I have an operator $\hat{O}=e^{i s (\hat{a}^\dagger+\hat{a})}=e^{i2s\hat{x}}$, where $\hat{a}^\dagger, \hat{a}$ are the creation/annihilation operators (or $\hat{x}$ is the real quadrature operator) for the SHO with real $s$, operating on a coherent state $|\alpha\rangle$. The result should be a coherent state $|\beta\rangle$ possibly with some complex phase, but I am unsure how to complete the calculation.

In other words, I have $\hat{O}|\alpha\rangle=e^{i\phi}|\beta\rangle$, and I am trying to compute $\phi,\beta$ with no luck so far.

  • $\begingroup$ Have you learned about displacement operators? If you recognize that $\hat{O}$ is a displacement operator, then you can use some displacement operator identities to understand how $\hat{O}$ acts on $|\alpha\rangle$. $\endgroup$ Commented Jan 13, 2017 at 6:26
  • $\begingroup$ You could try and calculate $\hat a \left( \hat O |\alpha \rangle \right)$. $\endgroup$
    – Noiralef
    Commented Jan 13, 2017 at 12:21
  • $\begingroup$ I was thinking about trying to frame $\hat{O} $ as a displacement operator, but the fact both operators have the same sign in the exponential would seem to make that difficult $\endgroup$
    – QtizedQ
    Commented Jan 13, 2017 at 14:44
  • 1
    $\begingroup$ Just reexamined the definition of the displacement operator and realized it doesn't include the $i$ in the exponential, which means the displacement operator approach should work. Thank you for bringing that up, it's been a little while since I did quantum optics. $\endgroup$
    – QtizedQ
    Commented Jan 13, 2017 at 14:48
  • $\begingroup$ Had the same exact double-take when comparing $\hat{O}$ to the definition of the displacement operator. Nice one! $\endgroup$ Commented Jan 13, 2017 at 23:58

1 Answer 1


$\hat{O}$ is actually an instance of the displacement operator $D(\beta)=e^{\beta\hat{a}^\dagger-\beta^*\hat{a}}$ with $\beta=is$. From here, $\hat{O}|\alpha\rangle$ is easy to compute by utilizing some common identities associated with the displacement operator.

We will use $|\alpha\rangle=D(\alpha)|0\rangle$, where $|0\rangle$ is the $n=0$ number state. Additionally, we we use $D(\alpha)D(\beta)=D(\alpha+\beta)e^{\frac{\alpha\beta^*-\alpha^*\beta}{2}}$

If we apply these identities, we find (where $\text{Re}(\cdot)$ is the real part of $\cdot$)

$$D(is)|\alpha\rangle = D(is)D(\alpha)|0\rangle = e^{i \text{Re}(s\alpha)}D(is+\alpha)|0\rangle=e^{i \text{Re}(s\alpha)}|is+\alpha\rangle$$


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