What is the expectation value of the number operator on the generalized squeeze state?

Consider a generalized squeezed state defined by $|\alpha,γ \rangle$ to be $D(\alpha)S(γ)|0\rangle$ . So, the expectation value of $\hat{N}$ (which is $a^\dagger a$) would be $$\langle 0 | S^\dagger(γ)D^\dagger(\alpha)\hat{N}D(\alpha)S(γ)|0\rangle = \langle 0|S^\dagger(γ)D^\dagger(\alpha)a^\dagger a D(\alpha)S(γ)|0\rangle \, .$$

I do not know where to go from here, given that \begin{align} S^\dagger(γ)a S(γ) &= \cosh (γ) a − \sinh (γ) a^\dagger \\ D^\dagger(\alpha) a D(\alpha) &= a + \alpha \, . \end{align}

If $$D^\dagger(\alpha)aD(\alpha)=a+\alpha\, , \tag{1}$$ then take transpose conjugate to find $$D^\dagger(\alpha) a^\dagger D(\alpha)=a^\dagger + \alpha^*\, . \tag{2}$$
Next, note that $$D^\dagger(\alpha) a^\dagger aD(\alpha)= D^\dagger(\alpha) a^\dagger D(\alpha)D^\dagger(\alpha) aD(\alpha) \tag{3}$$ since $D D ^\dagger =\hat 1$. You can then use (1) and (2) in (3), sandwich the result of (3) in $S^\dagger$ and $S$, and then evaluate the result between $\langle 0\vert$ and $\vert 0\rangle$. You might have to use $$S(\gamma)S^\dagger(\gamma)=\hat 1\, .$$
• I.e., just some judicious multiplications by $1$ here and there ;-). Dec 1, 2017 at 20:12
• @EmilioPisanty A large amount of physics reduces to writing $1$ or $\hat 1$ in a judicious way. Dec 1, 2017 at 20:13
• @NameNotFoundException You're misreading the transformation under $S(\gamma)$. It should read $$S^\dagger(γ)a S(γ) = \cosh(γ) a − \sinh (γ) a^\dagger.$$ Cf. e.g. Wikipedia's article on the squeeze operator. Dec 2, 2017 at 11:20
As an alternative to ZeroTheHero's answer, once you've conjugated $$D^\dagger(\alpha)aD(\alpha)=a+\alpha\, , \tag{1}$$ to get $$D^\dagger(\alpha) a^\dagger D(\alpha)=a^\dagger + \alpha^* , \tag{2}$$ if you want to avoid writing a bunch of operator-inverse pairs, you can just multiply on the left by $D(\alpha)$ to turn both of those into $$aD(\alpha)=D(\alpha)(a+\alpha)\, \tag{1'}$$ and $$a^\dagger D(\alpha)=D(\alpha)(a^\dagger + \alpha^* ) \tag{2'}$$ (with similar versions for $S(\gamma)a$ and $S(\gamma)a^\dagger$), and you can work your way left (or right) from there.