In the context of one exercise, I came across the need to act an operator in front of an exponential of another operator. In terms of equations, what I mean is something like:

$$\hat{a}|c\rangle=\hat{a}\exp\left(-i\int d^3p \hat{a}^{\dagger}\right)|0\rangle$$

Since this is the first time I've seen this, I'm not sure how I could act the operator $\hat{a}$ in front of the exponential, or if there's a way to "switch" the operator so it acts directly on state (I cannot use, for example, BCH relation since the operator is not sandwiched between exponentials). Could you offer me some tips to understand this method?

PS: If it helps, $|c\rangle$ is a coherent state.

  • $\begingroup$ Use \, not /, for LaTeX. $\endgroup$ – knzhou Sep 20 '18 at 16:53
  • $\begingroup$ Sorry I'm working on the edit. There it is. $\endgroup$ – Charlie Sep 20 '18 at 16:54
  • $\begingroup$ see my answer for a more general discussion but the coherent state $\vert c\rangle$ is an eigenstate of $\hat a$ with eigenvalue $c$, so this should make your specific calculation easier. $\endgroup$ – ZeroTheHero Sep 20 '18 at 17:53
  • $\begingroup$ Thank you. Indeed I know coherent states are eigenstates of $\hat{a}) but I had to work on an explicit proof by acting and seeing what I got. I've done that when the coherent state is expressed as a sum but never as an exponential, so that's why I was at a loss. $\endgroup$ – Charlie Sep 20 '18 at 18:52

The key is to write $$ \hat B e^{-\hat A}=e^{-\hat A}e^{\hat A} \hat B e^{-\hat A} $$ and use the equality \begin{align} e^{\hat A} \hat B e^{-\hat A}&= \left(1+\hat A+\frac{1}{2!}\hat A^2+\ldots\right)\hat B\left(1-\hat A+\frac{1}{2!}\hat A^2+\ldots\right)\, , \tag{1}\\ &= \hat B+[\hat A,\hat B]+\frac{1}{2!}[\hat A,[\hat A,\hat B]]+\ldots \tag{2} \end{align} which you can prove by expanding (1) explicitly, combining the resulting terms to obtain the series given in (2). In your specific case the series should terminate.

  • $\begingroup$ Thank you, I don't know why it never ocurred to me to insert the identity there, but in that way I can prove it. Actually I've already proved the BCH relation, so it only rests to find the commutators. $\endgroup$ – Charlie Sep 20 '18 at 18:55

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