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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?

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Well, for one, you don't need absolute value bars around μ since anything squared is positive. The rest is beyond me... –  Matt Nov 20 '11 at 14:24
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That's a Comment, not an Answer. Only your newness saves you from a Down-vote. Also it's wrong, since $\mu$ may be complex in this example. –  Peter Morgan Nov 20 '11 at 14:55
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I fixed the inversion (did I get it right?) You fix the part after "So the question is, ...". How does that come out? –  Peter Morgan Nov 20 '11 at 15:06
    
Thanks for the fix. Unfortunately it doesn't change anything as you'll see from my upcoming edit. –  Dan Piponi Nov 20 '11 at 15:11
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I converted @Matt's answer to a comment. –  David Z Nov 20 '11 at 18:16
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1 Answer 1

up vote 3 down vote accepted

No, OP's calculation is correct. In more detail, the paper states on page 323 (apparently assuming that $\mu$ and $\nu$ are real numbers), that the result is

$$ \mu^2 + 2 \nu^2 ~=~ 2\mu^2 -1~.$$

The first expression is correct, and corresponds to OP's $3\mu^2-2$. The second expression is wrong. In other words, the paper makes a mistake in the very last step while reducing with $\nu^2=\mu^2-1$.

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All that effort put into correctly formatting/typesetting the question and you answer it this briefly? ;-) –  Nic Nov 23 '11 at 13:11
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Well, I believe OP did not entirely waste his time. I wish that all questions would be formatted as nicely as his. –  Qmechanic Nov 23 '11 at 15:11
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I was taking LaTeX notes already. It was just a copy and paste :-) Plus look at it this way: I tend to assume papers, having been peer-reviewed, are accurate. More so if they are also pedagogical so the work is well established. From an information theory perspective, a response contradicting such a paper, even if short, actually has a high information content. –  Dan Piponi Dec 2 '11 at 1:49
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