I consider a quantum harmonic oscillator and regard $a$ and $a^\dagger$ is ladder operators. Let $|0\rangle$ be a vacuum, and a coherent state $|\alpha\rangle$ is defined as the eigenstate of the annihilation operator; $$a|\alpha\rangle=\alpha|\alpha\rangle,$$ $$|\alpha\rangle=\exp{(\alpha a^†-\alpha^*a)}|0\rangle$$ Then how can I calculate the following expression? $$\langle\alpha|\exp N|\alpha\rangle=?$$ where $N=a^†a$.

  • $\begingroup$ latex tip: use \langle and \rangle instead of < and > $\endgroup$ Mar 18, 2020 at 8:52
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    $\begingroup$ are you familiar with the representation of the coherent state as a sum of eigenstates of $N$? (it can be directly derived from the definitions you gave of $|\alpha\rangle$) $\endgroup$
    – user245141
    Mar 18, 2020 at 8:59
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    $\begingroup$ have you tried expanding $\exp N$ as a series and trying to compute $\langle \alpha|(a^\dagger a)^n|\alpha \rangle$ for $n\in \mathbb N$? $\endgroup$ Mar 18, 2020 at 8:59

1 Answer 1


You can do this in following way: Expansion of $|\alpha \rangle $ in $|n\rangle$ given by $$|\alpha \rangle=\langle0|\alpha\rangle\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n !}}|n\rangle$$ where $$\langle0|\alpha\rangle=exp(-\frac{1}{2}|\alpha|^2)=r(say)$$ Now $$e^N|\alpha \rangle=r\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n !}}e^N|n\rangle=r\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n !}}e^n|n\rangle$$ Thus $$\langle\alpha|e^N|\alpha\rangle=rr^*\sum_{n=0}^{\infty}\frac{\alpha^n\alpha^{*n}}{n!}e^n$$


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