Eigenvectors of "squeezed" amplitude operator

[carllacan]

Homework Statement

Prove that the states $$|z, \alpha \rangle = \hat S(z)\hat D(\alpha) | 0 \rangle $$ $$|\alpha, z \rangle = \hat D(\alpha) \hat S(z)| 0 \rangle $$
are eigenvectors of the squeezed amplitude operator $$ \hat b = \hat S(z) \hat a \hat S ^\dagger (z) = \mu \hat a + \nu \hat a ^\dagger $$, with μ, ν and z being complex numbers and where $$\hat D(\alpha) = e^{\alpha \hat a ^\dagger - \alpha ^* \hat a }$$ is the displacement operator and $$ \hat S(z) = e ^{ \frac{z^*}{2} \hat a ^2 - \frac{z}{2} \hat a ^{\dagger 2}}$$ is the compression operator.

Homework Equations

The Attempt at a Solution

For the first one I've tried $$ \hat b \hat S(z)\hat D(\alpha) | 0 \rangle = \hat S(z) \hat a \hat S ^\dagger (z) \hat S(z)\hat D(\alpha) | 0 \rangle = \hat S(z) \hat a | \alpha \rangle $$, but I can't get farther than that. I've tried and reordering the exponentials, writing them as taylor series, and writing the coherent state α as a sum of number states n. but I haven't got nowhere.

I have the feeling that the solution is much easier than what I am doing. Could anyone point me in the right direction?

Thank you for your time.

 

[blue_leaf77]

$$ \hat b \hat S(z)\hat D(\alpha) | 0 \rangle = \hat S(z) \hat a \hat S ^\dagger (z) \hat S(z)\hat D(\alpha) | 0 \rangle = \hat S(z) \hat a | \alpha \rangle $$

Just a few more (if not one) steps needed. Remember that the coherent state is an eigenstate of $\hat{a}$.

 

[carllacan]

Just a few more (if not one) steps needed. Remember that the coherent state is an eigenstate of $\hat{a}$.

Yes, I realized that I already had the solution just after I went to bed. I'm retarded, haha. And the other state is easy to prove once you have this one. Using D(α)S(z) = S(z)D(μα+να*) it's easy to see that it's an eigenvector with eigenvalue μα+να*.

Thanks for your help.