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Hartree-Fock Theory - Chemistry

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<strong>Hartree</strong>-<strong>Fock</strong> <strong>Theory</strong><br />

I. He atom revisited<br />

H = h(1) + h(2) + e 2 r 12<br />

(1)<br />

Consider a more general form for ψ (1, 2), i.e.,<br />

⎛ αβ − βα<br />

ψ (1, 2) = φ 1s (1)φ 1s (2) ⎜<br />

⎝ 2<br />

⎞<br />

⎟ (2)<br />

⎠<br />

Seek best form for φ 15 . <strong>Hartree</strong> theory can be shown that optimum φ 15<br />

satisfies<br />

⎡<br />

⎤<br />

⎢ h(1) + φ (2) e2<br />

φ(2) − ε<br />

r 1<br />

⎥<br />

⎣ ⎢<br />

12 ⎦ ⎥ φ 1s (1) = 0<br />

⎡<br />

h(2) + φ(1) e 2 ⎤<br />

⎢<br />

φ(1) − ε<br />

r 2<br />

⎥<br />

⎣ ⎢<br />

12 ⎦ ⎥ φ 1s (2) = 0<br />

These are coupled integro-differential equations. Difficult to solve<br />

directly. Instead, use an iterative method. Describe this later.<br />

(3a)<br />

(3b)<br />

First, rewrite (3) as<br />

[ h eff (1) − ε 1 ]φ 1s ( r 1 ) = 0 (4a)<br />

[ h eff (2) − ε 2 ]φ 1s (r 2 ) = 0 (4b)<br />

where<br />

h eff (i) = −h2<br />

2m ∇ 2 i − Ze2<br />

r i<br />

+ φ 1s (j) e2<br />

r 12<br />

φ 1s ( j) . (5)<br />

<strong>Hartree</strong>-<strong>Fock</strong> <strong>Theory</strong><br />

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The expectation value of H is, as usual,<br />

E = H = E (0) 1s +E (0) 1s + φ 1s (1)φ 1s (2) e2<br />

φ<br />

r 1s (1) φ 1s (2) (6)<br />

12<br />

= −Z 2<br />

2 − Z 2<br />

2 + J 1s,1s = −Z 2 + J 1s,1s<br />

, (7)<br />

where J 1s,1s ≡ 1s(1)1 s (2) e2<br />

r 12<br />

1s(1)1 s (2) , (8)<br />

but where φ 1s here is determined from eq. (4), i.e., it is not the<br />

hydrogenic 1s wavefunction, nor even the one with the orbital exponent<br />

optimized.<br />

More details<br />

Define the effective potential u i eff (r i ) by<br />

u i eff (r i ) ≡ −Ze2<br />

r i<br />

+ φ(2) e2<br />

r 12<br />

φ(2) . (9)<br />

Make u i eff (r i ) spherically symmetric by averaging over angles.<br />

So, if in general φ( r i ) = R i (r i )Υ li m i<br />

( θ i ,φ i ) (10)<br />

⎡<br />

⎢<br />

⎣ ⎢<br />

⎡<br />

⎢<br />

⎣ ⎢<br />

−h 2<br />

2m<br />

−h 2<br />

2m<br />

we have<br />

⎛ d 2<br />

2<br />

dr + 2 d ⎞<br />

⎜<br />

⎝ 1 r 1 dr<br />

⎟<br />

1 ⎠<br />

+ l 1( l 1 +1 )h 2 ⎤<br />

2<br />

+ u eff<br />

2mr 1 − ε 1<br />

⎥<br />

1 ⎦ ⎥ R 1 (r 1 ) = 0<br />

⎛ d 2<br />

2<br />

dr + 2 d ⎞<br />

⎜<br />

⎝ 2 r 2 dr<br />

⎟<br />

2 ⎠<br />

+ l 2( l 2 + 1 )h 2 ⎤<br />

2<br />

+ u eff<br />

2mr 2 − ε 2<br />

⎥<br />

2 ⎦ ⎥ R 2 (r 2 ) = 0<br />

The solution to these equations is obtained by an iterative method, as<br />

follows.<br />

(11a)<br />

(11b)<br />

<strong>Hartree</strong>-<strong>Fock</strong> <strong>Theory</strong><br />

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Start with<br />

φ (0) 1s (i) = hydrogenic 1s orbital<br />

Calculate u eff ,0 (i), then solve as the uncoupled equations using a<br />

numerical method (see below for another approach). Call ε i<br />

(1)<br />

and φ (1) 1s (i)<br />

the 1 st iterate. Use these to obtain u eff ,1 (i) and re-solve the eigenvalue<br />

equations to get ε (2) i<br />

and φ (2) 1s ( i). Continue to iterate until the process<br />

converges, i.e.,<br />

ε i (n) = ε i (n−1) ± δ i<br />

where δ i<br />

is a small value, the convergence threshold.<br />

<strong>Hartree</strong>-<strong>Fock</strong> <strong>Theory</strong><br />

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