# Normalising squeezed position eigenket?

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

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.
• @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. – mike stone Apr 19 '19 at 15:44
• @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. – mike stone Apr 19 '19 at 16:21