I'm working my way through A Squeezed State Primer, filling in details along the way.
Let $a$ and $a^\dagger$ be the usual annihilation and creation operators with $[a,a^\dagger]=1$ and $|n\rangle=\frac{1}{\sqrt{n}}(a^\dagger)^n|0\rangle$.
With $\mu$ and $\nu$ complex numbers, define \begin{eqnarray*} b &=& \mu a+\nu a^\dagger \\ b^\dagger &=& \mu^\ast a^\dagger+\nu^\ast a \end{eqnarray*}
Choose $\mu$ and $\nu$ so that $b$ and $b^\dagger$ satisfy $[b,b^\dagger]=1$, ie. $|\mu|^2-|\nu|^2=1$. So $b$ and $b^\dagger$ give a set of states 'isomorphic' to the usual number eigenstates.
Define generalised number states $|n'\rangle$ by \begin{eqnarray*} b|0'\rangle &=& 0\\ |n'\rangle &=& \frac{1}{\sqrt{n}}({b^\dagger})^n|0'\rangle \end{eqnarray*} (So the prime as attached to the state, not the $n$.)
With $N'={b^\dagger}b$ we have $\langle n'|N'|n'\rangle=n$.
Inverting the relationship between $a$ and $b$: \begin{eqnarray*} \mu^\ast b &=& |\mu|^2a+\mu^\ast\nu{a^\dagger}\\ \nu{b^\dagger} &=& \nu\mu^\ast{a^\dagger}+|\nu|^2a\\ \end{eqnarray*} \begin{eqnarray*} a &=&(|\mu|^2-|\nu|^2)a &=& \mu^\ast b-\nu{b^\dagger} \\ a^\dagger &=& (|\mu|^2-|\nu|^2){a^\dagger} &=& \mu{b^\dagger}-\nu^\ast b \\ \end{eqnarray*}
So the question is, what are the expected number of quanta in the $|n'\rangle$ states? I think I can compute this via:
\begin{eqnarray*} \langle n'|N|n'\rangle &=& \langle n'|{a^\dagger} a|n'\rangle \\ &=& \langle n'|(\mu{b^\dagger}-\nu^\ast b)(\mu^\ast b-\nu{b^\dagger})|n'\rangle \\ &=& \langle n'||\mu|^2{b^\dagger} b+|\nu|^2b{b^\dagger}-\mu\nu({b^\dagger})^2-\mu^\ast\nu^\ast b^2|n'\rangle \\ &=& \langle n'||\mu|^2{b^\dagger} b+|\nu|^2b{b^\dagger}|n'\rangle \\ &=& \langle n'||\mu|^2N'+|\nu|^2(N'+1)|n'\rangle \\ &=& n(|\mu|^2+|\nu|^2)+|\nu|^2 \\ &=& n(2|\mu|^2-1)+|\mu|^2-1 \\ \end{eqnarray*}
Is that right?
For $n=1$ I get $3|\mu|^2-2$.
Page 323 of the paper appears to say it's $2\mu^2-1$ but I may be misunderstanding it. Where is my error?
