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It is a standard textbooks remark that, in the usual notation, $$ \langle\Psi,\mathcal{N}\psi\rangle\;=\;\sum_\alpha\langle\Psi,a^*(f_\alpha)a(f_\alpha)\psi\rangle\,, $$ where $\mathcal{N}$ is the number operator on (the bosonic or fermionic) Fock space $\mathfrak{F}_\pm$, $\Psi\in\mathfrak{F}_\pm$, $a(f)$ is the (bosonic or fermionic) creation operator relative to the one-body state $f$, and $(f_\alpha)_\alpha$ is an orthonormal basis of the Hilbert space which the Fock space il built on. ($\mathcal{N}$ is defined the usual way by $(\mathcal{N}\Psi)^{(n)}:=n\Psi^{(n)}$, etc.)

Could anyone help me in proving this identity? Thanks!

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I am not sure whether you are still interested in an answer, but nevertheless I am going to provide one.

In principle there come two different approaches to my mind. The first one requires a bit more effort but does not use occupation numbers (if you are not familiar with them). Let $\mathfrak{h}$ be the one-particle Hilbert space the Fock space is builded from. Moreover, for a one-particle operator $h$ we define $h_j=\mathbb{1}\otimes\cdots\otimes\mathbb{1}\otimes h\otimes\mathbb{1}\otimes\cdots\otimes\mathbb{1}$ ($h$ at the $j^\text{th}$ position). We need the theorem:

Let $h:D(h)\to\mathfrak{h}$ be symmetric and let $\{\psi_i\}_i\subset D(h)$ be an onb of $\mathfrak{h}$. Then, \begin{align*} \bigoplus_{N=1}^\infty\sum\limits_{j=1}^{N}h_j=\sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}\langle\psi_i,h\psi_j\rangle a^*(\psi_i)a(\psi_j) \end{align*} where the left-hand side is defined on $\bigcup_{n=0}^\infty\bigoplus_{N=1}^nP_\pm D\left(\sum_{j=1}^{N}h_j\right)$ $(n=0\to\mathbb{C})$ and \begin{align*} P_+&=\frac{1}{N!}\sum_{\sigma\in S_N}U_\sigma,\\ P_-&=\frac{1}{N!}\sum_{\sigma\in S_N}\operatorname{sign}(\sigma)U_\sigma \end{align*} (cf. Reed Simon 1, Hilbert space chapter). Observe that the theorem holds true in both cases, fermionic and bosonic.

Proof: We start with distinguishable particles described by the same Hilbert space $\mathfrak{h}$ and the same one-particle Hamiltonian $h$: \begin{align*} \sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}\langle\psi_i,h\psi_j\rangle a^*(\psi_i)a(\psi_j)\varphi_1\otimes\dots\otimes\varphi_N&=N\sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}\langle\psi_i,h\psi_j\rangle\langle\psi_j,\varphi_1\rangle\psi_i\otimes\varphi_2\otimes\dots\otimes\varphi_N\\ &=Nh_1\varphi_1\otimes\dots\otimes\varphi_N \end{align*} for $\varphi_1,\dots,\varphi_N\in D(h)$. Using the symmetrizing respectively antisymmetrizing property of $P_\pm$, we obtain $P_\pm Nh_1P_\pm=P_\pm\sum_{j=1}^{N}h_jP_\pm=\sum_{j=1}^Nh_jP_\pm$ and hence \begin{align*} \sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}\langle\psi_i,h\psi_j\rangle a^*_\pm(\psi_i) a_{\pm}(\psi_j)P_\pm=\sum\limits_{j=1}^{\infty}\sum\limits_{i=1}^{\infty}\langle\psi_i,h\psi_j\rangle P_\pm a^\dagger(\psi_i) a(\psi_j)P_\pm=\bigoplus_{N=1}^\infty\sum\limits_{j=1}^{N}h_jP_\pm. \end{align*}

From this theorem, your assertion follows immediately by setting $h_j=\mathbb{1}$.

The second way of proving your claim is using the fact that the occupation number representation provides an orthonormal basis for the fermionic (bosonic) Fock space and to show the equation for this basis vectors which is not really complicated. If you want to know about this approach, please let me know.

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  • $\begingroup$ Thanks a lot!! Would you also be so kind as to explain to me the other method? You were extremely clear. Thanks! $\endgroup$ – TullioRicci87 Oct 29 '17 at 17:18
  • $\begingroup$ I am glad that my answer helps. To which extent are you familiar with the occupation number representation? $\endgroup$ – julian Oct 29 '17 at 18:20

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