How much can we currently squeeze light?

Question

In quantum optics, the single-mode squeezed vacuum is the state $$\hat S(z)|0\rangle=\exp\left(\frac{1}{2}\left(z^*\hat a_{\mathbf k}^2-z\hat a_{\mathbf k}^{\dagger 2}\right)\right)|0\rangle$$ where $z=re^{i\theta}$ is a complex number, $\hat a_{\mathbf k}$ is the annihilation operator of the mode $\mathbf{k}$, and $|0\rangle$ denotes the vacuum state.

My questions are:

• In current state-of-the-art experiments, what is the largest value that can be achieved for $r$?
• What is the wavevector $|\mathbf k|$ (or equivelantly the wavelength $\lambda=2\pi/|\mathbf k|$) of the light used in these experiments?
• What is the typical linewidth $\Delta \lambda$? (After all, no light source is truly monochromatic.)

Please provide a reference if possible.

Answer 1

Current state-of-the-art experiments show squeezing of more than 15 dB [1], which corresponds to $r=1.73$ (see below for calculation). Be aware that this is the measured squeezing, not the initial squeezing which was around 27 dB ($r=3.10$). The initial squeezing value solely depends on the pump parameter, not on the actual losses in the experiment, and is of no practical relevance, but probably more accurate if you are really interested in the $r$-parameter. In principle, it is easy to achieve arbitrarily high $r$-parameters as one approaches lasing threshold. The challenge lies in mitigating losses (preventing decoherence) such that you can actually measure your reduced variances.

The experiment was done with infrared light (1064 nm), the used laser was a Mephisto by Coherent. Specs report a line-width of about 1 kHz [2] (though I do not know how relevant that is here; also we tend to think more in terms of noise spectra than in terms of line-widths).

Rather high squeezing at other wavelengths has been shown (e.g. 12.3 dB at 1550 nm [3]), the challenge is to find suitable materials, i.e. very low losses and high enough non-linearity at the respective wavelengths.

15 dB squeezing means a variance $V_\text{dB}$ of 15 dB in the detected quadrature below the shot noise variance of that quadrature (which is set to 1). The variance in terms of $r$ is $V=e^{-2r}$. Thus, $V=10^{-V_\text{dB}/10}=e^{-2r}$ and then $r=-\log(10^{-V_\text{dB}/10})/2$.