If a signal level is below thermal noise level ( so the signal is dropped down under the noise-sea) can you recover this signal again ??? I don't think so..
I suggest you look up the lock-in amplifier, as that is one method of signal recovery, and the one I personally know best, having used one extensively during my Ph.D.
https://en.wikipedia.org/wiki/Lock-in_amplifier
The key is that you must have prior knowledge about the signal you want to detect.
There will always be a bandwidth where the thermal noise power is less than the signal. That might mean you have to make the bandwidth of your detector 1 m Hz, but that is possible to do, even if the frequency stability of the signal you want to detect is poor and changes far more than 1 m Hz. The key is to ensure the centre frequency of the detector tracks the signal frequency you want to recover. That is possible to do in a lock-in amplifier, but it does require that you have a clean reference signal, which is at the frequency you want to recover.
Consider what happens if you multiply a large noise voltage by a square wave whose amplitude varies from +1 V to -1 V. Then you low-pass filter the output. The longer the time constant of the filter, the smaller the output will be, since you are just low-pass filtering noise. The fact you have multiplied the noise by a square wave does not change this.
Now consider what happens if you multiply your weak signal (a sine wave) by that same square wave. The frequency of the sine and square wave are the same, but it does not matter what they are. It does not matter if they are constantly changing. As I said before, you must have prior knowledge about the signal you want to detect. In this case, we know the frequency it will be, but we don't know the amplitude. That's what we want to measure.
The easiest to consider is that whenever the sine wave is positive, the square wave is positive, but whenever the sine wave is negative, so is the square wave. Now if you low-pass filter the output, you get a DC voltage which will approach the signal voltage. The longer the time-constant of the filter, the longer it will take the DC voltage to stabilize. (It is normally considered stable after 5 times the time-constant of the filter, though it does depend on the filter characteristics).
It should be clear that if you then put both the signal and the noise into this multiplier, the noise will average to zero, but the signal will average to a steady DC voltage. So you recover the signal. You will need to amplify it to measure it, but that is no big deal. The key is the noise averages to zero, but the signal does not.
The longer the time-constant of your filter, the longer the measurement takes to stabilize, but the better the weak signal recover. I think the Standard lock-in I used had filters with time constants of several thousand seconds, so to measure a really weak signal, you might have to wait hours to measure it. But other filters have time-constants in the µs or ms range.
Of course, if the sine and square wave are 90 degrees out of phase, and you multiply them together, the average of that will also be zero, so you don't get any DC voltage from your signal. However, you get around that problem by doing the multiplication by two square waves, one displaced 90 degrees in phase to the other. So you get two signals, X and Y, then you calculate
sqrt(X^2 + Y^2)
That will always be positive.
Although I've not looked recently, Stanford Research and EG & G were the main manufacturers of lock-in amplifiers. I used a DSP controlled Stanford Research unit extensively, but I've also used analogue instruments from EG & G, and made one myself with a a few op-amps and FET switches. Modern units use DSPs for the filters, but the older analogue units use RC and/or LC filters. The better the unit, the more it is able to recover weak signals.
Lock-in amplifiers are regularly used in research labs. They have many uses.
Dave