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
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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 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
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