# Why are 2 ways to calculate the energy of the ground state?

Setup

I found in Szabo, Quantum chemistry page 88, these two ways to calculate the energy of the ground state:

$$E_0= \sum_a^N \langle a|h|a\rangle + \frac{1}{2} \sum_a^N \sum_b^N \langle ab||ab\rangle$$

and

$$E_0= \sum_a^N \langle a|h|a\rangle + \sum_a^N \sum_{b>a}^N \langle ab||ab\rangle$$

Question

Why?

• Looks like by the basal state the OP actually meant the ground state. – Wildcat Apr 20 '15 at 17:16
• Yes, thank you to enlightenme and in that moment I didn't remember the correct word in english – Another.Chemist Apr 20 '15 at 17:17
• And a rather short parenthetical comment: try to avoid temptation to understand the meaning of an expression without following its derivation. Try to follow the whole Szabo's derivation of the HF equations and you will understand the matter way better. – Wildcat Apr 20 '15 at 18:07
• Yes @wildcat that procedure is the obvious way and... also, as one day you suggest me... try to improve the functional analysis skills which I don't have well developed them. Of course, I want to improve those skills, but now I'm running out of time. Thanks again to you and all that have lending me a hand. – Another.Chemist Apr 20 '15 at 22:29

Both formulas have exactly the same meaning. They are just two different ways of not double counting the Coulomb and the exchange interactions. In principle, the goal can be achieved in two ways:

• double counting in the first place and then dividing by two (first formula);
• counting wisely so that there is no double counting in the first place (second formula).

Obviously, for any practical purposes when you one does not want to double count one would almost always choose the second procedure, but the first representation in some cases provides more apparent picture when dealing with the formulas analytically.

Where does all this come from?

The whole derivation of the expression for the ground state Hartree-Fock energy basically starts from writing down the expectation value of the ground state electronic energy for the ground state Slater determinant $\Phi_{0}$ $$\langle \Phi_{0} \mid \hat{H}_{\mathrm{e}} \mid \Phi_{0} \rangle \, ,$$ in terms of the individual spin-orbitals of which $\Phi_{0}$ is constructed. And in doing so we use the so-called Slater rules and it happens so that the electronic Hamiltonian $\hat{H}_{\mathrm{e}}$ contains only two different kinds of operators:

• one-electron operators of the form $$\hat{F} = \sum\limits_{i=1}^{N}\ \hat{f}(i)\,,$$
• and two-electron operators of the form $$\hat{G} = \frac 12 \sum\limits_{i=1}^{N} \sum\limits_{j=1\atop{j\neq i}}^{N}\ \hat{g}(i,j)\,.$$

Already at this stage an expression for any two-electron operator $\hat{G}$ could as well be written as $$\hat{G} = \sum\limits_{i=1}^{N} \sum\limits_{j>i}^{N}\ \hat{g}(i,j)\,.$$ The mathematical meaning is exactly the same: we do not want to double count pairwise interactions, so that if, say $\hat{g}(1,2)$ is already counted, $\hat{g}(2,1)$ has to be excluded.

Note carefully that we also have to exclude physically meaningless self-interactions, such as, say, $\hat{g}(1,1)$, which is done in the first case by an additional restriction on summation $j\neq i$ and in the second case by the strict inequality in the lower limit of summation $j>i$.

With respect to the exclusion of this non-physical self-interactions note now that they it is missing from the HF energy expression being written in the first form. But that is perfectly fine since each and every Coulomb self-interaction is perfectly canceled by the corresponding exchange self-interaction. Thus, you do not necessarily need the above mentioned additional restriction of the form $j\neq i$ ($b \neq a$ this time), although you could include it if you would like to, $$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b=1\atop{b \neq a}}^N \langle ab||ab\rangle \, ,$$ or adopting a rather usual convention for such summations, as follows, $$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b \neq a}^N \langle ab||ab\rangle \, .$$

• Thank Wildcat so much for your answer and for sharing some of your time. But, I still don't understand the details of why those expressions are different. – Another.Chemist Apr 20 '15 at 17:09
• They are not different, they are the same. These are two different mathematical ways to express the same idea: we don't want to double count pairwise interactions. – Wildcat Apr 20 '15 at 17:11
• ap.... dough ... slater determinant, only take the upper triangle of the determinant. – Another.Chemist Apr 20 '15 at 17:14
• @beginner, not, it has nothing to do with the Slater determinant. Give me a few minutes, I'll try to elaborate a little bit more. – Wildcat Apr 20 '15 at 17:16
• beginner, what @Wildcat is saying is that for a pair (i,j) interaction you only have to count it just one time. Note that (i,j) interaction is the same that (j,i) interaction. So if the interaction (i,j) is $I_{i,j}$, it is true that: $I_{i,j} = I_{j,i} = (1/2) ( I_{i,j} + I_{j,i})$ – user1420303 Apr 20 '15 at 22:35