Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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
I converted @Matt's answer to a comment. – David Z Nov 20 '11 at 18:16
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 $\nu^2=\mu^2-1$.

share|cite|improve this answer
All that effort put into correctly formatting/typesetting the question and you answer it this briefly? ;-) – Nic Nov 23 '11 at 13:11
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.