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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}

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2 Answers 2

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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 $$ S(\gamma)S^\dagger(\gamma)=\hat 1\, . $$

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  • $\begingroup$ I.e., just some judicious multiplications by $1$ here and there ;-). $\endgroup$ Dec 1, 2017 at 20:12
  • $\begingroup$ @EmilioPisanty A large amount of physics reduces to writing $1$ or $\hat 1$ in a judicious way. $\endgroup$ Dec 1, 2017 at 20:13
  • $\begingroup$ @ZeroTheHero Using your advice, I've solved what D†(α)a†aD(α) is, and I have the S† and S operators around it. I tried solving for what that would be but I'm ending up trying to use cosh(γa) and sinh(γa†) on the |0> state and trying to complex conjugate them(which I'm lost at doing). The answer might be looking at me in my face, but a little help would be appreciated. $\endgroup$
    – user350378
    Dec 2, 2017 at 5:29
  • $\begingroup$ @NameNotFoundException I don’t think the operators are inside the arguments of the hyperbolic functions. $\endgroup$ Dec 2, 2017 at 10:44
  • $\begingroup$ @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. $\endgroup$ Dec 2, 2017 at 11:20
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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.

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