0
$\begingroup$

The simplest two body interaction term for fermions is

$$H = \sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l$$

and I'm trying to determine the symmetries on $U$. Unfortunately I keep getting weird sign errors. The first symmetry comes from Hermiticity. To have $H$ be Hermitian, we need

$$H = H^\dagger = \sum_{ijkl} U_{ijkl}^\dagger a_l^\dagger a_k^\dagger a_j a_i$$

Then relabel the indices $i\leftrightarrow l$, $j\leftrightarrow k$:

$$H = \sum_{ijkl} U_{lkji}^\dagger a_i^\dagger a_j^\dagger a_k a_l$$

This should indicate that $U_{lkji}^\dagger = U_{ijkl}$. Along similar lines,

$$H = \frac{H + H}{2} = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l + \sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l}{2}$$

Relabel the indices in the second one as $k\leftrightarrow l$:

$$H = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l + \sum_{ijkl} U_{ijlk} a_i^\dagger a_j^\dagger a_l a_k}{2}$$

then apply the anticommutation relation:

$$H = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l - \sum_{ijkl} U_{ijlk} a_i^\dagger a_j^\dagger a_k a_l}{2} =\frac{\sum_{ijkl} (U_{ijkl}-U_{ijlk}) a_i^\dagger a_j^\dagger a_k a_l}{2}$$

Thus suggests that $U_{ijkl}$ is antisymmetric under the last two indices. Similarly, it should be antisymmetric under the first two indices. Unfortunately, a number of online sources seem to suggest that it should be symmetric, for instance http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project3#step_3two-electron_integrals -- here $\langle{\mu\sigma|\lambda\rho\rangle} = U_{\mu\sigma\lambda\rho}$, unless I'm somehow very sorely mistaken. Another reason is that $U_{ijkl}$ will often get contract with the density matrix $D_{kl}$ which is symmetric, and so would vanish if it wasn't antisymmetric.

Is it correct that the symmetries necessary of $U$ are the Hermitian symmetry given above, and antisymmetry in the (12) or (34) pairs? Or is it symmetric? Or neither? Thank you.

$\endgroup$
0
$\begingroup$

$U_{ijkl}$ is anti-symmetric in $(12)$ and $(34)$, your derivation is correct.

The page you're referring to (as well as many others, indeed) contains a mistake. See the correct definitions here.

We define $$ \langle ij|kl\rangle=\int\psi_i^*(\vec{r}_1)\psi_j^*(\vec{r}_2)\hat{h}\psi_k(\vec{r}_1)\psi_l(\vec{r}_2)\,d^3r_1d^3r_2 $$

  1. For general complex wavefunctions, the symmetry of $\langle ij|kl\rangle$ is 4-fold: $$ \langle ij|kl\rangle = \langle ji|lk\rangle = \langle kl|ij\rangle = \langle lk|ji\rangle $$
  2. For real functions, it's 8-fold due to: $$ \langle ij|kl\rangle = \langle ji|kl\rangle $$
  3. For $U_{ijkl}$, only the anti-symmetric part of $\langle ij|kl\rangle$ matters: $$ U_{ijkl} = \langle ij||kl\rangle = \langle ij|kl\rangle -\langle ij|lk\rangle $$
$\endgroup$
  • $\begingroup$ I figured out where my confusion was coming from. The sources that claim it is symmetric are working "2-electron orbitals", that is, treating them as bosons with a Fermi exclusion principle. Exchanging bosons will lead to it being even, obviously. $\endgroup$ – Alex Meiburg Mar 1 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.