Number operator on wavefunction

[doggydan42]

Homework Statement

Consider the state
$$\psi_\alpha = Ne^{\alpha \hat a^\dagger}\phi_0, $$
where $\alpha$ can be complex, and $N = e^{-\frac{1}{2}|\alpha|^2}$ normalizes $\psi_\alpha$.

Find $\hat N \psi_\alpha$.

Homework Equations

$$\hat N = \hat a^\dagger \hat a$$
$$\hat a\phi_n = \sqrt{n}\phi_{n-1}$$
$$\hat a^\dagger = \sqrt{n+1}\phi_{n+1}$$
$$\hat a \psi_\alpha = \alpha \psi_\alpha$$
$$\phi_n = \frac{1}{\sqrt{n!}}(\hat a^\dagger)^n \phi_0$$

The Attempt at a Solution

First applying $\hat a$ gives
$$\hat N \psi_\alpha = \hat a ^\dagger \alpha \psi_\alpha$$
Expanding the exponential gives
$$\alpha N \sum_{n=0}^\infty \frac{\alpha^n}{n!}(\hat a ^\dagger)^{n+1}\phi_0 = N \sum_{n=0}^\infty \frac{\alpha^{n+1}}{n!}(\hat a ^\dagger)^{n+1}\phi_0 = N \sum_{n=1}^\infty \frac{\alpha^{n}}{(n-1)!}(\hat a ^\dagger)^{n}\phi_0 = N \sum_{n=1}^\infty \frac{\alpha^{n}}{n!}n(\hat a ^\dagger)^{n}\phi_0$$

Because at n = 0 the sum becomes 0,
$$\hat N \psi_\alpha = N \sum_{n=0}^\infty \frac{\alpha^{n}}{n!}n(\hat a ^\dagger)^{n}\phi_0$$

From here, I am unsure of how to simplify the summation.

 

[hilbert2]

There's several typos or errors in the Relevant equations part... I also think that the commutator of $\hat{a}$ and $\hat{a}^\dagger$ may be needed in this problem.

Also, note that $\frac{1}{n!} = \frac{1}{\sqrt{n!}\sqrt{n!}}$, which allows you to convert the $\phi_0$ in the last equation to $\phi_n$.