Hartree-Fock Theory - Chemistry
Hartree-Fock Theory - Chemistry
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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