lecture 30: linear variation theory the material in this lecture covers the following in atkins

23
ure 30: Linear Variation Theory material in this lecture covers the following in A 14.7 Heteronuclear diatomic molecules (c) The variation principle Lecture on-line Linear Variation Theory (PowerPoin Linear Variation Theory (PDF) out for this lecture

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Lecture 30: Linear Variation Theory The material in this lecture covers the following in Atkins. 14.7 Heteronuclear diatomic molecules (c) The variation principle Lecture on-line Linear Variation Theory (PowerPoint) - PowerPoint PPT Presentation

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Lecture 30: Linear Variation Theory The material in this lecture covers the following in Atkins.

14.7 Heteronuclear diatomic molecules (c) The variation principle

Lecture on-line Linear Variation Theory (PowerPoint) Linear Variation Theory (PDF)

Handout for this lecture

The Linear Variation Method

We have a Hamiltonian H with the eigenfunctions ψna nd eigenvalues En give n by t he SWE

Hψn = Enψn

Let u s l ook a t th e groundstat e ψ1 with t he energ y E1 .

W e would lik e to fi nd a wavefuncti on Φ1 fo r which the

energy

W = ∫ Φ

1

∗H Φ

1d τ / ∫ Φ

1

∗Φ

1d τ

i s clos e to E1

The linear variation method

We write Φ1 as a ttriri aall wavefuncti on in te rms of a

line ar combinati on of kno wn function s {fi }

tha t depen ds on th e same variable s a s ψ1 a nd hav e the

sa meboundary conditions

Φ1 = ∑j=1

j=n Cjfj

Or

Φ1 = C1f1 +C2f2 +....+Cjfj + Cnfn

The linear variation method

We shall now vary all the coefficients {Cj ,j=1,n}

in such a way that W has the smallest possible

value.

That is ,we shall find the absolute minima of the

function W(C1,C2,C3,.....,Ci,..Cn).

Let the values of the coefficients {C1,C2,C3,.....,Ci,..Cn}

at the minimum  be given by

{C11 ,C

12,C

13, ...,C

1j ,..C

1n }

The linear variation method

What do we know about this particular coefficients ?

We know that if we look at the derivatives of W

δ W

δ Ci

= W’(C1,C2,..... ,Cn)

The n ’W (C11 ,C

12,C

13, ...,C

1j ,..C

1n ) = 0

W e shall us e thi s fac t to fi nd { C11 ,C

12,C

13, ...,C

1j ,..C

1n}

The linear variation method

First let us substitute the expression for Φ1 i nto the

expressi on fo r W. The denomenato r of I1 is

⌡⎮⌠

Φ1 ∗ Φ1 dτ = ⌡⎮⌠

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n Cj*fj*

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑k=1

k=n Ckfk

W e shal l assume tha t a ll functi ons {fj j=1,n} ar e rea l and

tha t al l coefficient s {Cj=1,n} are real

The linear variation method

After multiplication of the two parantheses

⌡⎮⌠

Φ1 ∗ Φ1 dτ =∑j=1

j=n ∑k=1

k=n Cj Ck ⌡

⎮⌠

fjfkdτ

Le t usintroduce : ⌡⎮⌠

fjfkdτ = Sjk

Thu s : ⌡⎮⌠

Φ1 ∗ Φ1 dτ =∑j=1

j=n ∑k=1

k=n Cj Ck Sjk

The linear variation method

For the numerator in the expression for I1 we have

⌡⎮⌠

Φ1∗ H Φ1dτ = ⌡⎮⌠

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n Cj*fj* H

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑k=1

k=n Ckfk

O r aft er multiplicati on ofparanthesis

⌡⎮⌠

Φ1∗ H Φ1dτ = =∑j=1

j=n ∑k=1

k=n Cj Ck ⌡

⎮⌠

fj H fkdτ

The linear variation method

We shall now introduce

⌡⎮⌠

 

 

 fj H fkdτ = Hjk

We know that H is Hermetian

⌡⎮⌠

 

 

 fj* H fkdτ = ⌡⎮⌠

 

 

 fk (H fj)*dτ (1)

We shall also assume that it is real H = (H )*.

Thus since {fi=1,n} are real functions it follows from (1)

Hjk = Hkj

The linear variation method

We have

⌡⎮⌠

 

 

 Φ1∗ H Φ1dτ = =∑j=1

j=n  ∑k=1

k=n  Cj Ck Hkj

We can now write W =

∑j=1

j=n  ∑k=1

k=n   Cj Ck Hkj  

∑j=1

j=n  ∑k=1

k=n   Cj Ck Skj

The linear variation method

Or

W⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj = ∑

j=1

j=n ∑k=1

k=n C j Ck Hkj

It is important t o observ e tha (t C1,C2,..,Ci,.. Cn) are

independe nt variables

W e shal l no w differentiat e wi th respect t o one of

them,say C i , on bothside s of th e equation.

The linear variation method

We have

δ W

δ Ci ⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj + W

δδCi

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj =

δ

δ Ci j = 1

j = n

k = 1

k = n

∑ Cj

Ck

Hkj

Let u s no w loo k a t δ

δCi⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj =

The linear variation method

Since we differentiate a sum by differentiating

each term from rules for differentiating a product

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n

δδCi( C j Ck Skj ) =

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n [

δC jδCiCk Skj +

δCk δCi Cj Skj]

Since (C1,C2,..,Ci,.. Cn) ar e independe nt variables

δCjδCi = 0 i f ≠i j

δCjδCi = 1 i f i= j ;

δCjδCi = δij

The linear variation method

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

 ∑j=1

j=n  ∑k=1

k=n   [δjiCk Skj +δkiCj Skj] = ∑

k=1

k=n  CkSik + ∑

j=1

j=n  CjSjk

= ∑k=1

k=n  CkSik + ∑

k=1

k=n  CkSki = 2 ∑

k=1

k=n  CkSik

Since : Sik = ⌡⎮⌠

 

 

  fi fk dτ = ⌡⎮⌠

 

 

  fk fi dτ = Ski

Thus δ

δCi⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

 ∑j=1

j=n  ∑k=1

k=n   Cj Ck Skj  = 2 ∑

k=1

k=n  CkSik

The linear variation method

Now by replacing Skj with Hkj

δδCi

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Hkj = 2 ∑

k=1

k=n CkHik

Thu s from: δ W

δ Ci ⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj

+ WδδCi

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj = W

δδCi

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j CkH kj

The linear variation method

we get by substitution

δ W

δ Ci

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑j=1

j=n ∑k=1

k=n C j Ck Skj + 2 W

⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑k=1

k=n CkSik =

2 ∑k=1

k=n CkHik

This equation is satisfie d fo r al {l Cj ,j=1,n}. The optimal

set for whic h W i s a t aminimu mmus t i n additi on satisfy

δ W

δ Ci

= 0

The linear variation method

They must thus satisfy

2 W⎝⎜⎜⎜⎛

⎠⎟⎟⎟⎞

∑k=1

k=n CkSik = 2 ∑

k=1

k=n CkHik

Or b y combini ng terms: ∑k=1

k=n Ck [ Hik –2WSi k ] = 0

writte n out

C1 [ Hi1 –2WSi1 ]+C2 [ Hi2 –2WSi2 ] ,.Cn [ Hin –2WSi n ]

= 0

The linear variation method

We have obtained this set of equations from

δ W

δ Ci

= 0

Howev er i= 1,2,3,.. ..,nW e ca n a s a consequenc e obtai n th e se t of nequations

∑k=1

k=n Ck [ Hik -WSi k ] = 0 =i 1,n

The linear variation method

C1[ H11 -WS11 ]+C2 [ H12 - WS12 ] ,...Cn [ H1n - WS1n ]

= 0

C1[ H21 - WS21 ]+C2 [ H22 - WS22 ] ,...Cn [ H2n - WS2n ]

= 0

C1[ H31 -WS31 ]+C2 [ H32 -WS32 ] ,...Cn [ H3n -WS3n ] =

0

.......

C1 [ Hi1 -WSi1 ] +C2 [ Hi2 -WSi2 ] ,...Cn [ Hin -WSin ] = 0

C1 [ Hn1 -WSn1 ] +C2 [ Hn2 -WSn2 ] ,...Cn [ Hnn -WSnn ]

= 0

This is a set of n homogeneous equations

The linear variation method

This set of equations has only non-trivial solutions

provided that the secular determinant

[ H11 -WS11 ] [ H12 - WS12 ] [ H1n - WS1n ]

[ H21 - WS21 ] [ H22 - WS22 ] [ H2n - WS2n ]

[ H31 -WS31 ] [ H32 -WS32 ] [ H3n -WS3n ] = 0

.......

[ Hi1 -WSi1 ] [ Hi2 -WSi2 ] [ Hin -WSin ]

....

[ Hn1 -WSn1 ] [ Hn2 -WSn2 ] [ Hnn -WSnn ]

The linear variation method

By expanding out the determinant we will obtain an n-

order polynomial in W.This polynomial has n-roots

I[1]1 <I

[2]1 < I

[3]1 < .....,<I

[i]1 <..I

[n]1

The lowest root I[1]1

is the approximation to the

actual groundstate

energy E1E

1

E

E

3

E

i

E

n

I

1

[1]

I

1

[2]

I

1

[3]

I

1

[i]

I

1

[n]

The linear variation method

Further I[i]1 is an approximation to Ei. In all cases I

[i]1

larger than or equal Ei. Having obtained the roots we

can now find {Ci,i=1,n) by substituting into the set of

equationsC1[ H11 -WS11 ]+C2 [ H12 - WS12 ] ,...Cn [ H1n - WS1n ] = 0

C1[ H21 - WS21 ]+C2 [ H22 - WS22 ] ,...Cn [ H2n - WS2n ] = 0

..

C1 [ Hi1 -WSi1 ] +C2 [ Hi2 -WSi2 ] ,...Cn [ Hin -WSin ] = 0

....

C1 [ Hn1 -I1Sn1 ] +C2 [ Hn2 -I1Sn2 ] ,...Cn [ Hnn -I1Snn ] = 0

where W = I[i]1 i=1,n

The linear variation method

What you should learn from this lecture1. You should understand that in linear variationtheory the trial wavefunction is written as a linear combination of KNOWN functionswhere the relative contribution from each functionis optimized.

2. You should know how the set of linear equationsare generated and why the secular determinant must bezero and how this is used to determine theenergies.

3. You should be able to derive the equation in the case where n=2