Finding the quadrature variance of a superposition of squeezed coherent states How do you find the quadrature variance of a state $$\lvert x\rangle =\lvert a,b\rangle +\lvert a,-b\rangle$$ where $\lvert a,b\rangle = D(a) S(b) \lvert 0\rangle$? 
$\lvert x\rangle$ is a superposition of squeezed coherent states.
 A: Starting with the expression for $S(z)$ we have:
$\hat{S(b)} = e^{b\hat{a}^\dagger \hat{a}^\dagger - b^* \hat{a}\hat{a}}$
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:$
$\hat{S(b)} \hat{a} \hat{S(b)}^\dagger = \hat{a}cosh(r) + \hat{a}^\dagger e^{i\theta} sinh(r)$
And a simmilar expression for $\hat{a}^\dagger$.
We can also do the same thing for $\hat{D(a)} = e^{a\hat{a}^\dagger - a^* \hat{a}}$ . 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:
$\hat{a} \hat{D(a)} \hat{S(b)} \lvert 0 \rangle = \hat{D(a)} (\hat{a} + a) \hat{S(b)} \lvert 0 \rangle$
$= a\hat{D(a)} \hat{S(b)}  \lvert 0 \rangle + e^{i\theta} 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,b\rangle$:
$$ \hat{a} \lvert a,b\rangle = a\lvert a,b\rangle + e^{i\theta} \sinh(r)D(a) S(b) \hat{a}^\dagger \lvert 0\rangle $$
$$ \hat{a}^\dagger \lvert a,b\rangle = a^* \lvert a,b\rangle + \cosh(r)D(a) S(b) \hat{a}^\dagger \lvert 0\rangle $$
...
Using the definitions of quadrature, and their respective variances:
$$\sigma^2_q = \langle q^2\rangle_x - \langle q\rangle^2_x = \frac{1}{2} \left[\langle x \rvert (\hat{a}^2 + \hat{a}\hat{a}^{\dagger} +\hat{a}^{\dagger} \hat{a}  + \hat{a}^{\dagger2})  \lvert x \rangle - \langle x \rvert (\hat{a} + \hat{a}^{\dagger}) \lvert x \rangle ^2\right] $$
