klein gordon equation - What is the relationship between vibration of the field and quantum fluctuation?

Consider a free field like the KG equation.

I see that why $$\tilde \phi(\mathbf{p},t)$$ a momentum-dependent quantity, is an oscillator, vibrating at a frequency because when we apply the Fourier transform to the KG equation we have:

$$(\frac{\partial^2}{\partial t^2}+ p^2+m^2)\tilde \phi(\mathbf{p},t)=0$$ which is the equation of an oscillator vibrating at frequency $\sqrt{(p^2+m^2)}$.

But is this oscillation or vibration the same notion of quantum fluctuation, or are they related?

Answer

I'll be very bold and try to be straight to the point.

No they are not the same thing in the sense that their nature is very different although related.

Think of a usual harmonic oscillator whether it is quantum mechanical or classical, the system will oscillate with some frequency that is independent of the quantum (or classical) nature of the system.

Now, because a harmonic oscillator is effectively a confined system, there is kind of a typical confining "length scale" that will affect the quantum behaviour. In particular it imposes a rough bound on the uncertainty in position (or diplacement field or whatever) which in turn makes the uncertainty on the momentum of this field non zero as well (in virtue of the commutation relations they satisfy).

The fact that this confinement generates a non zero momentum is called a zero point fluctuation and gives rise to a non zero energy ground state (that is for ever oscillating).

Note that the stronger the "spring constant" the more it can fluctuate via this uncertainty mechanism.