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}

  • 2
    $\begingroup$ I think you need to define your $P_\pm$. There are many projection operators out there! $\endgroup$
    – mike stone
    Commented May 11, 2021 at 12:29
  • $\begingroup$ you are absolutely right, i added it $\endgroup$
    – uzizi_1
    Commented May 11, 2021 at 13:15
  • $\begingroup$ You should maybe also add the definition of creation/annihilation that you are using $\endgroup$ Commented May 11, 2021 at 14:04
  • $\begingroup$ i did, thanks!! $\endgroup$
    – uzizi_1
    Commented May 11, 2021 at 16:04

1 Answer 1


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 $a^\dagger(\varphi)$.

  • $\begingroup$ thank you very much! $\endgroup$
    – uzizi_1
    Commented May 12, 2021 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.