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For the creation and annihilation operator $a(\varphi)$ and $a^\dagger(\varphi)$ and the orthogonal projections i would like to understand why the following holds \begin{equation} P_\pm a(\varphi) P_\pm = a(\varphi) P_\pm \,\neq \,P_\pm a(\varphi) \end{equation} and

\begin{equation} P_\pm a^\dagger(\varphi) P_\pm = P_\pm a^\dagger(\varphi) \neq \,a^\dagger(\varphi) P_\pm \end{equation}

I know how $a(\varphi)$ and $a^\dagger(\varphi)$ are defined, but I am not sure how to the projection operators act on them.

Edit: We defined the projection operators in the following way: \begin{equation} P_+ := \frac{1}{N!} \sum\limits_{\sigma \in S_N} U_\sigma \quad\text{and}\quad P_- := \frac{1}{N!} \sum\limits_{\sigma \in S_N} \text{sgn}(\sigma) U_\sigma \end{equation} and annihilation and creation operator as \begin{equation} (a^\dagger(\varphi)\Psi)_N := \sqrt{N}\varphi\otimes \psi_{N-1} \end{equation}

\begin{equation} (a(\varphi)\Psi)_N := \sqrt{N+1}\langle \varphi, \psi_{N+1}\rangle_1 \end{equation}

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    $\begingroup$ I think you need to define your $P_\pm$. There are many projection operators out there! $\endgroup$
    – mike stone
    May 11 at 12:29
  • $\begingroup$ you are absolutely right, i added it $\endgroup$
    – uzizi_1
    May 11 at 13:15
  • $\begingroup$ You should maybe also add the definition of creation/annihilation that you are using $\endgroup$ May 11 at 14:04
  • $\begingroup$ i did, thanks!! $\endgroup$
    – uzizi_1
    May 11 at 16:04
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First of all, the range of $P_\pm a^\dagger(\varphi)$ is a subspace of the (anti)-symmetric states (who are stable under $P_\pm$), while this is not true of $a^\dagger(\varphi)P_\pm $. For example, if $\psi\neq \varphi$ is another $1$-particle state, then : $$a^\dagger(\varphi)P_\pm \psi =a^\dagger(\varphi) \psi = \sqrt{2} \varphi \otimes \psi$$

is not symmetric.

To prove the other part, let $\psi_0,\psi_1,\ldots, \psi_n$ be $1$-particle states. Then :

\begin{align} P_\pm a^\dagger(\psi_0) P_\pm \psi_1 \otimes \ldots \psi_n &= \frac{\sqrt{n+1}}{(n+1)!n!}\sum_{\sigma \in \mathfrak S_{n+1}} \sum_{\sigma' \in \mathfrak S_n}\varepsilon^\pm(\sigma)\varepsilon^\pm(\sigma')U_\sigma (\psi_0 \otimes \psi_{\sigma'(1)} \otimes \ldots \psi_{\sigma'(n)}) \\ &= \frac{\sqrt{n+1}}{(n+1)!n!}\sum_{\sigma \in \mathfrak S_{n+1}} \varepsilon^\pm(\sigma\circ\sigma') (\psi_{\sigma(0)} \otimes \psi_{\sigma\circ\sigma'(1)} \otimes \ldots \psi_{\sigma\circ\sigma'(n)}) \\ &= \frac{\sqrt{n+1}}{(n+1)!}\sum_{\sigma \in \mathfrak S_{n+1}} \sum_{\sigma' \in \mathfrak S_n} \varepsilon^\pm(\sigma) (\psi_{\sigma(0)} \otimes \psi_{\sigma(1)} \otimes \ldots \psi_{\sigma(n)}) \\ &= P_\pm a^\dagger(\psi_0) \psi_1 \otimes \ldots \psi_n \end{align}

where we consider $\sigma' \in \mathfrak S_n \subset \mathfrak S_{n+1}$ by setting $\sigma'(0)=0$.

The results for $a(\varphi)$ are just the adjoint of the ones for $a^\dagger(\varphi)$.

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  • $\begingroup$ thank you very much! $\endgroup$
    – uzizi_1
    May 12 at 12:15

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