# A calculation about coherent state

1. Jul 6, 2014

### Robert_G

Hi, there. The following is copied from a book " atom-photon interaction " by Prof. Claude Cohen-Tannoudji, Page 415.

I do not understand it at all. I do know some thing about the coherent state.
such as

$a |\alpha\rangle = \alpha |\alpha\rangle$

and

$|\alpha\rangle = e^{-|\alpha|^2/2} \sum_n \frac{\alpha^n}{\sqrt{n!}}|n\rangle$

But I don't understand what's going on here.

(1) Why the coherent state is $|\alpha \exp(-i\omega_L t)\rangle$?

(2) $\langle N \rangle = \alpha^2$ means $\langle \alpha \exp(-i\omega_L t)| N |\alpha \exp(-i\omega_L t)\rangle = \alpha^2$?

(3) Where does the $\cos(\omega_L t)$ come from in the first equation?

2. Jul 6, 2014

### Staff: Mentor

The exponential is just the time-evolution with the laser frequency.

Write N in terms of a and $a^\dagger$ and let them operate on the different sides, that should work.

It is composed of the two exponentials on the left side, but I didn't check in detail which part comes from what.

3. Jul 6, 2014

### Robert_G

In a book called quantum optics by M.O.Scully, the eigen-state of the operator $a$ is written as $|\alpha\rangle$, Here, in the book by Claude. the eigen-state is $|\alpha \exp(-i\omega_L t)\rangle$. This is the same thing. The latter is just using the form of $\rho e^{i \theta}$ to represent the complex eigen-value of the $a$.

am i right?

4. Jul 8, 2014

### Staff: Mentor

Probably, yes.