# How to calculate expectation value of exponentiation of number operator for coherent state?

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 $$N=a^†a$$.

• latex tip: use \langle and \rangle instead of < and > – user2723984 Mar 18 '20 at 8:52
• 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 $|\alpha\rangle$) – user245141 Mar 18 '20 at 8:59
• have you tried expanding $\exp N$ as a series and trying to compute $\langle \alpha|(a^\dagger a)^n|\alpha \rangle$ for $n\in \mathbb N$? – user2723984 Mar 18 '20 at 8:59

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 $$\langle\alpha|e^N|\alpha\rangle=rr^*\sum_{n=0}^{\infty}\frac{\alpha^n\alpha^{*n}}{n!}e^n$$