In standard quantum chemistry books (e.g., Szabo Ostlund), Hartree-Fock is usually introduced from a first quantized picture. Given molecular orbitals $\psi_a(\mathbf{r})$ that are expanded in terms of atomic orbitals $\phi_\mu(\mathbf{r})$, such that

$$ \psi_a(\mathbf{r}) = \sum_{\mu}C_{a\mu}\phi_\mu(\mathbf{r})$$

the density matrix is introduced as

$$ P_{\mu\nu} = 2\sum_{a=1}^{N/2} C_{a\mu}^* C_{a\nu} $$

where the factor of 2 accounts for spin, and N is the number of electrons. This makes sense. However, I've also (usually in physics textbooks) seen a second-quantized approach to Hartree-Fock where the central variational object is given by $P_{\alpha\beta} = \langle \Omega\vert c^\dagger_{\alpha}c_\beta\vert\Omega\rangle$, where $\Omega$ is the true ground state. I assume this is equal to the density matrix, but I am having a difficult time proving it and was wondering if someone could help. Changing the first quantized notation into second quantization gives

$$ \rho(\mathbf{r}) = \sum_{a=1}^N \langle\mathbf{r}\vert a\rangle\langle a\vert \mathbf{r}\rangle \\ = \sum_{a=1}^N\sum_{\mu,\nu} \langle\mathbf{r}\vert \mu\rangle\langle\mu\vert a\rangle\langle a\vert\nu\rangle\langle\nu\vert\mathbf{r}\rangle\\ = \sum_{\mu\nu}\left( \sum_{a=1}^N \langle\mu\vert a\rangle\langle a\vert\nu\rangle \right)\langle\mathbf{r}\vert \mu\rangle \langle\nu\vert\mathbf{r}\rangle\\ \Rightarrow P_{\mu\nu} = \sum_{a=1}^N \langle\mu\vert a\rangle\langle a\vert\nu\rangle $$

How should I proceed from here?

  • $\begingroup$ Two things: By $\langle c^\dagger_{\alpha}c_\beta\rangle$ do you really mean $\langle c_{\alpha} | c_\beta\rangle$? If not, what does it mean? Also in your last line you have an identity, so this reduces to $P_{\mu\nu} = \langle \mu | \nu \rangle$. Is this what you're looking for? $\endgroup$
    – Jacob A
    Aug 30 at 5:51
  • $\begingroup$ Updated the question with more details. I meant $\langle\Omega\vert c_\alpha^\dagger c_\beta\vert\Omega\rangle$, as in the expectation value of the creation-anihilation pair. And the last line isn't an identity, because it's a sum over all the electrons, not all the states in general. $\endgroup$
    – user147177
    Aug 30 at 16:33

I figured this out, it's just a change of basis.

$$ \langle f_\mu^\dagger f_\nu \rangle = \langle\, \vert c_{a_N}\dotsc_{a_1}\, f_{\mu}^\dagger f_\nu \, c_{a_1}^\dagger\dots c_{a_N}^{\dagger}\vert\, \rangle\\ = \sum_{i,j} \langle \mu\vert a_i\rangle\langle a_j\vert\nu\rangle \langle\,\vert c_{a_N}\dots c_{a_1}\, c_{a_i}^{\dagger} c_{a_j}\, c_{a_1}^\dagger\dots c_{a_N}^\dagger\vert\,\rangle\\ =\sum_{a=1}^{N}\langle\mu\vert a\rangle\langle a\vert\nu\rangle = \sum_{a=1}^N C_{\mu a} C_{\nu}^* $$


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.