Confused2 said:
LC oscillator - is it quantised?
If yes and the oscillation is maintained by a transistor then what is the transistor amplifying?
I would say that a pure LC oscillator is quantized, in the sense that it has stationary states and so on (but probably, the best description is using coherent states). However, from the moment you have losses and a transistor amplifier, this becomes an interacting system. You can try to treat it perturbatively if the losses (= interactions) are weak.
Now, as an exercise, you should calculate what are the classical amplitudes corresponding to a particular LC, say, with a frequency of 100 MHz. You'll find them to be VERY VERY small. I would guess that thermal noise is already much much bigger.
Let's do the calculation:
the energy quantum of a 100 MHz oscillator is h x nu ~ 6.6 10^(-26) J
On a 1pF capacitor, for instance, that corresponds to (E = C V^2/2)
0.36 microvolt. So the ground state of this oscillator will be such that the maximum voltage on the capacitor equals 0.36 microvolt.
Now, the thermal energy that should be present in this oscillator, at room temperature, is 1/2 kT = 0.5 1.38 10^(-23) 300 = 2.1 10^(-21) J, which is already much higher than the lowest quantum. Even at liquid helium temperature (4K) we have 2.7 10^(-23) J of thermal noise. It corresponds to 7.5 microvolts of thermal noise on the 1 pF capacitor.
So the quantization of "classical" electronic circuits is usually completely swamped by thermal noise considerations.