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I am working my through the Strawberry Fields documentation & the section on state teleportation states:

Here, qumodes $q1$ and $q2$ are initially prepared as (the unphysical) infinitely squeezed vacuum states in momentum and position space respectively,

$\begin{split}&{|0\rangle}_x \sim \lim_{z\rightarrow\infty} S(z){|0\rangle}\\ &{|0\rangle}_p \sim \lim_{z\rightarrow-\infty} S(z){|0\rangle}=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty {|x\rangle}~dx\end{split}$

Related: Quantum teleportation over continuous variables?

Additionally, Constructing finite dimensional codes with optical continuous variables mentions "superpositions of an infinite number of infinitely squeezed states" in the introduction.

My primary question is, what is an infinitely squeezed state & how are they used in practice?

Additionally, what is meant by unphysical? Does this mean purely mathematical?

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  • $\begingroup$ what do you mean with "what is" here? Because a possible answer is that it's the state defined as in your formula. Or are you asking what is $S(z)$ here? Did you have a look at the relevant Wikipedia page? $\endgroup$ – glS May 1 at 13:57
  • $\begingroup$ Do you know what a finitely squeezed state is? $\endgroup$ – Norbert Schuch May 1 at 19:27
  • $\begingroup$ @glS I guess I don't understand the math well enough to have an intuition of what is happening in either formula. The Wikipedia page has this image (en.wikipedia.org/wiki/Squeezed_coherent_state#/media/…) which is the model I had in my head for finite cases. I'm not sure about the inifinite cases however. $\endgroup$ – meowzz May 2 at 2:01
  • $\begingroup$ @NorbertSchuch I think I have a rudimentary idea. $\endgroup$ – meowzz May 2 at 2:03
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The idea of squeezing arises when discussing the state of a quantum harmonic oscillator (e.g. a bosonic system). Such systems differ from simpler qudit systems in that, even when only a single mode is being considered, the system is infinitely dimensional.

A common way to describe these systems is via pairs of non-commuting observables, often the "position" and "momentum" operators $\hat x$ and $\hat p$. For an arbitrary pair of observables $\hat A,\hat B$, the corresponding uncertainties are bounded by $\sigma_A^2\sigma_B^2\ge\frac14|\langle[A,B]\rangle|^2$. Whenever a state is such that $\sigma_A<\frac12|\langle[A,B]\rangle|$ (or the same holds for $\sigma_B$) we talk of a squeezed state.

More formally, a squeezed state can be written by having a squeezing operator $$S(\xi)\equiv\exp\left[\frac12(\xi a^{\dagger 2}-\xi^*a^2)\right],\quad \xi\in\mathbb C$$ act on some other state. For example, squeezed vacuum states have the form $S(\xi)|0\rangle$.

The higher the amount of squeezing, the more the uncertainty of one observable is smaller and the other one is larger. This can be pictured in the phase-space representation of the state as a stretching of the function in some direction.

The limit of infinite squeezing corresponds to the uncertainty of one observable being zero and the other one being infinite. Think position eigenstates corresponding to infinite uncertainty over the momentum. Now, are such states physical? Not really: you can never generate a really infinitely-squeezed state. But one can generate enough squeezing that in a given application you can simplify the model by assuming infinite squeezing. It's just an approximation, which can be useful depending on the circumstances. See this review to read more about squeezing.

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  • $\begingroup$ Thanks for the insightful answer. "The limit of infinite squeezing corresponds to the uncertainty of one observable being zero and the other one being infinite" makes sense to me, however I was reading through this paper (arxiv.org/abs/quant-ph/0008040) & it mentions states that are "infinitely squeezed in both position and momentum". Could you possibly comment on that? $\endgroup$ – meowzz 2 days ago
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    $\begingroup$ it says those are "nonnormalizable states". In other words, they don't represent physical states, they are just mathematical tools $\endgroup$ – glS 2 days ago
  • $\begingroup$ Got it. In that state would you not be able to measure position and momentum? $\endgroup$ – meowzz yesterday
  • $\begingroup$ @meowzz you wouldn't be able to "measure" anything about them, as they are not physical states $\endgroup$ – glS yesterday
  • $\begingroup$ Thanks, just wanted to confirm. $\endgroup$ – meowzz yesterday

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