Squeeze operator If $\phi(x)$ is an arbitrary normalized function, and $S$ the squeeze operator,
$$
S=e^{\frac{\mu\cdot h}{2\pi}(a^{\dagger2}-a^{2})}
$$
with $\mu \in \mathbb R$.
How can I find the value and the interpretation of
$$
S\phi(x)
$$
?
 A: Let us assume that the operators are normalized so that $a^\dagger a$ has eigenvalues $0,1,2,3,\dots $. The prefactor in the exponent is $\mu h/2\pi=\mu\hbar$ must therefore be dimensionless, too. So $\mu$ has to have the units of the inverse action.
Now, if $a=(x+ip)/ \sqrt 2$ – let's simplify the "masses etc." and set $\hbar=1$, then
$$a^{\dagger 2} - a^2 = \frac{ixp+ipx + ipx + ixp}{ 2} = 2ixp+1 $$
Note that this operator is anti-Hermitian. Using $p=-i\partial / \partial x$, the operator has the form
$$2x \frac{\partial}{\partial x} + 1 $$
So it differentiates the wave function with respect to $x$ and multiplies the result by $x$. Effectively, it's twice a generator of translations in the $x$-direction that is weighted by $x$. Equivalently, it is the generator of translations in the variable $\log x$. The term $1$ is there just to change the overall real normalization of the wave function so that it remains normalized.
Clearly, the argument $x$ gets scaled by the factor $\exp(2\mu \hbar)$ and
$$[S \phi] (x) = exp(\mu\hbar) \phi(\exp(2\mu\hbar)x) $$
The wave function is simply scaled in the horizontal $x$-direction by that factor. Note that in the momentum representation, due to the inverse relationship of $x,p$ (or because we could run the derivation above with the $x,p$ interchanged but the sign in $S$ would then be the opposite one), the scaling goes in the opposite direction:
$$[S \tilde \phi] (p) = exp(-\mu\hbar) \tilde \phi(-\exp(2\mu\hbar)p) $$
The derivation was only valid for a real $\mu$ – when $S is manifestly unitary (the exponential of an anti-Hermitian operator).
