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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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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