- #1

CharlieCW

- 56

- 5

## Homework Statement

Show that the coherent state ##|c\rangle=exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle## is an eigenstate of the anhiquilation operator ##a_{\vec{p}}##. Express it in terms of the states of type ##|\vec{p}_1...\vec{p}_N\rangle##

## Homework Equations

$$a^{\dagger}_{\vec{p}_i}|\vec{p}_1...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle$$

$$a_{\vec{p}}|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_N\rangle$$

$$e^{A}=1+A+\frac{A^2}{2!}+...$$

## The Attempt at a Solution

I tried applying directly the operator:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle$$

Since the other operator is in an exponential, the only thing I could think about is expanding the exponential:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}(1+\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}}+\frac{1}{2!}(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})^2+...)|0\rangle$$

However, I don't know if I can just introduce the anhiquilation operator randomly into the integral, since it depends on ##\vec{p}## and the integral is an integral on ##d^3p##. Moreover, I have an integral squared which I'm not sure how to reduce. Could you give some advice?