If I have a 2 mode squeezed state of light, emitted from some nonlinear parametric process, then the more it is squeezed, the larger the variance of one quadratures $X$, while the variance of the other quadrature $Y$ decreases. Their combined product still satisfy the uncertainty relation

$$\Delta X^2 \Delta Y^2 \ge 1/16$$

Now let's say we are able to approach infinity with the amount of squeezing of one of the quadratures (let's say $X$). Then $\Delta X^2$ approaches zero, while $\Delta Y^2$ approaches $\infty$. But, in order to have an infinite variance, we would need to make electric field measurements with infinite amplitude. Does this therefore mean that in order to have a large amount of squeezing, we need to have a large photon number?


There are basically two main ways to squeeze light:

  1. AM, when you have an amplitude (you modify the amplitude) squeezed light, you have light with an intensity noise below the shot noise level

  2. FM, you do frequency doubling, that is you have two photons incoming into the crystal and you have one photon with doubled frequency coming out

Squeezed light is used to reduce the photon counting noise (shot noise) in optical high-precision measurements, most notably in laser interferometers.

In quantum physics, light is in a squeezed state,[1] if its electric field strength Ԑ for some phases ϑ {\displaystyle \vartheta } has a quantum uncertainty smaller than that of a coherent state. The term squeezing thus refers to a reduced quantum uncertainty.


  1. AM

Now when you modify the amplidute (to squeeze light), you will get higher amplitude as you say.

For amplitude squeezed light the photon number distribution is usually narrower than the one of a coherent state of the same amplitude resulting in sub-Poissonian light, whereas its phase distribution is wider. The opposite is true for the phase-squeezed light, which displays a large intensity (photon number) noise but a narrow phase distribution. Nevertheless, the statistics of amplitude squeezed light was not observed directly with photon number resolving detector due to experimental difficulty.


  1. FM

Now when you modify the frequency, you use SHG, frequency doubling. You will have smaller intensity. In this case the amplitude is reduced.


Now to answer your question it is very important how you define amplitude for an EM wave.

For electromagnetic radiation, the amplitude of a photon corresponds to the changes in the electric field of the wave. However, radio signals may be carried by electromagnetic radiation; the intensity of the radiation (amplitude modulation) or the frequency of the radiation (frequency modulation) is oscillated and then the individual oscillations are varied (modulated) to produce the signal.

So the answer to your question is that the amplitude is reduced by squeezing the light.

  • $\begingroup$ I have 2 questions. 1. What about an SPDC process of 1 photon coming in and 2 photons coming out with a lower frequency? 2. The quadratures of a photon are both electric field intensity measurements. Meaning, for a squeezed state if one quadrature has a very small quantum uncertainty, the other quadrature (which is an electric field measurement) must have a very large uncertainty. How can the amplitude be reduced if the second quadrature has a very large uncertainty? Especially in the case where the squeeze parameter approaches infinity. $\endgroup$
    – user41178
    Jun 26 '19 at 17:25
  • $\begingroup$ @user41178 what you state, SPDC is the opposite of frequency doubling (in terms of changing intensity). With frequency doubling, the amplitude decreases, but with SPDC, the amplitude increases. As you say, the SPDC increases the intensity on average. $\endgroup$ Jun 26 '19 at 19:42

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