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spaghetti3451
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Homework Statement
In a coherent state ##|\alpha\rangle##, letting ##P(n)## denote the probability of finding ##n^{\text{th}}## harmomic oscillator state. Show that
$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)=|\alpha|^{2}}$$
Homework Equations
The Attempt at a Solution
A coherent state ##\alpha\rangle## is defined as ##D(\alpha)|\alpha\rangle = \exp(\alpha a^{\dagger}-\alpha^{*}a)|\alpha\rangle## so that
$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)= \sum\limits_{n}n\ |\langle n|\alpha\rangle|^{2}}$$
Therefore, it is prudent to evaluate ##|\alpha\rangle## in terms of ##|n\rangle## as follows:
##\displaystyle{|\alpha\rangle = \exp\ (\alpha a^{\dagger}-\alpha^{*}a)|0\rangle}##
##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \exp\ (\alpha a^{\dagger})|0\rangle}##
##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)##
##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)##
##\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}\right) |n'\rangle}.##
Am I correct so far?