ch 4. using quantum mechanics on simple systems ms310 quantum physical chemistry - discussion of...
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Ch 4. Using Quantum Ch 4. Using Quantum Mechanics on Simple SystemsMechanics on Simple Systems
MS310 Quantum Physical Chemistry
- Discussion of constrained and not constrained - Discussion of constrained and not constrained particle motion particle motion ex) free particle, In 2-D or 3-D boxes, In vice versa ex) free particle, In 2-D or 3-D boxes, In vice versa
- Continuous energy spectrum of Q.M free particleContinuous energy spectrum of Q.M free particle
- Discrete energy spectrum and preferred position of - Discrete energy spectrum and preferred position of Q.M particles in the box (Quantized energy levels)Q.M particles in the box (Quantized energy levels)
MS310 Quantum Physical Chemistry
4.1. The free particle
Free particle : no forcesClassical 1-dimension, no forces : 02
2
dt
xdmmaF
Solution : x = x0 + v0tx0 ,v0 : initial condition, constants of integrationExplicit value : must be known initial condition
How about the free particle in Q.M?
Time-independent Schrödinger Equation in 1-dimension is
)()()()(
2 2
22
xExxVdx
xd
m
MS310 Quantum Physical Chemistry
constant V(x) : can choose the reference V(x)=0 (absolute potential reference doesn’t exist) → reduced to )(
2)(22
2
xEm
dx
xd
Solution is given by
ikxxmEi
ikxxmEi
eAeAx
eAeAx
)/2(
)/2(
2
2
)(
)(
Use these notations2/2
2,2
mE
pkmEp
Obtain Ψ(x,t) : multiply each e-i(E/ )tℏ or equivalently e-iωt (E = ℏω)
Eigenvalue : not quantized(all energy allowed : k is
continuous variable)m
kE
2
22
ikxikx eAeAxxx )()()(
MS310 Quantum Physical Chemistry
Plane wave cannot be localized. → cannot speak about the position of particle.
Then, what about probability of finding a particle? → also cannot calculate (wave function cannot be normalized in interval -∞ < x < ∞)
However, if x is ‘restricted’ to the interval –L ≤ x ≤ L then
L
dx
dxeeAA
dxeeAA
dxxx
dxxxdxxP L
L
ikxikx
ikxikx
L
L
2)()(
)()()(
*
*
P(x) : independent of x → no information about position
MS310 Quantum Physical Chemistry
)()()(ˆ
)()()(ˆ
xkekAeAx
ixx
ixp
xkekAeAx
ixx
ixp
ikxikxx
ikxikxx
What about the momentum of particle?
ψ+(x) : state of momentum + k(positive direction)ℏ ψ-(x) : state of momentum – k(negative direction)ℏ
4.2 The particle in a One-dimensional box
MS310 Quantum Physical Chemistry
1-dimensional box : particle in the range 0<x<a only impenetrable : infinite potential
V(x) = 0 for 0 < x < a = ∞ for x ≥ a , x ≤ 0
MS310 Quantum Physical Chemistry
Schrödinger Equation is changed by
)(])([2)(
22
2
xExVm
dx
xd
If ψ(x) ≠ 0 outside the box, then value of ψ’’(x) becomes infinite because value of V(x) is infinity outside the box.
However, 2nd derivative exists and well-behaved → ψ(x) must be 0 outside the box boundary condition : ψ(0) = ψ(a) =0
Inside the box : same as the free particle
We can write the solution by the sin and cos.
kxBkxA
kxAAikxAA
kxikxAkxikxA
eAeAx ikxikx
cossin
sin)(cos)(
)sin(cos)sin(cos
)(
): (noticesin)( nxa
nAxn
From the consideration of boundary conditions
a
nk
nnka
kaAa
kxAx
B
a
,3,2,1,0,
0sin)(
sin)(
00)0(
0)()0(
2
0
2
0
2
0
22
2
1
2
12cos1
2
1
sin)()(1
aAxAdxa
xnA
dxa
xnAdxxx
aa
a
nn
)sin(2
)( , 2
a
xn
ax
aA n
Normalization
2
2222
22
2
22
82
)()(2
)(2
)(ˆ
ma
hn
a
n
mE
xExa
n
mx
dx
d
mxH
n
nnnnn
Energy of the particle
‘Quantization’ arises by the boundary condition
MS310 Quantum Physical Chemistry
Particle is ‘quantized’, n : quantum number
Ground state : n=1However, energy of n=1 is not zero : zero point energy(ZPE)
particle in a box : ‘stationary’ wave(not a traveling wave)
Also, n increase → # of node increase → wave vector k increase because
Finally, what about a classical limit? → same as result of C.M(same probability in everywhere)
2
222
1 82 ma
h
amEn
2/22
mEp
k
MS310 Quantum Physical Chemistry
graph of ψn(x) and ψn*(x)ψn(x)
MS310 Quantum Physical Chemistry
Graph of ψn2(x) / [ψ1
2(x)]max
n increase : large energyLower resolution : cannot precise measure → near to C.MResult of Q.M ‘approach’ to the C.M when classical limit
MS310 Quantum Physical Chemistry
4.3 Two- and Three- dimensional boxes
Boundary condition : similar to 1-dimensional box
V(x, y, z) = 0 for 0 < x < a, 0 < y < b, 0 < z < c = ∞ otherwise
Inside the box, Schrödinger Equation is given by
),,(),,()(2 2
2
2
2
2
22
zyxEzyxzyxm
Solving by separation of variable
And equation is changed by
)()()(
))()()()()()()()()((2 2
2
2
2
2
22
zZyYxEX
zZdz
dyYxXyY
dy
dzZxXxX
dx
dzZyY
m
MS310 Quantum Physical Chemistry
Edz
zZd
zZdy
yYd
yYdx
xXd
xXm )
)(
)(
1)(
)(
1)(
)(
1(
2 2
2
2
2
2
22
Divide both side by X(x)Y(y)Z(z)
E : independent to coordinate → E = Ex + Ey + Ez and original equation(PDE) reduced to three ODEs.
)()(
2),(
)(
2),(
)(
2 2
22
2
22
2
22
zZEdz
zZd
myYE
dy
yYd
mxXE
dx
xXd
m zyx
Solution of each equation is already given.
c
zn
b
yn
a
xnNzyx zyx
nnn zyx
sinsinsin),,(
And energy is given by )(8 2
2
2
2
2
22
c
n
b
n
a
n
m
hEEEE zyxzyx
MS310 Quantum Physical Chemistry
Normalization
2
000
2
0 00
2
0
2
0
2
0
22
8
1
2
1
2
1
2
1
2cos1
2
12cos1
2
12cos1
2
1
sinsinsin1
abcNzyxN
dzc
xndx
b
yndx
a
xnN
dzc
zndy
b
yndx
a
xnNd
cba
a cz
b yx
cz
b yax
nn
...)3,2,1:,,(
sinsinsin8
),,(,8
zyx
zyxnnn
nnnc
zn
b
yn
a
xn
abczyx
abcN
zyx
MS310 Quantum Physical Chemistry
If total energy is sum of independent terms → wave function is product of corresponding functions
Solution has a three quantum numbers : nx, ny, nz
→ more than one state may have a same energy : energy level is degenerate and # of state is degeneracy
ex) if a=b=c, energy of (2,1,1), (1,2,1), and (1,1,2) is same.
in this case, state (2,1,1), (1,2,1) and (1,1,2) is degenerate and degeneracy of the level is 3.
2-dimensional box problem : similar to 3-dimensional problem(end-of-chapter problem)
2
2222
2
2
2,1,1 4
3)211(
8 ma
h
ma
hE
4. Using the postulate to understand the particle in the box and vice versa
MS310 Quantum Physical Chemistry
Postulate 1 : The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
We see the postulates of Q.M using the particle in a box.
Ex) 4.2 ψ(x) = c sin (πx/a) + d sin (2πx/a) a. Is ψ(x) an acceptable wave function of particle in a box? b. Is ψ(x) an eigenfunction of the total energy operator Ĥ? c. Is ψ(x) normalized?
MS310 Quantum Physical Chemistry
Sol) a. Yes. ψ(x) = c sin (πx/a) + d sin (2πx/a) satisfies the boundary condition, ψ(0) = ψ(a) = 0 and well-behaved function. Therefore, ψ(x) is acceptable wave function.
b. No.
Result of Ĥψ(x) is not ψ(x) multiplied by constants. Therefore, ψ(x) is not a eigenfunction of the total energy operator.
c. No
)2
sin4sin(2
)2
sinsin(2
)(ˆ2
22
2
22
a
xd
a
xc
maa
xd
a
xc
dx
d
mxH
dxa
x
a
xdccddx
a
xddx
a
xc
dxa
xd
a
xc
aaa
a
2sinsin)(
2sin||sin||
|2
sinsin|
0
**
0
22
0
22
2
0
MS310 Quantum Physical Chemistry
Third integral becomes zero because of orthogonality.
)|||(|2
]2
1[||]
2
1[||
2sin||sin||
220
20
2
0
22
0
22
dca
dc
dxa
xddx
a
xc
aa
aa
Therefore, ψ(x) is not normalized. However, the function
is normalized when |c|2+|d|2=1
Superposition state depends on time. Why?
Therefore, this state doesn’t describe the stationary state.
]2
sinsin[2
a
xd
a
xc
a
)()(]2
sinsin[2
),( // 21 tfxa
xde
a
xce
atx tiEtiE
MS310 Quantum Physical Chemistry
Then, what about a probability of particle in the interval?
Ex) 4.3probability of ground-state particle in the central third?
Sol) ground state : a
x
ax
sin2
)(1
609.0)]3
2sin
3
4(sin
46[
2sin
2)()(
3/2
3/
23/2
3/
1*1
aa
adx
a
x
adxxxP
a
a
a
a
Probability of finding a particle in central third is 60.9%.
However, we cannot obtain this result by one individual measurement. We can only predict the result of large number of experiment(60.9%).
MS310 Quantum Physical Chemistry
Postulate 3: In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.
Postulate 4 : If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by
dxtxtx
dxtxAtx
a
),(),(
),(ˆ),(
*
*
We can understand the two postulates together.
MS310 Quantum Physical Chemistry
2) wave function is not an eigenfunction of operator. → each measurement gives different value
Ex) normalized superposition state
2
222
1
111
22
12
22
1
82
)()(2
)(2
)(ˆ
ma
h
amE
xExam
xdx
d
mxH
1) wave function is an eigenfunction of operator. → all measurement gives same value and it is average value
Ex) ground state of particle in a box
1||||],2
sinsin[2
)( 22 dca
xd
a
xc
ax
MS310 Quantum Physical Chemistry
]2
sinsin2
sinsin[2
]2
sin||sin|[|2
)2
sinsin](2
[)2
sinsin(2
)()](2
)[()(ˆ)(
00
1*
2*
0
22
2
0
21
2
2
22
0
**
2
22**
aa
aa
a
dxa
x
a
xEcddx
a
x
a
xdEc
a
dxa
xEddx
a
xEc
a
a
xd
a
xc
dx
d
ma
xd
a
xc
a
dxxxVdx
d
mxdxxHxE
Last two integrals are zero by orthogonality and final result is
22
12
0
22
2
0
21
2 ||||]2
sin||sin|[|2
EdEcdxa
xEddx
a
xEc
aE
aa
where2
22
8ma
hnEn
MS310 Quantum Physical Chemistry
However, result of individual experiment is only E1 or E2 by the postulate 3. How can represent the result?
→ by the postulate 4, result of the large number of individual experiment, probability of E1 is |c|2 and probability of E2 is |d|2, and the ‘average value’ of energy <E> = |c|2E1| + |d|2E2.
More generally, we can think about this case
Ψ(x) = cΨ1(x) + dΨ2(x) + 0(Ψ3(x) + Ψ4(x) + …)
All coefficient except Ψ1(x) and Ψ2(x) is zero. Therefore, no other energy is measured except the E1 and E2.
MS310 Quantum Physical Chemistry
0)0sin(sin2
cossin2
)](sin[sin2
)(ˆ)(
sin2
)(
222
02
00
*
na
nidx
a
xn
a
xn
a
ni
dxa
xn
dx
di
a
xn
adxxpxp
a
xn
ax
a
aa
Now, consider the momentum.
We know and can calculate the average value of Momentum of nth state.
dx
dipx ˆ
In the Q.M, momentum of particle : cannot be zero(energy E = p2 / 2m cannot be zero in Q.M)
→ average of two superposition state is zero!
We can rewrite the wave function by complex form.
(use the )i
eex
ixix
2sin
MS310 Quantum Physical Chemistry
axinaxinaxin
axinaxinaxin
axinaxin
ea
ne
dx
diep
ea
ne
dx
diep
i
ee
aa
xn
ax
///
///
//
ˆ
ˆ
)2
(2
sin2
)(
In the case of momentum, two probability of positive momentum and negative momentum is same. Therefore, the average value seems to zero.
MS310 Quantum Physical Chemistry
Ex) 4.4Particle in the ground state. a. Is wave function the eigenfunction of position operator? b. calculate the average value of the position <x>.
Sol) a. position operator :
a
x
ax
sin2
)(
)(sin2
sin2
)(,ˆ xca
x
ac
a
xxa
xxxx
Therefore, wave function is not an eigenfunction of position operator.
b. expectation value is calculated by postulate 4.
2]
8)0
84[(
2]
4
2sin
)(8
2cos
4[
sin2
sinsin2
)(ˆ)(
2
2
2
22
02
2
0 0
2
0
*
aaaa
aa
a
xx
a
a
xx
dxa
xx
adx
a
xx
a
x
adxxxxx
a
a aa
Average value of particle is half, the expected position.
MS310 Quantum Physical Chemistry
- The motion of particle which is not constrained shows continuous energy spectrum however, the particle in a box has a discrete energy spectrum.
- The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
- In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.
Summary Summary
MS310 Quantum Physical Chemistry
If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by
dxtxtx
dxtxAtx
a
),(),(
),(ˆ),(
*
*