What are the units of the squeezing parameter?

[Sciencemaster]

TL;DR Summary

The action of the squeezing operator is a function of sinh(r) and cosh(r) is produced, where r is the squeezing parameter. What are the units of r?

As I understand it, when the squeezing operator acts on an annihilation/creation operator, a function of sinh(r) and cosh(r) is produced, where r is the squeezing parameter. I've been reading some papers that say that up to '15 dB of squeezing' have been produced in a laboratory. Does this mean that dB are the units of the squeezing parameter? If not, what is, and how high can the squeezing parameter feasibly be made with current technology?

 

[vanhees71]

https://en.wikipedia.org/wiki/Squeeze_operator

Of course $r$ must be dimensionless. Otherwise to put it as an argument into the hyperbolic functions wouldn't make any sense!

 

[Sciencemaster]

That makes sense. But in that case, what value for the squeeze parameter would result in one dB of squeezing?

 

[strangerep]

I've been reading some papers that say that up to '15 dB of squeezing' have been produced in a laboratory. [...]

Since we're not mind readers here, it's usually wise to post links to the paper(s) one has been reading.

 

[Sciencemaster]

Since we're not mind readers here, it's usually wise to post links to the paper(s) one has been reading.

My bad, here are links to the papers I was referring to:

https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.117.110801
https://dspace.mit.edu/bitstream/handle/1721.1/79427/850467045-MIT.pdf?sequence=2
https://arxiv.org/pdf/2005.09891.pdf

Sorry about the paywall on the first one....

 

[Cthugha]

For squeezed light, the spread of one of the quadratures Q is reduced, while the spread in the orthogonal quadrature P is enhanced.

If one does the math, one finds that the quadrature variance along Q is $(\Delta Q)^2=\frac{1}{2}e^{-2r}$, while the variance along P is $(\Delta P)^2 =\frac{1}{2} e^{2r}$, when choosing the definition such that the quadrature variances of a vacuum state are $\frac{1}{2}$ along each quadrature. Note that these are the minimum values. The antisqueezing along the P direction is often larger than that.

If I remember correctly, the 15 dB squeezing is in noise power (not in amplitude), so the experimentally observed value for the squeezed quadrature variance should be around $(\Delta Q)^2=\frac{1}{2} 10^{-1.5}$. You should be able to calculate the squeezing parameter from that.
If the squeezing was instead reported for amplitudes, you need to consider the standard deviation instead of the variance.

 

[Sciencemaster]

For squeezed light, the spread of one of the quadratures Q is reduced, while the spread in the orthogonal quadrature P is enhanced.

If one does the math, one finds that the quadrature variance along Q is $(\Delta Q)^2=\frac{1}{2}e^{-2r}$, while the variance along P is $(\Delta P)^2 =\frac{1}{2} e^{2r}$, when choosing the definition such that the quadrature variances of a vacuum state are $\frac{1}{2}$ along each quadrature. Note that these are the minimum values. The antisqueezing along the P direction is often larger than that.

If I remember correctly, the 15 dB squeezing is in noise power (not in amplitude), so the experimentally observed value for the squeezed quadrature variance should be around $(\Delta Q)^2=\frac{1}{2} 10^{-1.5}$. You should be able to calculate the squeezing parameter from that.
If the squeezing was instead reported for amplitudes, you need to consider the standard deviation instead of the variance.

Alright, so in this case, we would find $\frac{1}{2} 10^{-1.5}=e^{-2r}$, so $r=-\frac{1}{2}\ln(\frac{10^{-1.5}}{2})=2.074$. I might be misremembering, but I think there isn't a 1/2 factor in the squeezed quadrature variance, which would mean that $(\Delta Q)^2=10^{-1.5}$. If this is the case, the squeeze parameter would instead be $r=-\frac{1}{2}\ln(10^{-1.5})=1.726$.

Either way, your answer is very helpful and greatly appreciated.

 

[Cthugha]

The factor of 1/2 is the quadrature variance of the vacuum state, so there is no possibility to avoid it. One can redefine the quadratures such that this variance becomes 1 or 0.25, but in any case this redefinition needs to be applied in both equations. Squeezing is always measured relative to vacuum noise. If $(Q_{vac})^2$ denotes the quadrature variance of the vacuum state, you will always get:
$(\Delta Q)^2= (Q_{vac})^2 \times 10^{-1.5}$ and $(\Delta Q)^2= (Q_{vac})^2 \times e^{(-2r)}$ for the two equations.

 

[Sciencemaster]

Alright, that makes sense. Thank you for your help!