Computational Physics

9.3 An example: quantum molecular dynamics for the hydrogen molecule 275

Taking the derivative with respect toXwe find

d

dX

〈1s,α,A

∣∣

∣∣−^1

2

∇^2

∣∣

∣∣1s,β,B〉

=− 4 σ^2 XSα,A;β,B+[ 3 σ− 2 σ^2 X^2 ]

d

dX

Sα,A;β,B. (9.36)

The Coulomb matrix element is given by

〈1s,α,A

∣∣

∣∣

∣

∑

c

1

rc

∣∣

∣∣

∣

1s,β,B〉=θ

∑

c

Sα,A;β,BF 0 (tc) (9.37)

withθ= 2

√

(α+β)/π,tc=(α+β)(PC)^2 wherePis the point

RP=

αRA+βRB

α+β

, (9.38)

PQ=RP−RQ, andCis the position of the nucleus. The sum

∑

cis over the two
nuclei.F 0 was given inSection 4.8– its derivative is given by

F 0 ′(t)=

e−t−F 0 (t)

2 t

(9.39)

fort =0, andF 0 ′( 0 )=− 1 /3. Taking the derivative, we obtain for two basis
functions centred on the same nucleus:

d

dX

〈1s,α,A

∣∣

∣∣

∣

∑

c

1

rc

∣∣

∣∣

∣

1s,β,B〉= 2 θSα,A;β,BF 0 ′(t)X(α+β) (9.40)

witht=(α+β)X.
For basis functions centred on different nuclei, we have
d
dX

〈1s,α,A

∣∣

∣∣

∣

∑

c

1

rc

∣∣

∣∣

∣

1s,β,B〉=θ

d

dX

(Sα,A;β,B)

∑

c

[F 0 (t 1 )+F 0 (t 2 )]

+ 2

θ

α+β

Sα,A;β,B[F 0 ′(t 1 )α^2 +F 0 ′(t 2 )β^2 ]X. (9.41)

where

t 1 =

α^2 X^2

α+β

; (9.42a)

t 2 =

β^2 X^2

α+β

. (9.42b)

Finally the four-electron matrix element is given by

〈α,A;γ,C|g|β,B;δ,D〉=ρSα,A;β,BSγ,C;δ,DF 0 (t) (9.43)