0
$\begingroup$

A coherent state $|\alpha\rangle$ is defined as $D(\alpha)|\alpha\rangle = \exp(\alpha a^{\dagger}-\alpha^{*}a)|\alpha\rangle$ so that

$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)= \sum\limits_{n}n\ |\langle n|\alpha\rangle|^{2}}$$


In a coherent state $|\alpha\rangle$, letting $P(n)$ denote the probability of finding $n^{\text{th}}$ harmonic oscillator state. How can you prove the following relation?

$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)=|\alpha|^{2}}$$


Let me start off anyway:

$$\displaystyle{|\alpha\rangle = \exp\ (\alpha a^{\dagger}-\alpha^{*}a)|0\rangle}$$

$$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \exp\ (\alpha a^{\dagger})|0\rangle}$$

$$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)$$

$$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)$$

$$\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}\right) |n'\rangle}.$$

How do you proceed next?

$\endgroup$

1 Answer 1

2
$\begingroup$

$$ \langle {\hat n} \rangle = \langle \alpha | a^\dagger a | \alpha \rangle = \langle 0| D^\dagger(\alpha) \;a^\dagger a\; D(\alpha) |0 \rangle = \\ = \langle 0| \left[ D^\dagger(\alpha) \;a^\dagger\; D(\alpha)\right] \; \left[ D^\dagger(\alpha)\; a \; D(\alpha)\right] |0 \rangle = \\ = \langle 0| \left[ \;a^\dagger + \alpha^* \right] \; \left[ a \; + \alpha\;\right] |0 \rangle = |\alpha|^2 $$ where use was made of the unitarity of $D(\alpha)$, $D(\alpha) D^\dagger(\alpha) = I$, and $$ D^\dagger(\alpha)\; a \; D(\alpha) = a + \alpha\;\; \;\;\;\text{(and h.c.)} $$ follows straightforwardly from the Baker–Campbell–Hausdorff formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.