TITLE: Precise definition of exchange
QUESTION [0 upvotes]: I was wondering about how could one define the exchange interaction in the most general way. 
I am aware of the definition of exchange for the Coulomb interaction, which looks basically like $\int \phi^\ast_1(r)\phi^\ast_2(r^\prime) \dfrac{1}{\vert r-r^\prime \vert} \phi_2(r^\prime) \phi_1(r),$ and I have also seen, for example, how one defines the exchange energy for Hund-like terms in second-quantized hamiltonians, $\hat{n}_{\alpha,s}\hat{n}_{\beta,s^{\prime}}$ (here $\hat{n}_{\alpha,s}=\hat{c}^\dagger_{\alpha,s}\hat{c}_{\alpha,s}$ is the number operator for state $\alpha$ with spin $s$) as $\langle\hat{c}^\dagger_{\alpha,s} \hat{c}_{\beta,s}\rangle \langle \hat{c}^\dagger_{\beta,s^\prime}\hat{c}_{\alpha,s^\prime} \rangle. $
My first doubt is that I don't understand at all how these two definitions relate to each other, although it is clear that they share the logic of "twisting the indeces".
Secondly, I was wondering whether it would make sense to define a Hartree and an exchange contribution for very general terms like $\hat{c}^\dagger_i\hat{c}^\dagger_j\hat{c}_k\hat{c}_l.$ 
I believe that, intuitively at least, the exchange are just those non-classical terms which arise due to the antisymmetry of the wave-function when computing expected values over the lowest-energy Slater determinant. Right?
Thus, given a Hamiltonian, I thought of trying to define "the exchange of $\hat{c}^\dagger_i\hat{c}^\dagger_j\hat{c}_k\hat{c}_l$" as the diference:
$$\langle \Phi \vert \hat{c}^\dagger_i\hat{c}^\dagger_j\hat{c}_k\hat{c}_l \vert \Phi \rangle -\langle \bigotimes_{
n=1}^{N}\phi_{n} \vert \hat{c}^\dagger_i\hat{c}^\dagger_j\hat{c}_k\hat{c}_l \vert \bigotimes_{n=1}^{N} \phi_n  \rangle,$$
where $\text{Alt}\left( \bigotimes_{
n=1}^{N}\vert \phi_{n} \rangle \right),$ where the $\vert  \phi_n \rangle$ are the Hartree-Fock spin-orbitals (or they could be the Kohn-Sham spin-orbitals, if you want. The point is just we define the exchange in a precise way in the context of a one-electron approximation).
This seemed to me like a fine candidate for a pretty general definition of exchange, but the problem is, of course, that the action of operators $\hat{c}^{\dagger}_{i}, \hat{c}_j$ is not well defined for states which are not antisymmetrized. Is there any way around this difficulty so that we can indeed define an exchange term in a meaningful way for the expected values of so general operators?  Perhaps something in the line of Wick's theorem?

REPLY [1 votes]: A 2-body operator in second quantization $\hat{O}$ has the form (assuming spin-independent interactions)
$$\dfrac{1}{2}\sum_{a,b,c,d,s,s^\prime}O_{a,b}^{c,d}\hat{c}^\dagger_{a,s}\hat{c}^\dagger_{b,s^\prime}\hat{c}_{d,s^\prime}\hat{c}_{c,s},$$
where 
$$O_{a,b}^{c,d}=(a,b\vert \hat{O} \vert c,d)=\int dr dr^\prime O(r,r^\prime) \phi_{a,s}^\ast(r)\phi_{b,s^\prime}^\ast(r^\prime)\phi_{c,s}(r)\phi_{d,s^\prime}(r^\prime ).$$
The Hartree part has to do with pairing up the terms with the same variable and spin, $\phi_{a,s}(r)$ with $\phi_{c,s}(r)$ and $\phi_{b,s^\prime}(r^\prime)$ with $\phi_{d,s^\prime}(r^\prime).$ The exchange part correspond to the "twisted" pairing in which we associate the $a$ with the $d$ and the $b$ with the $c.$
Using Wick/Isserlis theorem, valid for expected values over Slater determinants, one has:
$$\langle \hat{c}^\dagger_{a,s}\hat{c}^\dagger_{b,s^\prime}\hat{c}_{d,s^\prime}\hat{c}_{c,s} \rangle= \langle \hat{c}^\dagger_{a,s}\hat{c}_{c,s} \rangle \langle \hat{c}^\dagger_{b,s^\prime}\hat{c}_{d,s^\prime} \rangle-\langle \hat{c}^\dagger_{a,s}\hat{c}_{d,s} \rangle \langle \hat{c}^\dagger_{b,s^\prime}\hat{c}_{c,s^\prime} \rangle.$$
From the considerations above, is clear that the first terms should be defined as the Hartree contribution and the second one as the exchange part.
{"set_name": "stack_exchange", "score": 0, "question_id": 356596}