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, 2021 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, 2021 at 16:33

1 Answer 1


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

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


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