Chem 373- Lecture 30: Linear Variation Theory

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Transcript of Chem 373- Lecture 30: Linear Variation Theory

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 The Linear Variation Method ^ We have a Hamiltonian H with the eigenfunctions nnd eigenvalues En given by the SWE ^ Hn = Enn Let us look at the groundstate 1 with the energy E1 . We would like to find a wavefunction 1 nergy for which the

= H 1d / 1d 1 1is close to E1

The linear variation methodWe write 1 as a t r i a l wavefunction in terms of a inear combination of known functions {fi } hat depends on the same variables as 1 and have the ame boundary conditions j=n 1 =

Cjfj

j=1

Or 1 = 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 alue. That is ,we shall find the absolute minima of the unction W(C1,C2,C3,.....,Ci,..Cn). et the values of the coefficients {C1,C2,C3,.....,Ci,..Cn} t the minimum be given by

1 1 1 1 1 C1 ,C2,C3, ...,Cj ,..Cn }

The linear variation methodWhat do we know about this particular coefficients ? We know that if we look at the derivatives of W

W = W(C1,C2,.....,Cn) Ci1 1 1 1 1 Then W(C1 ,C2,C3, ...,Cj ,..Cn ) = 0 1 1 1 1 1 We shall use this fact to find { C1 ,C2,C3, ...,Cj ,..Cn}

The linear variation methodirst let us substitute the expression for 1 into the xpression for W. The denomenator of I1 is j=n k=n Cj*fj* Ckfk j=1 k=1

1 1 d =

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

The linear variation methodAfter multiplication of the two parantheses j=n k=n 1 d = Cj Ck fjfkd 1 j=1 k=1

Let us introduce : fjfkd = Sjk

Thus :

1 1 d = Cj Ck Sjk j=1 k=1

j=n k=n

The linear variation methodFor the numerator in the expression for I1 we have k=n j=n ^ ^ Cj*fj* H Ckfk 1 H 1d = j=1 k=1 Or after multiplication of paranthesis

^ H 1d = = Cj Ck fj H fkd 1 ^ j=1 k=1

j=n k=n

The linear variation methodWe shall now introduce

^ fj H fkd =

Hjk

^ We know that H is Hermetian

^ fj* H fkd =

^ fk (H fj)*d (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 methodWe have j=n k=n

^ 1 H 1d = = Cj Ck Hkj j=1 k=1

j=n k=n

Cj Ck Hkj

j=1 k=1 We can now write W = j=n k=n

Cj Ck Skj

j=1 k=1

The linear variation methodOr j=n k=n j=n k=n W Cj Ck Skj = Cj Ck Hkj j=1 k=1 j=1 k=1 It is important to observe that (C1,C2,..,Ci,.. Cn) are ndependent variables We shall now differentiate with respect to one of hem,say Ci ,on both sides of the equation.

The linear variation methodWe have j=n k=n j=n k=n W + W Cj Ck Skj = Cj Ck Skj Ci Ci j=1 k=1 j=1 k=1 j= n k = n C j Ck H kj Ci j=1 k =1

j=n k=n Let us now look at Ci Cj Ck Skj = j=1 k=1

The linear variation methodince we differentiate a sum by differentiating ach term from rules for differentiating a product j=n k=n ( Cj Ck Skj ) = Ci j=1 k=1 j=n k=n [Cj Ck Skj +Ck Cj Skj] Ci Ci j=1 k=1 Since (C1,C2,..,Ci,.. Cn) are independent variables Cj Ci Cj Ci Cj Ci = ij

= 0 if ij

= 1 if i=j

;

The linear variation method k=n j=n j=n k=n [jiCk Skj +kiCj Skj] = CkSik + CjSjk j=1 k=1 k=1 j=1 k=n

k=n CkSik +

k=n CkSki = 2

CkSik

k=1

k=1

k=1

Since : Sik =

fi fk d =

fk fi d = Ski

k=n j=n k=n hus Ci Cj Ck Skj = 2 CkSik k=1 j=1 k=1

The linear variation method

Now by replacing Skj with Hkj k=n j=n k=n Cj Ck Hkj = 2 CkHik Ci k=1 j=1 k=1 j=n k=n W hus from: Cj Ck Skj Ci j=1 k=1 j=n k=n j=n k=n WCi Cj Ck Skj = WCi Cj CkH kj j=1 k=1 j=1 k=1

The linear variation method we get by substitution k=n j=n k=n W Cj Ck Skj + 2 W CkSik = Ci j=1 k=1 k=1 k=n 2

CkHik

k=1 his equation is satisfied for all {Cj ,j=1,n}. The optimal et for which W is at a minimum must in addition satisfy

W =0 Ci

The linear variation methodhey must thus satisfy k=n k=n 2 W CkSik = 2 CkHik k=1 k=1 k=n Or by combining terms: written out C1 [ Hi1 2WSi1 ]+C2 [ Hi2 2WSi2 ] ,.Cn [ Hin 2WSin ] 0

Ck [ Hik 2WSik ] = 0

k=1

The linear variation methodWe have obtained this set of equations from

W =0 CiHowever i= 1,2,3,....,n We can as a consequence obtain the set of n equations k=n

Ck [ Hik -WSik ] = 0

i=1,n

k=1

The linear variation methodC1[ 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 ] = ...... 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 methodhis set of equations has only non-trivial solutions rovided that the secular determinant [ H11 -WS11 ] [ H12 - WS12 ] [ H1n - WS1n ] H21 - WS21 ] [ H22 - WS22 ] [ H31 -WS31 ] ...... [ Hi1 -WSi1 ] ... [ Hn1 -WSn1 ] [ Hn2 -WSn2 ] [ Hnn -WSnn ] [ H32 -WS32 ] [ Hi2 -WSi2 ] [ H2n - WS2n ] [ H3n -WS3n ] [ Hin -WSin ] =0

The linear variation method By expanding out the determinant we will obtain an norder polynomial in W.This polynomial has n-rootsI En I[n]

[1] [2] [3] [i] [n] I1