry 3

Equating the coefficients for 1205852 120585 and 1205850 = 1 on both sides ofthe last equation (12) gives

119889119886 (119891)

119889119891= (1 minus 4119886

2

(119891)) (13)

119889119887 (119891)

119889119891= (120582 minus 4119886 (119891) 119887 (119891)) (14)

119889119888 (119891)

119889119891= (2119886 (119891) minus 120582119892 minus 119887

2

(119891)) (15)

respectively Computing (13) is integrated to give

119886 (119891) =1

2coth (2119891) (16)

Equation (14) is the linear first-order ordinary differentialequation to give [5 6]

120582 =119889119887 (119891)

119889119891+ 2 coth (2119891) 119887 (119891)

119887 (119891) = 119890minusint 2 coth(2119891)119889119891

(int120582119890int 2 coth(2119891)119889119891

119889119891 + 119860)

119887 (119891) =120582

2coth (2119891) + 119860csch (2119891)

(17)

Equation (15) is integrated with respect to 119891 as119889119888 (119891)

119889119891= (coth (2119891) minus 120582119892 minus (120582

2coth (2119891) +119860csch (2119891))

2

)

119888 (119891) = int coth (2119891) 119889119891 minus int120582119892119889119891

minus int(120582

2coth (2119891) + 119860csch (2119891))

2

119889119891

119888 (119891) =1

8(minus2119891120582 (4119892 + 120582) + (4119860

2

+ 1205822

) coth (2119891)

+ 4 ln (sinh (2119891))

+2119860120582csch (119891) sech (119891) minus ln119861) (18)

where 119861 120582 and 119892 are constants Substituting 119886(119891) 119887(119891) and119888(119891) of (16) (17) and (18) respectively into (11) we mayrewrite density matrix asP119906

=119861

radicsinh (2119891)

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

)) coth (2119891)

+(120585119860 +119860120582

2) csch (2119891) minus

119891120582

4(4119892 + 120582) ])

(19)

Thus the Taylor series expansion of sinh(2119891) coth(2119891) about119891 rarr 0 is given by

lim119891rarr0

sinh (2119891) asymp 2119891 lim119891rarr0

coth (2119891) asymp 1

2119891 (20)

we obtain the density matrix final form

P119906asymp

119861

radic2119891

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

))1

2119891

+(120585119860 +119860120582

2)1

2119891minus120582119891

4(4119892 + 120582)])

(21)

The initial condition is

P119906= 120575 (119909 minus ) for 119891 = 0 (22)

or

P119906= radic

119898120596

ℎ120575 (120585 minus 120585) for 119891 = 0 (23)

For low 119891 = (ℎ1205962) 120573 = ℎ1205962119896119861119879 (high temperature) the

particle should act almost like a free particle as its probablekinetic energy is so high Therefore we expect that for low119891 the density matrix for a harmonic oscillator will be givenapproximately by [7]

P119906119891119903119890

(120585 120585 119891) asymp radic119898120596

4120587ℎ119891119890minus(120585minus

120585)2

4119891

(24)

Comparing the value of 119860 119861 in (21) with (24) identifiesthe solution as

119861

radic2119891= radic

119898120596

4120587ℎ119891997888rarr 119861 = radic

119898120596

2120587ℎ

1205852

minus 2120585 120585 + 1205852

= 1205852

+ (120582 + 2119860) 120585

+ (1198602

+1205822

4+ 119860120582 minus 4119892120582119891

2

minus 1205822

1198912

)

119860 = minus( 120585 +120582

2)

(25)

Substituting (25) into (19) and setting 120585 = 120585 produce

P119906(120585 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus[(120585+(1205822))

2

((cosh(2119891)minus1) sinh(2119891))minus(1205821198914)(120582+4119892)]

(26)

Using the identity of trigonometric tanh(1199092) =

(cosh(119909) minus 1) sinh(119909) and setting 120585 = radic119898120596ℎ119909 into

4 Advances in Physical Chemistry

(26) we can rewrite the canonical density matrix or chargebond-order matrix [8] as

P119906(119909 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus(1198981205964ℎ)[(2119909+radicℎ119898120596120582)

2 tanh(119891)minus(120582119891ℎ119898120596)(120582+4119892)]

(27)

The obtained results are neatly summarized in Figure 1 Itis the probability density for finding the system at 119909 we noteas a Gaussian distribution in 119909 Let us define four new thevariables 120574 120578 120583 and 120601 as function of 119891 by

120574 (119891) = radic119898120596

2120587ℎ sinh (2119891) 120578 (119891) =

119898120596

4ℎtanh (119891)

120583 (119891) =120582119891

4(120582 + 4119892) 120601 = radic

119898120596120582

(28)

we can rewrite the density matrix completely in terms of thenew variable (120574 120578 120583 120601) as

P119906(119909 119891) asymp 120574119890

minus(120578(2119909+120601)2

minus120583)

(29)

The expectation value of 1199092 in the density matrix P119906(119909 119891)

can be determined from the definition of the expectationvalues as

⟨1199092

⟩ =intinfin

minusinfin

1199092P119906(119909 119891) 119889119909

intinfin

minusinfin

P119906(119909 119891) 119889119909

=120574radic120587119890

120583

1612057832

2radic120578

120574radic120587119890120583(1 + 2120578120601

2

)

⟨1199092

⟩ =ℎ

2119898120596coth(

ℎ120596120573

2) +

41198981205961205822

(30)

The average of the potential energy (119864119904) is thus

⟨119864119904⟩ =

1

21198981205962

⟨1199092

⟨119864119904⟩ =

ℎ120596

4coth(

ℎ120596120573

2) +

ℎ120596

81205822

(31)

Thekinetic energy per unit length 120591119871(119909 119891) is defined from

the canonical density matrix P119906(119909 119891) as [3]

120591119871(119909 119891) = minus

ℎ2

2119898

1205972

1205971199092P119906(119909 119891) (32)

By substituting (29) into (32) we can write the kinetic energyper unit length completely as follows

120591119871(119909 119879) =

ℎ120596

2tanh( ℎ120596

2119896119861119879)radic

119898120596

2120587ℎ sinh (ℎ120596119896119861119879)

times 119890(120582ℎ1205968119896

119861119879)(120582+4119892)

times [

[

2 minus119898120596

ℎtanh( ℎ120596

2119896119861119879)(2119909 + radic

119898120596120582)

2

]

]

times 119890minus(1198981205964ℎ) tanh(ℎ1205962119896

119861119879)(2119909+radicℎ119898120596120582)

2

(33)

The obtained results from (33) are summarized in Figure 2Now the Helmholtz free energy of the harmonic oscilla-

tor asymmetric potential system (119865New) is given by

119865New = minus1

120573lnQ

119873(120573) (34)

or

Q119873(120573) = 119890

minus120573119865New (35)

The density matrix of the system in the integrationrepresentation will be given by

119890minus120573119865new = int

infin

minusinfin

P119906(119909 120573) 119889119909 (36)

Substituting (29) into (36) we obtain for the partitionfunction of the particle in harmonic oscillator asymmetricpotential system

119890minus120573119865new = int

infin

minusinfin

120574119890minus(120578(2119909+120601)

2

minus120583)

119889119909

=120574

2radic120587

120578

=119890(1205821198914)(120582+4119892)

2 sinh (119891)

(37)

Thus we have the following results

119865new =1

120573ln (2 sinh (119891)) minus

120582119891

4120573(120582 + 4119892) (38)

This is the Helmholtz free energy for a one-dimensionalharmonic oscillator asymmetric potential (119865new) as alreadyderived When we write the Helmholtz free energy in newvariable is 119866new = 2119865newℎ120596 119891 = ℎ1205961205732 into (38) we obtainthat

119866new =1

119891ln (2 sinh (119891)) minus 120582

4(120582 + 4119892) (39)

One may summarize the numerical results of Helmholtzfree energy for harmonic oscillator asymmetric potential by

Advances in Physical Chemistry 5

x

05

10

15

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

120588u(x120573)

(a) Density Matrix 120582 = 1 119892 = 1

05

10

15

20

x

155 100minus5minus10minus15

1205821 = 01

1205822 = 20

1205823 = 35

120588u(x120573)

(b) Density Matrix 119892 = 1 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

g1 = 01

g2 = 25

g3 = 55

120588u(x120573)

(c) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

x

155 100minus5minus10minus15

g1 = minus01

g2 = minus45

g3 = minus85

120588u(x120573)

(d) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

1205821 = minus01

1205822 = minus40

1205823 = minus75

120588u(x120573)

(e) Density Matrix 119892 = 1 119879 = 15

Figure 1 Plot of the canonical density matrix of a single particle for harmonic oscillator asymmetric potential system

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

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CatalystsJournal of

Page 3: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Advances in Physical Chemistry 3

Equating the coefficients for 1205852 120585 and 1205850 = 1 on both sides ofthe last equation (12) gives

119889119886 (119891)

119889119891= (1 minus 4119886

2

(119891)) (13)

119889119887 (119891)

119889119891= (120582 minus 4119886 (119891) 119887 (119891)) (14)

119889119888 (119891)

119889119891= (2119886 (119891) minus 120582119892 minus 119887

2

(119891)) (15)

respectively Computing (13) is integrated to give

119886 (119891) =1

2coth (2119891) (16)

Equation (14) is the linear first-order ordinary differentialequation to give [5 6]

120582 =119889119887 (119891)

119889119891+ 2 coth (2119891) 119887 (119891)

119887 (119891) = 119890minusint 2 coth(2119891)119889119891

(int120582119890int 2 coth(2119891)119889119891

119889119891 + 119860)

119887 (119891) =120582

2coth (2119891) + 119860csch (2119891)

(17)

Equation (15) is integrated with respect to 119891 as119889119888 (119891)

119889119891= (coth (2119891) minus 120582119892 minus (120582

2coth (2119891) +119860csch (2119891))

2

)

119888 (119891) = int coth (2119891) 119889119891 minus int120582119892119889119891

minus int(120582

2coth (2119891) + 119860csch (2119891))

2

119889119891

119888 (119891) =1

8(minus2119891120582 (4119892 + 120582) + (4119860

2

+ 1205822

) coth (2119891)

+ 4 ln (sinh (2119891))

+2119860120582csch (119891) sech (119891) minus ln119861) (18)

where 119861 120582 and 119892 are constants Substituting 119886(119891) 119887(119891) and119888(119891) of (16) (17) and (18) respectively into (11) we mayrewrite density matrix asP119906

=119861

radicsinh (2119891)

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

)) coth (2119891)

+(120585119860 +119860120582

2) csch (2119891) minus

119891120582

4(4119892 + 120582) ])

(19)

Thus the Taylor series expansion of sinh(2119891) coth(2119891) about119891 rarr 0 is given by

lim119891rarr0

sinh (2119891) asymp 2119891 lim119891rarr0

coth (2119891) asymp 1

2119891 (20)

we obtain the density matrix final form

P119906asymp

119861

radic2119891

times exp(minus[(1205852

2+120585120582

2+1

8(4119860

2

+ 1205822

))1

2119891

+(120585119860 +119860120582

2)1

2119891minus120582119891

4(4119892 + 120582)])

(21)

The initial condition is

P119906= 120575 (119909 minus ) for 119891 = 0 (22)

or

P119906= radic

119898120596

ℎ120575 (120585 minus 120585) for 119891 = 0 (23)

For low 119891 = (ℎ1205962) 120573 = ℎ1205962119896119861119879 (high temperature) the

particle should act almost like a free particle as its probablekinetic energy is so high Therefore we expect that for low119891 the density matrix for a harmonic oscillator will be givenapproximately by [7]

P119906119891119903119890

(120585 120585 119891) asymp radic119898120596

4120587ℎ119891119890minus(120585minus

120585)2

4119891

(24)

Comparing the value of 119860 119861 in (21) with (24) identifiesthe solution as

119861

radic2119891= radic

119898120596

4120587ℎ119891997888rarr 119861 = radic

119898120596

2120587ℎ

1205852

minus 2120585 120585 + 1205852

= 1205852

+ (120582 + 2119860) 120585

+ (1198602

+1205822

4+ 119860120582 minus 4119892120582119891

2

minus 1205822

1198912

)

119860 = minus( 120585 +120582

2)

(25)

Substituting (25) into (19) and setting 120585 = 120585 produce

P119906(120585 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus[(120585+(1205822))

2

((cosh(2119891)minus1) sinh(2119891))minus(1205821198914)(120582+4119892)]

(26)

Using the identity of trigonometric tanh(1199092) =

(cosh(119909) minus 1) sinh(119909) and setting 120585 = radic119898120596ℎ119909 into

4 Advances in Physical Chemistry

(26) we can rewrite the canonical density matrix or chargebond-order matrix [8] as

P119906(119909 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus(1198981205964ℎ)[(2119909+radicℎ119898120596120582)

2 tanh(119891)minus(120582119891ℎ119898120596)(120582+4119892)]

(27)

The obtained results are neatly summarized in Figure 1 Itis the probability density for finding the system at 119909 we noteas a Gaussian distribution in 119909 Let us define four new thevariables 120574 120578 120583 and 120601 as function of 119891 by

120574 (119891) = radic119898120596

2120587ℎ sinh (2119891) 120578 (119891) =

119898120596

4ℎtanh (119891)

120583 (119891) =120582119891

4(120582 + 4119892) 120601 = radic

119898120596120582

(28)

we can rewrite the density matrix completely in terms of thenew variable (120574 120578 120583 120601) as

P119906(119909 119891) asymp 120574119890

minus(120578(2119909+120601)2

minus120583)

(29)

The expectation value of 1199092 in the density matrix P119906(119909 119891)

can be determined from the definition of the expectationvalues as

⟨1199092

⟩ =intinfin

minusinfin

1199092P119906(119909 119891) 119889119909

intinfin

minusinfin

P119906(119909 119891) 119889119909

=120574radic120587119890

120583

1612057832

2radic120578

120574radic120587119890120583(1 + 2120578120601

2

)

⟨1199092

⟩ =ℎ

2119898120596coth(

ℎ120596120573

2) +

41198981205961205822

(30)

The average of the potential energy (119864119904) is thus

⟨119864119904⟩ =

1

21198981205962

⟨1199092

⟨119864119904⟩ =

ℎ120596

4coth(

ℎ120596120573

2) +

ℎ120596

81205822

(31)

Thekinetic energy per unit length 120591119871(119909 119891) is defined from

the canonical density matrix P119906(119909 119891) as [3]

120591119871(119909 119891) = minus

ℎ2

2119898

1205972

1205971199092P119906(119909 119891) (32)

By substituting (29) into (32) we can write the kinetic energyper unit length completely as follows

120591119871(119909 119879) =

ℎ120596

2tanh( ℎ120596

2119896119861119879)radic

119898120596

2120587ℎ sinh (ℎ120596119896119861119879)

times 119890(120582ℎ1205968119896

119861119879)(120582+4119892)

times [

[

2 minus119898120596

ℎtanh( ℎ120596

2119896119861119879)(2119909 + radic

119898120596120582)

2

]

]

times 119890minus(1198981205964ℎ) tanh(ℎ1205962119896

119861119879)(2119909+radicℎ119898120596120582)

2

(33)

The obtained results from (33) are summarized in Figure 2Now the Helmholtz free energy of the harmonic oscilla-

tor asymmetric potential system (119865New) is given by

119865New = minus1

120573lnQ

119873(120573) (34)

or

Q119873(120573) = 119890

minus120573119865New (35)

The density matrix of the system in the integrationrepresentation will be given by

119890minus120573119865new = int

infin

minusinfin

P119906(119909 120573) 119889119909 (36)

Substituting (29) into (36) we obtain for the partitionfunction of the particle in harmonic oscillator asymmetricpotential system

119890minus120573119865new = int

infin

minusinfin

120574119890minus(120578(2119909+120601)

2

minus120583)

119889119909

=120574

2radic120587

120578

=119890(1205821198914)(120582+4119892)

2 sinh (119891)

(37)

Thus we have the following results

119865new =1

120573ln (2 sinh (119891)) minus

120582119891

4120573(120582 + 4119892) (38)

This is the Helmholtz free energy for a one-dimensionalharmonic oscillator asymmetric potential (119865new) as alreadyderived When we write the Helmholtz free energy in newvariable is 119866new = 2119865newℎ120596 119891 = ℎ1205961205732 into (38) we obtainthat

119866new =1

119891ln (2 sinh (119891)) minus 120582

4(120582 + 4119892) (39)

One may summarize the numerical results of Helmholtzfree energy for harmonic oscillator asymmetric potential by

Advances in Physical Chemistry 5

x

05

10

15

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

120588u(x120573)

(a) Density Matrix 120582 = 1 119892 = 1

05

10

15

20

x

155 100minus5minus10minus15

1205821 = 01

1205822 = 20

1205823 = 35

120588u(x120573)

(b) Density Matrix 119892 = 1 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

g1 = 01

g2 = 25

g3 = 55

120588u(x120573)

(c) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

x

155 100minus5minus10minus15

g1 = minus01

g2 = minus45

g3 = minus85

120588u(x120573)

(d) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

1205821 = minus01

1205822 = minus40

1205823 = minus75

120588u(x120573)

(e) Density Matrix 119892 = 1 119879 = 15

Figure 1 Plot of the canonical density matrix of a single particle for harmonic oscillator asymmetric potential system

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

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Page 4: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

4 Advances in Physical Chemistry

(26) we can rewrite the canonical density matrix or chargebond-order matrix [8] as

P119906(119909 119891) asymp radic

119898120596

2120587ℎ sinh (2119891)

times 119890minus(1198981205964ℎ)[(2119909+radicℎ119898120596120582)

2 tanh(119891)minus(120582119891ℎ119898120596)(120582+4119892)]

(27)

The obtained results are neatly summarized in Figure 1 Itis the probability density for finding the system at 119909 we noteas a Gaussian distribution in 119909 Let us define four new thevariables 120574 120578 120583 and 120601 as function of 119891 by

120574 (119891) = radic119898120596

2120587ℎ sinh (2119891) 120578 (119891) =

119898120596

4ℎtanh (119891)

120583 (119891) =120582119891

4(120582 + 4119892) 120601 = radic

119898120596120582

(28)

we can rewrite the density matrix completely in terms of thenew variable (120574 120578 120583 120601) as

P119906(119909 119891) asymp 120574119890

minus(120578(2119909+120601)2

minus120583)

(29)

The expectation value of 1199092 in the density matrix P119906(119909 119891)

can be determined from the definition of the expectationvalues as

⟨1199092

⟩ =intinfin

minusinfin

1199092P119906(119909 119891) 119889119909

intinfin

minusinfin

P119906(119909 119891) 119889119909

=120574radic120587119890

120583

1612057832

2radic120578

120574radic120587119890120583(1 + 2120578120601

2

)

⟨1199092

⟩ =ℎ

2119898120596coth(

ℎ120596120573

2) +

41198981205961205822

(30)

The average of the potential energy (119864119904) is thus

⟨119864119904⟩ =

1

21198981205962

⟨1199092

⟨119864119904⟩ =

ℎ120596

4coth(

ℎ120596120573

2) +

ℎ120596

81205822

(31)

Thekinetic energy per unit length 120591119871(119909 119891) is defined from

the canonical density matrix P119906(119909 119891) as [3]

120591119871(119909 119891) = minus

ℎ2

2119898

1205972

1205971199092P119906(119909 119891) (32)

By substituting (29) into (32) we can write the kinetic energyper unit length completely as follows

120591119871(119909 119879) =

ℎ120596

2tanh( ℎ120596

2119896119861119879)radic

119898120596

2120587ℎ sinh (ℎ120596119896119861119879)

times 119890(120582ℎ1205968119896

119861119879)(120582+4119892)

times [

[

2 minus119898120596

ℎtanh( ℎ120596

2119896119861119879)(2119909 + radic

119898120596120582)

2

]

]

times 119890minus(1198981205964ℎ) tanh(ℎ1205962119896

119861119879)(2119909+radicℎ119898120596120582)

2

(33)

The obtained results from (33) are summarized in Figure 2Now the Helmholtz free energy of the harmonic oscilla-

tor asymmetric potential system (119865New) is given by

119865New = minus1

120573lnQ

119873(120573) (34)

or

Q119873(120573) = 119890

minus120573119865New (35)

The density matrix of the system in the integrationrepresentation will be given by

119890minus120573119865new = int

infin

minusinfin

P119906(119909 120573) 119889119909 (36)

Substituting (29) into (36) we obtain for the partitionfunction of the particle in harmonic oscillator asymmetricpotential system

119890minus120573119865new = int

infin

minusinfin

120574119890minus(120578(2119909+120601)

2

minus120583)

119889119909

=120574

2radic120587

120578

=119890(1205821198914)(120582+4119892)

2 sinh (119891)

(37)

Thus we have the following results

119865new =1

120573ln (2 sinh (119891)) minus

120582119891

4120573(120582 + 4119892) (38)

This is the Helmholtz free energy for a one-dimensionalharmonic oscillator asymmetric potential (119865new) as alreadyderived When we write the Helmholtz free energy in newvariable is 119866new = 2119865newℎ120596 119891 = ℎ1205961205732 into (38) we obtainthat

119866new =1

119891ln (2 sinh (119891)) minus 120582

4(120582 + 4119892) (39)

One may summarize the numerical results of Helmholtzfree energy for harmonic oscillator asymmetric potential by

Advances in Physical Chemistry 5

x

05

10

15

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

120588u(x120573)

(a) Density Matrix 120582 = 1 119892 = 1

05

10

15

20

x

155 100minus5minus10minus15

1205821 = 01

1205822 = 20

1205823 = 35

120588u(x120573)

(b) Density Matrix 119892 = 1 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

g1 = 01

g2 = 25

g3 = 55

120588u(x120573)

(c) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

x

155 100minus5minus10minus15

g1 = minus01

g2 = minus45

g3 = minus85

120588u(x120573)

(d) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

1205821 = minus01

1205822 = minus40

1205823 = minus75

120588u(x120573)

(e) Density Matrix 119892 = 1 119879 = 15

Figure 1 Plot of the canonical density matrix of a single particle for harmonic oscillator asymmetric potential system

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

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Page 5: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Advances in Physical Chemistry 5

x

05

10

15

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

120588u(x120573)

(a) Density Matrix 120582 = 1 119892 = 1

05

10

15

20

x

155 100minus5minus10minus15

1205821 = 01

1205822 = 20

1205823 = 35

120588u(x120573)

(b) Density Matrix 119892 = 1 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

g1 = 01

g2 = 25

g3 = 55

120588u(x120573)

(c) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

x

155 100minus5minus10minus15

g1 = minus01

g2 = minus45

g3 = minus85

120588u(x120573)

(d) Density Matrix 120582 = 1 119892 = + 119879 = 15

05

10

15

20

x

155 100minus5minus10minus15

1205821 = minus01

1205822 = minus40

1205823 = minus75

120588u(x120573)

(e) Density Matrix 119892 = 1 119879 = 15

Figure 1 Plot of the canonical density matrix of a single particle for harmonic oscillator asymmetric potential system

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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International Journal ofPhotoenergy

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Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

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Journal of

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CatalystsJournal of

Page 6: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

6 Advances in Physical Chemistry

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

1205821 = 01

1205822 = 10

1205823 = 19

(a) Kinetic Energy 119892 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

1205821 = minus01

1205822 = minus16

1205823 = minus28

x

155 100minus5minus10minus15

(b) Kinetic Energy 119892 = 1 119879 = 5K

x

155 100minus5minus10minus15

002

004

006

008

010

012

120591L(xT

)

g1 = 01

g2 = 16

g3 = 28

(c) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

g1 = minus01

g2 = minus16

g3 = minus28

x

155 100minus5minus10minus15

(d) Kinetic Energy 120582 = 1 119879 = 5K

002

004

006

008

010

120591L(xT

)

x

T1 = 5KT2 = 10KT3 = 15K

155 100minus5minus10minus15

(e) Kinetic Energy 120582 = 1 119892 = 1

Figure 2 Illustration the kinetic energy per unit length under harmonic oscillator asymmetric potential system

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Inorganic ChemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

International Journal ofPhotoenergy

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

Physical Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom

Analytical Methods in Chemistry

Journal of

Volume 2014

Bioinorganic Chemistry and ApplicationsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

SpectroscopyInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Medicinal ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chromatography Research International

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Applied ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Theoretical ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Spectroscopy

Analytical ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Quantum Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Organic Chemistry International

ElectrochemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CatalystsJournal of

Page 7: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Advances in Physical Chemistry 7

0 2 4 6 8 100

1

2

3

4

5

1205821 = 01

1205822 = 10

1205823 = 60

T (K)

S new(T

)

(a) Entropy (119878new(119879)) 119892 = minus2

0

1

2

3

4

5

0 2 4 6 8 10T (K)

minus1

1205821 = minus001

12058221205823

= minus003

= minus005

S new(T

)

(b) Entropy (119878new(119879)) 119892 = minus4

0

2

4

0 2 4 6 8 10T (K)

minus2

minus4

g1 = 010

g2 = 350

g3 = 670

S new(T

)

(c) Entropy (119878new(119879)) 120582 = 005

0

1

2

3

4

5

S new

(T)

0 2 4 6 8 10T (K)

g1 = minus010

g2 = minus1850

g3 = minus4270

(d) Entropy (119878new(119879)) 120582 = 005

Figure 3 Illustration of the entropy(119878) of a single particle under harmonic oscillator asymmetric potential

Feynmans approach in Section 5 In Section 5 comparisonof Helmholtz free energy for harmonic oscillator asymmetricpotential obtain from Feynmans approach and the path-integral method 120581(119910) This work aims at computing theentropy (119878) of the harmonic oscillator asymmetric potentialusing the Helmholtz free energy in (38) Using the relation

119878new (119879) = minus120597119865new (119879)

120597119879= 119896119861(120597

120597119879(119879 ln (Q

119873(120573)))) (40)

the entropy of a system can be write as

119878new (119879) =119896119861ℎ120596

4 (119896119861119879)

cosh (ℎ1205962119896119861119879)

sinh (ℎ1205962119896119861119879)

minus 119896119861ln (2 sinh (ℎ1205962119896

119861119879))

minus119896119861120582120596ℎ

8 (119896119861119879)2(120582 + 4119892)

(41)

Equation (41) is plotted in Figure 3(a) assuming the valueof axes-shift potential 119892 equal minus2 and vary the value ofasymmetric potential 120582 From (41) are plotted in Figure 3assuming the value of asymmetric potential 120582 equal 005 andvary the value of axes-shift potential 119892 Figure 3 shows thatas 120582 increases the depth of the entropy decreases and itsminimummoves farther away the origin

4 Evaluate Helmholtz Free Energy via Path-Integral Method (120581(119910))

In the Helmholtz free energy the quantity of the thermody-namics of a given harmonic oscillator asymmetric potentialsystem is derived from its the path-integral method

120581 (119910) = radic6119898119896

119861119879

120587ℎ2intinfin

minusinfin

119890minus(6119898119896

119861119879(119910minus119911)

2

ℎ2

)

119881 (119911) 119889119911 (42)

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Inorganic ChemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

International Journal ofPhotoenergy

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

Physical Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom

Analytical Methods in Chemistry

Journal of

Volume 2014

Bioinorganic Chemistry and ApplicationsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

SpectroscopyInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Medicinal ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chromatography Research International

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Applied ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Theoretical ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Spectroscopy

Analytical ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Quantum Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Organic Chemistry International

ElectrochemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CatalystsJournal of

Page 8: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

8 Advances in Physical Chemistry

In this case of the harmonic oscillator asymmetric potentialis

119881 (119911) =1

21198961199112

+ 120591120582 (119911 minus 120593119892) (43)

where setting 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962) 120593 = radicℎ119898120596

and substituting (43) into (42) we can write classical 120581(119910) tosimply produce

120581 (119910) = Φintinfin

minusinfin

(1

21198961199112

+ 120591120582 (119911 minus 120593119892)) 119890minus120572(119911minus119910)

2

119889119911 (44)

where 120572 = 6119898120573ℎ2 Φ = radic6119898120573120587ℎ2 The final the path-integral method is

120581 (119910) =Φradic120587 (119896 + 2119896119910

2

120572 + 4120572120582120591 (119910 minus 120593119892))

412057232 (45)

In such case it is more useful to write the partition functionof 120581(119910) as

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573120581(119910)

119889119910 (46)

where Ω = 1198982120587ℎ119880 Substituting (45) into (46) can obtainthe partition function of path-integral method (120581(119910)) in caseof harmonic oscillator asymmetric potential

119890minus120573119865119888119897minus119870 = Ωint

infin

minusinfin

119890minus120573(Φradic120587(119896+2119896119910

2

120572+4120572120582120591(119910minus120593119892))412057232

)

119889119910

= Ω119890minus((120573radic120587Φ1198964120572

32

)minus(4119892120573120572120591120582120593412057232

))

times intinfin

minusinfin

119890minus120573radic120587Φ((2119896120572119910

2

+4120572120591120582119910)412057232

)

119889119910

119890minus120573119865119888119897minus119870 =

radic2 4radic120587Ω

radic120573119896Φradic120572

[119890minus(120573Φradic1205874119896120572

32

)(1198962

minus21205721205822

1205912

minus4119892119896120572120582120591120593)

]

(47)

Substituting 120572 = 6119898120573ℎ2 119896 = 1198981205962 120591 = radic119898120596ℎ (ℎ1205962)120593 = radicℎ119898120596 Φ = radic6119898120573120587ℎ2 and Ω = radic1198982120587ℎ119880 into (47)leads to

119890minus120573119865119888119897minus119870

=1

ℎ120573120596

times 119890minus(1205732

ℎ2

241198982

1205962

)(1198982

1205964

minus(31205822

1198982

1205963

120573ℎ)minus(121198921205821198982

1205962

120573ℎ))

minus120573119865119888119897minus119870

= minus ln (2119891) minus (1198912

6minus1198911205822

4minus 119892119891120582)

(48)

Accordingly the Helmholtz free energy the path-integralmethod (120581(119910)) is given by

119865119888119897minus119870

(119891) =1

120573[ln (2119891) + (

1198912

6minus1198911205822

4minus 119892119891120582)] (49)

Table 1 Comparison of the Helmholtz free energy of the harmonicoscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where 120582 is the value of

asymmetric potential 119892 is the value of axes-shift potential systemand 119879 is the temperature setting 119892 = 120582 = 1

119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Difference1 minus1167350290774 minus1166666666667 00585791892 minus3981008518269 minus3980922055573 00021719023 minus7813921629128 minus7813895954231 00003285794 minus12319532395568 minus12319521555626 00000879895 minus17327718009705 minus17327712457674 00000320416 minus22737227955451 minus22737224741848 00000141337 minus28480839348833 minus28480837324869 710640 times 10minus6

8 minus34510649356211 minus34510648000211 392922 times 10minus6

9 minus40790784085205 minus40790783132793 233487 times 10minus6

10 minus47293369220882 minus47293368526548 146814 times 10minus6

11 minus53996120765666 minus53996120243988 966139 times 10minus7

12 minus60880815552301 minus60880815150468 660032 times 10minus7

13 minus67932273353648 minus67932273053847 465256 times 10minus7

14 minus75137653101331 minus75137652848275 336789 times 10minus7

15 minus82485950683258 minus82485950477511 249433 times 10minus7

After introducing new variable of the Helmholtz free energypath-integral method (120581(119910)) as 119866

119888119897minus119870= 2119865

119888119897minus119870ℎ120596 119891 =

ℎ1205961205732 the Helmholtz free energy of the path-integralmethod (120581(119910)) is then transformed to

119866119888119897minus119870

(119891) =1

119891ln (2119891) +

119891

6minus1205822

4minus 119892120582 (50)

This is known as the Helmholtz free energy of the path-integral method (120581(119910)) We can calculate numerical of theHelmholtz free energy Feynmans (119866new(119891) approach inSection 3) and the Helmholtz free energy of the path-integralmethod (120581(119910))(119866

119888119897minus119870(119891) in Section 4) for the harmonic oscil-

lator asymmetric potential by mathematica program to giveTables 1 and 2 as shown in Section 5

5 The Numerical ResultIn this section we have evaluated numerical of theHelmholtzfree energy Feynmans (119866new(119891)) approach and theHelmholtzfree energy of the path-integral method (120581(119910))(119866

119888119897minus119870(119891)) for

the harmonic oscillator asymmetric potential is (39) and (50)By substituting 119891 = ℎ1205962119896

119861119879 into (39) and (50) we obtain

the Helmholtz free energy Feynmans (119866new(119879)) approach asfollows

119866new (119879) =2119896119861119879

ℎ120596ln(2 sinh( ℎ120596

2119896119861119879)) minus

120582

4(120582 + 4119892) (51)

and the Helmholtz free energy of the path-integral method(120581(119910))(119866

119888119897minus119870(119879))

119866119888119897minus119870

(119879) =2119896119861119879

ℎ120596ln( ℎ120596

119896119861119879) +

ℎ120596

12119896119861119879minus1205822

4minus 119892120582 (52)

For the beginning of the numerical shootingmethodwe needto input the parameters 120582 119892 and 119879 and (39) and (50) into

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Inorganic ChemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

International Journal ofPhotoenergy

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

Physical Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom

Analytical Methods in Chemistry

Journal of

Volume 2014

Bioinorganic Chemistry and ApplicationsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

SpectroscopyInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Medicinal ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chromatography Research International

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Applied ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Theoretical ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Spectroscopy

Analytical ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Quantum Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Organic Chemistry International

ElectrochemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CatalystsJournal of

Page 9: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Advances in Physical Chemistry 9

Table 2 Comparison of the Helmholtz free energy of the harmonic oscillator asymmetric potential obtained from Feynmans approach(119866new) path-integral method 120581(119910)(119866

119888119897minus119870(119879)) where we may vary the value of 119892 120582 119879

120582 119892 119879(119870) 119866new(119879) 119866119888119897minus119870

(119879) Differenceminus1 1 1 0832649709225 0833333333333 0082068555minus2 2 2 0268991481731 0269077944427 0032138119minus3 3 3 0186078370872 0186104045769 0013796943minus4 4 4 0930467604432 0930478444374 0001164993minus5 5 5 2672281990295 2672287542326 0000207763minus6 6 6 5512772044548 5512775258152 0000058294minus7 7 7 9519160651167 9519162675130 0000021262minus8 8 8 14739350643789 14739351999789 9199869 times 10

minus6

minus9 9 9 21209215914795 21209216867207 4490557 times 10minus6

minus10 10 10 28956630779118 28956631473452 2397841 times 10minus6

minus11 11 11 38003879234334 38003879756012 1372696 times 10minus6

minus12 12 12 48369184447699 48369184849532 8307624 times 10minus7

minus13 13 13 60067726646352 60067726962410 5261694 times 10minus7

minus14 14 14 73112346898668 73112347151725 3461208 times 10minus7

minus15 15 15 87514049316743 87514049522489 2351017 times 10minus7

themathematical program [9] Next we obtain the numericalvalues of 119866new(119879) and 119866119888119897minus119870(119879) for the harmonic oscillatorasymmetric potential as displayed in Tables 1 and 2

From Table 1 we show that magnitude of the Helmholtzfree energy decrease from minus11673 to minus824859 with increas-ing the temperature of system From Table 2 we illustratemagnitude of theHelmholtz free energy increase from 08326

to 875140 with increasing the temperature of systemNevertheless 119866

119888119897minus119870(119879) is not as useful as Tables 1 and 2

might indicate The first it can not be used when quantummechanics exchange effect exist The second it fails in itspresent form when the potential 119881(119910) has a very largederivative as in the case of hard-sphere interatomic potential(see Lennard-Jones potential [10])

6 ConclusionThiswork is interested in finding the canonical densitymatrixby Feynmanrsquos technique and kinetic energy per unit lengthand evaluation of theHelmholtz free energyThe results showthat themagnitude of the densitymatrix (P

119906) and the kinetic

energy per unit length (120591119871) vary according to the parameters

120582 119892 and 119879 respectively The magnitude of (P119906) and (120591

119871)

for these cases 120582 is positive with the graph of (P119906) and (120591

119871)

moving from right hand side to left hand side Moreover wefound that as the values of119892have increased and the amplitudeof (P

119906) and (120591

119871) has decreased

Conflict of InterestsThe authors declare that there is no conflict of interestsregarding the publication of this paper

AcknowledgmentsPiyarut Moonsri and Artit Hutem wish to thank the InstituteResearch and Development Chemistry Division and Physics

Division Faculty of Science and Technology PhetchabunRajabhat University Thailand This work is supported by theInstitute Research and Development Phetchabun RajabhatUniversity

References

[1] N H March and A M Murray ldquoRelation between Dirac andcanonical density matrices with applications to imperfectionsin metalsrdquo Physical Review Letters vol 120 no 3 pp 830ndash8361960

[2] I A Howard A Minguzzi N H March and M Tosi ldquoSlatersum for the one-dimensional sech2 potential in relation to thekinetic energy densityrdquo Journal of Mathematical Physics vol 45no 6 pp 2411ndash2419 2004

[3] N H March and L M Nieto ldquoBloch equation for the canonicaldensity matrix in terms of its diagonal element the Slater sumrdquoPhysics Letters A General Atomic and Solid State Physics vol373 no 18-19 pp 1691ndash1692 2009

[4] R K Pathria Statistical Mechanics Butterworth HeinemannOxford UK 1996

[5] G B Arfken and H J Weber Mathematical Methods forPhysicists Elsevier Academic Press New York NY USA 6thedition 2005

[6] K F Riley and M P HobsonMathematical Method for Physicsand Engineering CambridgeUniversity Press 3rd edition 2006

[7] R P Feynman Statistical Mechanics Addison-Wesley LondonUK 1972

[8] I N Levin Quantum Chemistry Pearson Prentice Hall 6thedition 2009

[9] R L Zimmerman and F I Olness Mathematica for PhysicsAddison-Wesley Reading Mass USA 2nd edition 2002

[10] A Hutem and S Boonchui ldquoNumerical evaluation of secondand third virial coefficients of some inert gases via classicalcluster expansionrdquo Journal of Mathematical Chemistry vol 50no 5 pp 1262ndash1276 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Inorganic ChemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

International Journal ofPhotoenergy

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

Physical Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom

Analytical Methods in Chemistry

Journal of

Volume 2014

Bioinorganic Chemistry and ApplicationsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

SpectroscopyInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Medicinal ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chromatography Research International

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Applied ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Theoretical ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Spectroscopy

Analytical ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Quantum Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Organic Chemistry International

ElectrochemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CatalystsJournal of

Page 10: Research Article Evaluation of Density Matrix and Helmholtz Free Energy ...downloads.hindawi.com/archive/2014/912054.pdf · Research Article Evaluation of Density Matrix and Helmholtz

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Inorganic ChemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

International Journal ofPhotoenergy

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Carbohydrate Chemistry

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Advances in

Physical Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom

Analytical Methods in Chemistry

Journal of

Volume 2014

Bioinorganic Chemistry and ApplicationsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

SpectroscopyInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Medicinal ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Chromatography Research International

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Applied ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Theoretical ChemistryJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Spectroscopy

Analytical ChemistryInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Quantum Chemistry

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Organic Chemistry International

ElectrochemistryInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CatalystsJournal of