Eigenvector of the lowering operator for complex eigenvalues

[naele]

Homework Statement

Show that, for all complexe numbers $\alpha, a$ has a unique eigenvector $|\alpha\rangle$ that is nothing else but the coherent state
$$\psi(x)=\frac{e^{-\frac{i}{2\hbar}\langle X\rangle\langle P\rangle}}{(\pi\ell^2)^{1/4}}e^{-\frac{(X-\langle X\rangle)^2}{2\ell^2}+\frac{i\langle P\rangle X}{\hbar}}$$
with
$$\alpha=\langle a \rangle=\frac{1}{\sqrt{2}}\left(\frac{\langle X\rangle}{\ell}+\frac{i\ell\langle P\rangle}{\hbar}\right)$$

Homework Equations

$$a=\frac{1}{\sqrt{2}}\left(\frac{X}{\ell}+\frac{i\ell P}{\hbar}\right)$$
$$\ell=\sqrt{2}\Delta X$$

The Attempt at a Solution

Ok, so I think I have a game plan. Since $a$ isn't Hermitian then its eigenvalues can be complex. So I should solve the eigenvalue problem for $a|\alpha\rangle=\alpha|\alpha\rangle$. But since I already showed that when the equality is valid in the Heisenberg inequality we get a gaussian like $\psi(x)$ so if I can show that the eigenvalue problem admits a differential equation similar to what I already showed then that should be sufficient?

 

[fzero]

Ok, so I think I have a game plan. Since $a$ isn't Hermitian then its eigenvalues can be complex. So I should solve the eigenvalue problem for $a|\alpha\rangle=\alpha|\alpha\rangle$. But since I already showed that when the equality is valid in the Heisenberg inequality we get a gaussian like $\psi(x)$ so if I can show that the eigenvalue problem admits a differential equation similar to what I already showed then that should be sufficient?

The equations for the inequality involve expectation values, so you can't really draw too many conclusions about phases from them. The eigenvalue equation can be written as a differential equation though, and solving it should be easy since you're given the form of the solution.