I want to find the effect of squeezing operator $S(r) = \exp \big[r(\hat{a}^2 - \hat{a}^{{\dagger}^2})\big]$ on $|q\rangle$ i.e. $S(r)|q\rangle$.

I proceed as follows:

$$S(r)\hat{q}|q\rangle = S(r)q |q\rangle $$ Also $$S(r)\hat{q}|q\rangle = S(r)\hat{q}S(r)^{\dagger}S(r)|q\rangle = e^r \hat{q}(S(r)|q\rangle) $$ using $S(r)\hat{q}S(r)^{\dagger} = e^r \hat{q}$. Combining above two equations yield $$\hat{q}(S(r)|q\rangle) = e^{-r} q (S(r)|q\rangle) $$ Hence, $S(r)|q\rangle$ is eigenket of $\hat{q}$ operator with eigenvalue $ e^{-r} q $ i.e. $$S(r)|q\rangle = N |e^{-r} q\rangle $$ where N is the normalization factor.

How to determine normalization factor $N$?

For more details see Chapter 8 of Introduction to Optical Quantum Information Processing By Pieter Kok, Brendon W. Lovett.

Another ref: P.No. 12, arxiv: 1212.5340

enter image description here


If $$ S(z) = \exp\{z (\hat a^\dagger)^2 -z^*\hat a^2\} $$ then $S^\dagger(z) = S(-z)=S^{-1}(z)$, so $S(z)$ is unitary. It does not change the nomalization of any state therefore. You should be careful however: Your
$|q\rangle$ is an eigenstate of the position operator, so it is not itself normalizable to start with.

  • $\begingroup$ S(r)|q> is normalized; however, |e^(-r) q> is not. Here, I want to find N which normalizes the same. $\endgroup$ – Mark Robinson Apr 19 '19 at 15:29
  • $\begingroup$ @Mark Robinson. I do not understand your comment. For any state $|f\rangle$, both $S(r) |f\rangle$ and $|f\rangle$ have the same normalization. If you had a nomalizable eigenstate localized near $q$ now you have one that has been moved to $e^{-r}q$, but it still has the same normalization. $\endgroup$ – mike stone Apr 19 '19 at 15:44
  • $\begingroup$ Yes S(r)|f> and |f> will have the same normalization, but R.H.S. involves |e^(-r) f>, thus requiring a normalization factor. $\endgroup$ – Mark Robinson Apr 19 '19 at 15:47
  • $\begingroup$ I have updated a link in the question. $\endgroup$ – Mark Robinson Apr 19 '19 at 15:52
  • 1
    $\begingroup$ @Mark Robinson. His eq 8.31 is nonsense since we cannot normalize $|e^{-r}q\rangle$. What he is trying to say, however, is some muddled combination of $\langle q|q'\rangle =\delta(q-q')$ and $\delta(e^{-r} x)= e^r\delta(x)$ so he thinks that $\langle e^{-r} q|e^{-r} q'\rangle = \delta(e^{-r}q-e^{-r}q')= e^r\delta(q-q')$. Again let me stress that this is nonsense. $\endgroup$ – mike stone Apr 19 '19 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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