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In many-body solid-state physics, the Coulomb interaction term in the Hamiltonian usually implies the momentum conservation law in indicies:

$$H_c=\frac{1}{2} \sum_{\mathbf{k},\mathbf{k}',\mathbf{q} \neq 0} V_{\mathbf{q}} a^{\dagger}_{\mathbf{k}'-\mathbf{q}} a^{\dagger}_{\mathbf{k}+\mathbf{q}}a_{\mathbf{k}'} a_{\mathbf{k}},$$

where $\mathbf{k},\mathbf{k}',\mathbf{q}$ are quasi-momenta and quantum numbers for the continuum spectrum of electron gas simultaneously.

In quantum chemistry textbooks, the Coulomb term usually looks like:

$$H_c=\sum_{i,j,k,l} V_{i,j,k,l} a^{\dagger}_{i} a^{\dagger}_{j}a_{k} a_{l}$$

Numbers $i,j,k,$ and $l$ are running over some discrete energy spectrum. Is it possible to state any conservation laws for quantum numbers $i,j,k,$ and $l$ in the expression above? Should not they obey any conservation law, selection rules or any additional restrictions? I will appreciate any references to textbooks or papers.

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Haven't read any quantum chemistry textbooks, but yes, the Coulomb Hamiltonian conserves momentum. How that looks in terms of a restriction on indices depends on how the indices are being used to label the states. –  Michael Brown May 3 '13 at 9:24
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Yikes! That's a tricky basis to discuss momentum conservation in (obviously since translations have a complicated action). So... it's probably best to start in position representation with something like $\int dr_1 dr_2 \psi^\dagger(r_1)\psi(r_1) V(r_1-r_2) \psi^\dagger(r_2)\psi(r_2)$, put in your orthonormal basis for the fields $\psi(r)\sim\sum_{nlms} c_{nlms}(r) a_{nlms}$ and work out the integrals from scratch. I don't know any reference where this is done already. Hopefully someone here does. But that's the idea of where these terms come from. –  Michael Brown May 3 '13 at 10:56
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@DaPhil I disagree, in condensed matter physics the natural basis to choose is plane (Bloch) waves because the free Hamiltonian is invariant under (discrete) translations. In quantum chemistry the "free" problem is obviously not translation invariant and momentum is not conserved. Choosing a basis of momentum eigenstates to make the translation invariance of the Coulomb potential manifest would be a bizarre convention. Far more important is the nuclear Coulomb potential, which is not translation invariant (in the Born-Oppenheimer approximation), so molecular orbitals are a natural basis set. –  Mark Mitchison May 3 '13 at 13:25
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@Mark Mitchison Thank you for a very clear point concerning the momentum conservation law. It will be nice to see it as an answer below. But, still, are there any restrictions or selection rules for indicies in the Coulomb term when we deal with a natural basis set or one has to sum over all possible combinations? –  freude May 3 '13 at 14:05
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You're welcome, hopefully Emilio's detailed response has answered your question. –  Mark Mitchison May 3 '13 at 18:44

1 Answer 1

up vote 4 down vote accepted

Quantum chemistry treats localized systems for which the conservation of total momentum is a true but not-very-useful fact; it therefore doesn't make much sense to incorporate it very explicitly into the notation. (Solid-state systems, on the other hand, are infinite lattices that are invariant under discrete translations, so that the total electron momentum will be a constant of the motion. The formalism is then adapted to this.)

In the quantum chemical formalism, the interaction coefficients are $$V_{ijkl}=\langle\phi_i\phi_j|\hat{v}|\phi_k\phi_l\rangle,$$ where the creation operator $a_i^\dagger$ creates an electron in the state $|\phi_i\rangle$. The pairwise Coulomb repulsion $\hat{v}$ is indeed translation invariant, in that it commutes with the total translation $\hat{U}=e^{i(\hat{\mathbf{p}}_1+\hat{\mathbf{p}}_2) \cdot \mathbf{r}}$ by any displacement $\mathbf{r}$. Thus the coefficient is also equal to $$V_{ijkl}=\langle\phi_i|e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} \otimes\langle\phi_j|e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} \cdot\hat{v}\cdot e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_k\rangle\otimes e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_l\rangle.$$

Unlike in solid-state systems, though, orbitals like $e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_k\rangle$ are not related in any way to the rest of the basis, other than in the single necessary expansion $$e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_k\rangle= \sum_j |\phi_j\rangle\langle\phi_j|e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_k\rangle.$$ With this you can formulate the global translation invariance as a condition on the $V_{ijkl}$: $$V_{ijkl}=\sum_{i',j',k',l'} \langle\phi_i|e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_{i'}\rangle \langle\phi_j|e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi{j'}\rangle \langle\phi_{k'}|e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_k\rangle \langle\phi_{l'}| e^{-i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_l\rangle V_{i'j'k'l'}.$$

The reason this looks so ugly is that there is as yet no selection rule on the matrix elements of the translation between the different basis functions, such as $\langle\phi_i|e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_{i'}\rangle$. In quantum chemical applications, the basis is localized around the nuclei and there will not be any such selection rule, so the above is the best you'll get. (In practice this is not a problem as you know $\hat{v}$ beforehand and use it to calculate the $V_{ijkl}$. If you want to postulate some coefficients then you do need to check the above relation for all displacements $\mathbf{r}$ or your hamiltonian will not be translation invariant.)

Note, though, that since the above formalism is completely general, you still have the option to choose a translation-invariant basis, for which $e^{i\hat{\mathbf{p}} \cdot \mathbf{r}} |\phi_{i}\rangle=e^{i{\mathbf{p_i}} \cdot \mathbf{r}} |\phi_{i}\rangle$, as in solid-state applications. In this case the matrix elements will simplify to delta functions and the coefficients will be forced into the first form you give.

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Thank you for the answer! –  freude May 3 '13 at 17:05

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