Finding the quadrature variance of a superposition of squeezed coherent states

Starting with the expression for $S(z)$ we have:

$hat{S(b)} = e^{bhat{a}^dagger hat{a}^dagger - b^* hat{a}hat{a}}$

Through the following BCH formula:

$e^BAe^{-B} = A + [B,A] + frac{1}{2!} [B,[B,A]] ...$

We can see that, since $ [hat{a}, hat{a}^dagger]=1:$ $hat{S(b)} hat{a} hat{S(b)}^dagger = hat{a}cosh(r) + hat{a}^dagger e^{itheta} sinh(r)$

And a simmilar expression for $hat{a}^dagger$.

We can also do the same thing for $hat{D(a)} = e^{ahat{a}^dagger - a^* hat{a}}$ . Here the commutator [B,A] will be a constant and the series will have only two non-zero terms.

$hat{D(a)} hat{a} hat{D(a)}^dagger = hat{a} + a$

Then we use the unitarity of these operators to achieve:

$hat{a} hat{D(a)} hat{S(b)} lvert 0 rangle = hat{D(a)} (hat{a} + a) hat{S(b)} lvert 0 rangle$

$= ahat{D(a)} hat{S(b)} lvert 0 rangle + e^{itheta} sinh(r) hat{D(a)} hat{S(b)} hat{a}^dagger lvert 0 rangle $

...

We have expressions for how $hat{a}$ and $hat{a}^dagger$ act on $ lvert a,brangle$:

$$ hat{a} lvert a,brangle = alvert a,brangle + e^{itheta} sinh(r)D(a) S(b) hat{a}^dagger lvert 0rangle $$

$$ hat{a}^dagger lvert a,brangle = a^* lvert a,brangle + cosh(r)D(a) S(b) hat{a}^dagger lvert 0rangle $$

...

Using the definitions of quadrature, and their respective variances:

$$sigma^2_q = langle q^2rangle_x - langle qrangle^2_x = frac{1}{2} left[langle x rvert (hat{a}^2 + hat{a}hat{a}^{dagger} +hat{a}^{dagger} hat{a} + hat{a}^{dagger2}) lvert x rangle - langle x rvert (hat{a} + hat{a}^{dagger}) lvert x rangle ^2right] $$