In QFT what is the frequency of oscillation in time of a scalar field mode? Does it depend on the number of particles in the mode?

Question

In classical field theory, for a given real scalar field $\phi$, each mode $\vec{k}$ vibrates in time at a frequency $\omega = \sqrt{\vec{k}^2 + \mu^2}$ with $\mu$ being the mass.

In quantum field theory, for a given real scalar field $\phi$, the energy $E$ of a mode $\vec{k}$ depends on how many quanta/particles of the mode there are, as in for $n$ quanta we get $ E = \omega (n + 1/2)$. But what is the definition of the time frequency $\omega$, are we still able to link it to the mode value $\vec{k}$ as we did classically?

And most importantly: does $\omega$ of a mode change with the number of particles occupying the mode?

Answer 1

The answer is unequivocally no. I will mostly repeat Srednicki chapter 3 here. Before quantizing, the field can be written as $$\phi(\vec{x},t) = \int\tilde{dk}\, a(\vec{k})e^{i(\vec{k}\cdot\vec{x}-\omega t)} +a^*(\vec{k})e^{-i(\vec{k}\cdot\vec{x}-\omega t)} $$ where $k$ and $x$ are 4-vectors of momentum and position and $\omega$ is the so called time frequency of oscillation. By plugging it in to Klein Gordon, $$ (-\partial^2+m^2)\phi = 0$$, we can find OP's dispersion relation $\omega_{\vec{k}} = +\sqrt{\vec{k}^2 + m^2}$. Here, the $a(\vec{k})$ is the Fourier transform of the field and we are taking only positive $\omega$ because we included both positive and negative exponents in the integral (and we have $a^*$ because $\phi$ is real). Upon quantization,

$$\hat \phi(\vec{x},t) = \int\tilde{dk}\, \hat a(\vec{k})e^{i(\vec{k}\cdot\vec{x}-\omega t)} + \hat a^{\dagger}(\vec{k})e^{-i(\vec{k}\cdot\vec{x}-\omega t)} $$ since all we are doing is promoting all the $a$ and $a^*$'s to operators and imposing a commutation relation. This is the terribly named second quantization. Furthermore, Srednicki solves for the energy after quantizing (the derivation is done in the book and it is long so please refer to it). Here is the result:

$$H = \int \tilde{dk} \,\omega_\vec{k} \hat a^\dagger (\vec{k}) \hat a(\vec{k}) + (E_0-\Omega_0)V$$

Answer: And thus, the time frequency of mode oscillation in the energy formula is manifestly independent of mode occupation. Look at the formula for $\omega_\vec{k}$, is there any dependence on the number operator $N = a^\dagger a$'s on it? No. In fact, omega is just some scalar dependent on $\vec{k}$ and particle mass $m$.

Answer 2

In this answer, I'll use units with $c=1$, but to be explicit I'll include factors of $\hbar$.

For uncharged, scalar particles in empty Minkowski space, the modes can be chosen to be eigenstates of the spatial momentum operator in each direction. This leads to the idea that you can label the models by the spatial momenta, $\vec{p}$, or equivalently the wavenumbers, $\vec{k}$.

The energy for one quantum with a given set of momenta $\vec{p}$ is fixed by the usual relativistic formula $(E^{(1)}_\vec{p})^2=p^2+m^2$, or in terms of frequency and wavenumbers the dispersion relation $\omega_\vec{k}^2=k^2+m^2/\hbar^2$. To summarize, $$ (E^{(1)}_\vec{p})^2=\vec{p}^2+m^2 = \hbar^2 \vec{k}^2 + m^2 = \hbar^2 \omega_\vec{k}^2 $$ Within this formula, you might recognize the famous equation $E^{(1)}_{\vec{p}} = \hbar \omega_\vec{k}$, but notice that the interpretation is that the equation holds for the energy of one quanta of momenta $\vec{p}$; it is not a general formula for energy of any number of quanta with different momenta (we'll get to that later).

Note that the energy $E^{(1)}_\vec{p}$ is a function of the momenta (much like the frequency $\omega_\vec{k}$ is a function of the wavenumbers).

The total energy in mode $\vec{p}$ (or mode $\vec{k}$) -- summing over all quanta -- depends on the number of quanta $n_{\vec{p}}$ in that mode, by the equation

$$ E^{\rm (tot)}_{\vec{p}} = \left(n_\vec{p} + \frac{1}{2}\right) E_{\vec{p}}^{(1)} = \left(n_\vec{p} + \frac{1}{2}\right) \hbar \omega_\vec{k} $$ The $\frac{1}{2}$ term is due to vacuum fluctuations, which are independent of the occupation number. Note that $\omega_\vec{k}$ does not depend on the occupation numbers.

The total energy in the field comes from a sum over the energies of each mode

$$ E^{\rm (tot)} = \sum_{\vec{p}} E^{(\rm tot)}_\vec{p} = \sum_{\vec{p}} \left(n_\vec{p} + \frac{1}{2}\right) E^{(1)}_{\vec{p}} = \sum_{\vec{k}} \left(n_\vec{k} + \frac{1}{2}\right) \hbar \omega_\vec{k} $$ The sum here can be interpreted literally if we are working in a box of a finite size (leading to discrete momenta), or else can be replaced with an integration over momenta if we are careful about the measure.