I'm having a problem when trying to work out an inner product.

First I have the inner product $\left<0|\psi\right>$, where $0$ is the fundamental state and $\left|\psi\right>=(a^\dagger - \alpha^*)\left|\alpha\right>$, where $\left|\alpha\right>$ is the coherent state, $\alpha$ is a complex number and $a^\dagger$ is the creation operator. When operating this inner product I get $$\left<0|\psi\right> = \left<0\right|(a^\dagger - \alpha^*)\left|\alpha\right> = 0 - \alpha^*\left<0|\alpha\right> = -e^{-\frac{|\alpha|^2}{2}}\alpha^*$$

which I think is correct, and then I have the product $\left<0|\psi_n\right>$, where $\left|\psi_n\right>=\frac{(a^\dagger - \alpha^*)}{\sqrt{n!}}\left|\alpha\right>$, so I can transform the inner product into $\left<0\right|e^{(a^\dagger - \alpha^*)}\left|\alpha\right>$, so this way I'll get all the powers in just one operation. Then using the Baker-Campbell-Hausdorff identity $e^{A + B + \frac{1}{2}[A,B]} = e^{A}e^{B}$ and since any operator commutes with a scalar I get $$\left<0\right|e^{(a^\dagger - \alpha^*)}\left|\alpha\right> = e^{-\alpha^*}\left<0\right|e^{a^\dagger}\left|\alpha\right>$$ which for the second term for the $a^\dagger$ will be 0 (and for any term except the first one), so I don't know if that would be the same as I got in the first inner product, because I think that I should only multiply the second term in the expansion of the $e^{-\alpha^*}$ with the second term of $e^{a^\dagger}$, which as I said is 0. I suppose I'm wrong in this last assumption, but i don't know why.

  • $\begingroup$ $\langle 0 | a^{\dagger} = 0$ $\endgroup$ – secavara Mar 8 '18 at 12:58
  • $\begingroup$ Wow, pretty embarassing mistake, I'm gonna correct it now, but anyway, the difference still exists. $\endgroup$ – Mr. Nobody Mar 8 '18 at 13:28
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    $\begingroup$ Well, at this point it does not make much sense to try to compute this matrix element with the exponential, since it is fairly inconvenient to compute the matrix element of the exponential and then pick the right terms in the expansion. But to see which terms contribute, you can always compute instead $\langle 0 | \mathrm{e}^{\epsilon\left(a^{\dagger}-\alpha^* \right)} | \alpha \rangle$ and pick the terms linear in $\epsilon$. Then take $\epsilon \rightarrow 1$. That way you'll realize there is no contradiction. And, again, I would not recommend this approach in the general case. $\endgroup$ – secavara Mar 8 '18 at 16:17

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