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 ψn

nd eigenvalues En given by the SWE

Hψn = Enψn

Let us look at the groundstate ψ1 with the energy E1 .

We would like to find a wavefunction Φ1 for which the

nergy

H d d= ∫ ∫1 1 1 1∗ ∗Φ Φ Φ Φτ τ/

is close to E1

The linear variation method

We write Φ1 as a tt rr iiaa ll wavefunction in terms of a

inear combination of known functions {fi }

hat depends on the same variables as ψ1 and have the

ame boundary 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 possiblealue.

That is ,we shall find the absolute minima of theunction W(C1,C2,C3,.....,Ci,..Cn).

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

t 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

δδWCi

= W’(C1,C2,.....,Cn)

Then W’(C11 ,C

12,C

13, ...,C

1j ,..C

1n ) = 0

We shall use this fact to find { C11 ,C

12,C

13, ...,C

1j ,..C

1n}

The linear variation method

irst let us substitute the expression for Φ1 into the

xpression for W. The denomenator of I1 is

⌡⌠

Φ1 ∗ Φ1 dτ = ⌡⌠

∑j=1

j=n Cj*fj*

∑k=1

k=n Ckfk

We shall assume that all functions {fj j=1,n} are real and

hat all coefficients {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

Let us introduce : ⌡⌠

fjfkdτ = Sjk

Thus : ⌡⌠

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

Or after multiplication of paranthesis

⌡⌠

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 )*.hus 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 Cj Ck Skj = ∑

j=1

j=n ∑k=1

k=n Cj Ck Hkj

It is important to observe that (C1,C2,..,Ci,.. Cn) are

ndependent variables

We shall now differentiate with respect to one ofhem,say Ci ,on both sides of the equation.

The linear variation method

We have

δδW

Ci

∑j=1

j=n ∑k=1

k=n Cj Ck Skj + W

δδCi

∑j=1

j=n ∑k=1

k=n Cj Ck Skj =

δδCi j 1

j n

k 1

k nC

=

=

=

=∑ ∑

j k kjC H

Let us now look at δ

δCi

∑j=1

j=n ∑k=1

k=n Cj Ck Skj =

The linear variation method

ince we differentiate a sum by differentiatingach term from rules for differentiating a product

∑j=1

j=n ∑k=1

k=n

δδCi( Cj Ck Skj ) =

∑j=1

j=n ∑k=1

k=n [

δCj δCiCk Skj +

δCk δCi Cj Skj]

Since (C1,C2,..,Ci,.. Cn) are independent variables

δCjδCi = 0 if i≠j

δCjδCi = 1 if 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

hus 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

δ

δCi

∑j=1

j=n ∑k=1

k=n Cj Ck Hkj = 2 ∑

k=1

k=n CkHik

hus from: δδWCi

∑j=1

j=n ∑k=1

k=n Cj Ck Skj

Wδ

δCi

∑j=1

j=n ∑k=1

k=n Cj Ck Skj = W

δδCi

∑j=1

j=n ∑k=1

k=n Cj CkH kj

The linear variation method

we get by substitution

δδWCi

∑j=1

j=n ∑k=1

k=n Cj Ck Skj + 2 W

∑k=1

k=n CkSik =

2 ∑k=1

k=n CkHik

his equation is satisfied for all {Cj ,j=1,n}. The optimal

et for which W is at a minimum must in addition satisfy

δδWCi

= 0

The linear variation method

hey must thus satisfy

2 W

∑k=1

k=n CkSik = 2 ∑

k=1

k=n CkHik

Or by combining terms: ∑k=1

k=n Ck [ Hik –2WSik ] = 0

written out

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

0

The linear variation method

We have obtained this set of equations from

δδWCi

= 0

However i= 1,2,3,....,nWe can as a consequence obtain the set of n equations

∑k=1

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

The linear variation method

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

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

0C1[ H31 -WS31 ]+C2 [ H32 -WS32 ] ,...Cn [ H3n -WS3n ] =

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

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

0

his is a set of n homogeneous equations

The linear variation method

his set of equations has only non-trivial solutionsrovided 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

s the approximation to theactual groundstate

energy E1 E1

E

E3

Ei

En

I1[1]

I1[2]

I1[3]

I1[i]

I1[n]

The linear variation method

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

[i]1

arger than or equal Ei. Having obtained the roots we

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

quations1[ H11 -WS11 ]+C2 [ H12 - WS12 ] ,...Cn [ H1n - WS1n ] = 0

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

.

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

...

1 [ 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