It is said in arXiv:1401.4118 that Squeezed vacuum can be represented in the Fock state basis as:

$$|\mathrm{SMSV}\rangle=\frac{1}{\sqrt{\cosh r}} \sum_{n=0}^{\infty}\left(-e^{i \phi} \tanh r\right)^{n} \frac{\sqrt{(2 n) !}}{2^{n} n !}|2 n\rangle$$

In my attempt to plot some pictures of the wavefunctions of squeezed light, when I compare it to vacuum, it appears to be smaller:

enter image description here

It appears as though the area of the squeezed light state is smaller than vacuum. Checking the integral in Mathematica, it appears as though these squeezed states aren't normalized.

Here is the code to produce such a state:

SetOptions[Plot, Frame -> True, Axes -> True, 
  LabelStyle -> {FontFamily -> "Arial", FontSize -> 30}, 
  ImageSize -> {200, 200}, Frame -> True, 
  FrameTicks -> {{None, None}, {{0}, None}}, 
  FrameLabel -> {{None, None}, {None, None}}, 
  GridLinesStyle -> LightGray, BaseStyle -> 12];

Energy[n_] := (2 n + 1) \[HBar]/2 \[Omega];

\[Psi][z_, n_] := 
  1/Sqrt[2^n n!] ((m \[Omega])/(\[Pi] \[HBar]))^(1/
      4) Exp[-((m \[Omega] z^2)/(2 \[HBar]))] HermiteH[n, 
    Sqrt[(m \[Omega])/\[HBar]] z];

m = 1;
\[Omega] = 1;
\[HBar] = UnitConvert[Quantity[1, "PlanckConstant"], "SIBase"];
\[HBar] = QuantityMagnitude[\[HBar]];
\[HBar] = 1;

squeezedstate[r_, \[Phi]_] := 
  1/Sqrt[Cosh[r]] Sum[
    Sqrt[Factorial[(2 n)]]/(2^n Factorial[n]) (-E^(I \[Phi]) Tanh[r])^
     n \[Psi][z, 2 n], {n, 0, 35}];

squeezed\[Phi]16 = Abs[squeezedstate[.1, 0]]^2 ; fig = 
 Plot[{0, Abs[\[Psi][z, n]]^2 /. n -> 0, 
   Abs[squeezed\[Phi]16]^2}, {z, -5, 5}, 
  PlotStyle -> { Directive[Black, Dashed], Directive[Red, Thick], 
    Filling -> {Axis, Axis}}, PlotRange -> All];

steveFancyGrid = 
 Grid[{{ Labeled[fig,  
     "|\!\(\*SubscriptBox[\(\[Psi]\), \
\(0\)]\)(E)\!\(\*SuperscriptBox[\(|\), \(2\)]\) vs \
|\[Zeta](E)\!\(\*SuperscriptBox[\(|\), \(2\)]\)" , Top, 
     LabelStyle -> Large]}}]

NormalizationCheck = 
 Integrate[Abs[squeezed\[Phi]16]^2, {z, -\[Infinity], \[Infinity]}]

I truncate the Fock-states at 35. I don't believe this is the issue, as it seems as though my plot and integrals converge for much smaller truncated photon numbers. (And for the case where $\zeta$ is smaller than 1, I think 35 is probably already overkill).

Any feedback would be appreciated.


1 Answer 1


I suspect coefficient $ \frac{\sqrt{(2 n) !}}{2^{n} n !}$ is incorrect. It should be $\sqrt{\frac{\Gamma \left(n+\frac{1}{2}\right)}{n!\Gamma\left(\frac{1}{2}\right)}}$.

I get the first few cofficients to be \begin{align} \left\{\frac{1}{\sqrt{\cosh (r)}},\frac{\sinh (r)}{\sqrt{2} \cosh ^{\frac{3}{2}}(r)},\frac{\sqrt{3} \sinh ^2(r)}{2 \cosh ^{\frac{5}{2}}(r)},\frac{\sqrt{\frac{15}{2}} \sinh ^3(r)}{2 \cosh ^{\frac{7}{2}}(r)},\frac{\sqrt{105} \sinh ^4(r)}{4 \cosh ^{\frac{9}{2}}(r)}\right\} \end{align} for $n=0,2,4,6,8$, and with this everything integrates ok.

The simplest way to check is to recall that squeezed states you are looking at are ${SU}(1,1)$ coherent states with $k=1/4$, so I just pulled the expression in Eq.(5.2.11) of

Perelomov, A., 2012. Generalized coherent states and their applications. Springer Science & Business Media

and remembered the correspondence \begin{align} \vert{\textstyle\frac{1}{4},\frac{1}{4}+m}\rangle \leftrightarrow \vert{2m}\rangle \end{align} between Fock states and $\mathfrak{su}(1,1)$ states in the $k=1/4$ representation.

Back to $\Gamma \left(n+\frac{1}{2}\right)$: it's possible to mess with this using the duplication formula: \begin{align} \Gamma(2z)=\frac{2^{2z-1} \Gamma(z)\Gamma(z+1/2)}{\sqrt{\pi}} \end{align} and thus eliminate the $\Gamma(n+1/2)/\Gamma(1/2)$ for a ratio which does not contain any fraction in the argument of the $\Gamma$ function; I presume something went slightly wrong in converting the ratio of $\Gamma$'s to factorial form.

Nota: to get accurate normalization the truncation of the sum will have to be done as a function of the squeezing parameter $r$. I checked for $r=0.1$ and $0.2$ using only 20 terms (i.e. up to $2n=40$) in the sum and I get $1$ in all cases. For any sizable value of $r$ you will need very many terms.

Here's my bit of code:

psi[n_] := Exp[-x^2/2] HermiteH[n, x]/Sqrt[2^n n! Sqrt[Pi]]
lastn = 20;
ssq[r_] := Sum[ 
   Tanh[r/2]^n  Sqrt[Gamma[n + 1/2]/(n! Gamma[1/2])] psi[2 n] , {n, 0,
NIntegrate[ssq[1/10]^2, {x, -Infinity, Infinity}]
NIntegrate[ssq[2/10]^2, {x, -Infinity, Infinity}]
NIntegrate[ssq[0.6]^2, {x, -Infinity, Infinity}]

Just noticed in my code I have the hyperbolic functions with argument $r/2$ but you have $r$ so be mindful of that... sorry.


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