An identity involving creation and destruction operators Given a orthonormal basis of one-electron wavefunctions and its associated set of creation and destruction operators, $\hat{c}^\dagger_{\nu\sigma},$ $\hat{c}_{\nu\sigma},$ I've seen in a paper the following identity stated without proof:
$$\sum_{\mu} \langle \hat{c}^\dagger_{\nu\sigma} \hat{c}_{\mu\sigma} \rangle \langle \hat{c}^\dagger_{\mu\sigma}\hat{c}_{\nu\sigma} \rangle= \langle \hat{c}^\dagger_{\nu\sigma} \hat{c}_{\nu\sigma} \rangle.$$
I've tried to prove it by inserting a resolution of the identity in a couple of different forms, but I haven't been able to really reach a satisfactory proof of the fact.

If anyone is interested, the paper is 

Molecular-orbital theory for chemisorption: The case of H on normal metals. F. J. García-Vidal et al. Phys. Rev. B 44, 11412 (1991), author eprint

and the equation I am struggling with is number (41) in the paper.
 A: This relation is true over Slater determinants and false in general, as hinted in the comments. Indeed, if a Slater determinants is build up from the single particle wavefunctions $\vert \phi_j \rangle$ (I will omit spin for simplicity; this is not relevant), we have:
$$\sum_{\nu} \langle \hat{c}^{\dagger}_\mu \hat{c}_\nu \rangle \langle \hat{c}^\dagger_{\nu} \hat{c}_\mu \rangle=\sum_{\nu}\sum_{j,k} \langle \phi_j \vert \hat{c}^{\dagger}_\mu \hat{c}_\nu  \vert \phi_j \rangle \langle \phi_k \vert \hat{c}^{\dagger}_\nu \hat{c}_\mu  \vert \phi_k \rangle.$$
Next expand the $\vert \phi_j \rangle$ in the basis of orthonormal wavefunctions (call them $\vert l_\alpha \rangle,$ for example) to which the operators $\hat{c}^\dagger,$$\hat{c}$ are associated; that is, write $\vert \phi_j \rangle=\sum_{\alpha} b_{j\alpha}\vert l_\alpha \rangle.$ Then the expresion aboves transform into:
$$\sum_{\nu}\sum_{j,k}b^\ast_{j\mu}b_{j\nu}b^\ast_{k\nu}b_{k\mu}=\sum_{j}b^\ast_{j\mu}b_{j\mu}=\langle \hat{n}_{\mu} \rangle,$$
because $\sum_{\nu}b^\ast_{j\nu}b_{k\nu}=\delta_{j,k}$ due to the fact $\langle l_j \vert l_k \rangle =\delta_{j,k}.$
As can be seen, the only idea is to use the fact that, given a one particle operator $\hat{t}$ extended to a N particle Fock space as $\sum \text{Id} \otimes\ldots\otimes \hat{t} \otimes \ldots \otimes \text{Id}$ (the sum means that the $\hat{t}$ is put at a different sloth at each term), it holds for Slater determinants that:
$$\langle \text{Alt} \bigotimes_{j=1}^{N} \vert \phi_j \rangle \vert \hat{t} \vert \text{Alt} \bigotimes_{j=1}^{N} \vert \phi_j \rangle \rangle=\sum_{j}^{N} \langle \phi_j \vert \hat{t} \vert \phi_j \rangle$$
