[mszep]

I need to calculate some of the more complicated self-energy terms from chapter 7 of Szabo and Ostlund's "Modern Quantum Chemistry", and I'm having trouble converting summations from spin orbitals to spatial orbitals.

Exercise 2.18 (page 85) is indicative of what I'm doing wrong, since I can't get the correct answer.

The exercise is to show that the leading correction to the Hartree-Fock ground state energy

[tex]
E_0^2 = \frac{1}{4} \sum_{abrs} \frac{|\langle ab || rs \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
[/tex]

where the summation is over spin orbitals ([itex]a[/itex], [itex]b[/itex] over occupieds, [itex]r[/itex], [itex]s[/itex] over virtuals), can be converted to the following sum over spatial orbitals:

[tex]
E_0^2 =
\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ \langle ab | rs \rangle (2 \langle rs | ab \rangle - \langle rs | ba \rangle)}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
[/tex]

The integrals are in "physicists notation", which means that

[tex]
\langle ij|kl \rangle =
\int \chi_{i}^*(\mathbf{x}_1) \chi_{j}^*(\mathbf{x}_2) r_{12}^{-1}
     \chi_{k}(\mathbf{x}_1) \chi_{l}(\mathbf{x}_2) \,\mathrm{d}\mathbf{x}_1 \,\mathrm{d}\mathbf{x}_2
[/tex]

and

[tex]
\langle ij||kl \rangle = \langle ij | kl \rangle - \langle ij | lk \rangle
[/tex]

where the [itex]\chi_i[/itex] are spin orbitals and the variables [itex]\mathbf{x}[/itex] comprise both spatial and spin coordinates.

The following is what I have, I'd appreciate any pointers to get to the right answer.

First, decompose the sum into the different possible permutations of spin for the different orbitals ([itex]a[/itex] and [itex]\bar{a}[/itex] are orbitals with opposite spin), and use the definition of [itex]\langle ab || rs \rangle[/itex]:

[tex]
\begin{align}
E_0^2 &=\frac{1}{4}\Bigg(
\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|rs \rangle - \langle ab|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}\bar{b}|rs \rangle - \langle \bar{a}\bar{b}|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}b|\bar{r}s \rangle - \langle \bar{a}b|s\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}b|r\bar{s} \rangle - \langle \bar{a}b|\bar{s}r \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle a\bar{b}|\bar{r}s \rangle - \langle a\bar{b}|s\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle a\bar{b}|r\bar{s} \rangle - \langle a\bar{b}|\bar{s}r \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|\bar{r}\bar{s} \rangle - \langle ab|\bar{s}\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle \bar{a}\bar{b}|\bar{r}\bar{s} \rangle - \langle \bar{a}\bar{b}|\bar{s}\bar{r} \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\Bigg)
\end{align}
[/tex]

Note that we do not need to consider terms with odd numbers of orbitals of either spin, since terms of the form [itex]\langle \bar{i}j|kl \rangle[/itex] and [itex]\langle i\bar{j}|\bar{k}\bar{l}\rangle[/itex] are zero.

Next, we use the additional rules [itex]\langle \bar{i}j|k\bar{l}\rangle[/itex] = [itex]\langle i\bar{j}|\bar{k}l \rangle[/itex] = [itex]\langle \bar{i}\bar{j}|kl \rangle[/itex] = [itex]\langle ij|\bar{k}\bar{l}\rangle[/itex] = [itex]0[/itex].

We see that in the above eight summations, the second and seventh cancel completely, while in the middle four, only one term is retained.

Then, expanding the squares and collecting terms, we get

[tex]
\begin{align}
E_0^2 &=\frac{1}{4}\Bigg(\quad
4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|rs \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
-4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ \langle ab|rs \rangle  \langle ab|sr \rangle }
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\\&\qquad\quad
+4\sum_{a,b=1}^{N/2} \sum_{r,s=N/2+1}^{K}
\frac{ |\langle ab|sr \rangle |^2}
{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\Bigg)
\end{align}
[/tex]

which is where I'm stuck (besides of course cancelling out the factor 4). Any help in moving towards the correct answer is massively appreciated!

[Corribus]

mszep, it took me a while to get around to fooling with your problem. I gave it the old college try but it's hard to jump right into someone else's notation midstream without any context to go on. I don't have a copy of that textbook. It's been ages since I did this kind of manipulation, so I'm afraid I don't have a whole lot of suggestions for you. You might try over at the physics forums (www.physicsforums.com). It is populated by more people with more recent knowledge about advanced theoretical/quantum chemistry.

[mszep]

Thanks for your response, and for taking the time to look at it; I might post it on physicsforums if I can't figure it out this weekend.

That said, the book is relatively easy to find online, and I'm more than happy to clarify any notation or context that's unclear...

[pm133]

*MOD Edit -- Monster quote removed*

Not sure if you got to the bottom of this or not but I have a belated Christmas present for you.
A colleague of mine dug out a website where someone worked out most of the answers to the Szabo and Ostlund book. It was a life saver for me. When he found it I almost cried with relief 
Here it is:-

http://claudiug.com/9780486691862/chapter.php?c=1&e=19

Enjoy!