In a few quantum physics papers I saw an operator proportional to this one: $$\hat{N}=\frac{(\hat{n}-1)!!}{\hat{n}!!},$$ where $\hat{n}=\hat{a}^{\dagger}\hat{a}$ and $!!$ is the double factorial. Any idea on how to apply such an operator on e.g. a Fock state $|n\rangle$ or a coherent state?


1 Answer 1


$$\frac{(\hat{n}-1)!!}{\hat{n}!!}|n\rangle =\frac{(n-1)!!}{n!!}|n\rangle$$ Regarding coherent states it is enough expanding them in terms of states with defined $n$ and using linearity.

All that immediately arises from the general spectral theory: If $\psi_a$ is an eigenvector of a self-adjoint operator $A$ with eigenvalue $a\in \mathbb R$ and $f$ is any (measurable, real or complex valued) function over $\mathbb R$, by definition $$f(A) \psi_a := f(a) \psi_a\:.$$

  • $\begingroup$ Could you expand on how that works? For the operator in the denominator do you expand that in a Taylor series around $n$? $\endgroup$
    – garyp
    Nov 18, 2016 at 19:51
  • $\begingroup$ Yes, for the numerator I agree, but I was concerned about the denominator. How do you exactly compute $\frac{1}{\hat{n}!!}|n\rangle$? $\endgroup$
    – Andreas K.
    Nov 18, 2016 at 19:55
  • $\begingroup$ Just replacing the operator for the number inside the ket. $\endgroup$ Nov 18, 2016 at 20:02
  • $\begingroup$ @garyp I answered your question completing my answer. Taylor series has nothing to do with this issue. $\endgroup$ Nov 18, 2016 at 20:10
  • $\begingroup$ Indeed it was :) $\endgroup$ Nov 18, 2016 at 20:11

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