# Amplitude of a single mode quantized light field in a cavity

1. When a single mode light field in a cavity is quantized, how is the amplitude of the field obtained? Is the dependence of the amplitude on parameters like cavity volume and frequency of radiation obtained from purely dimensional considerations?

2. On a broader note, where does the physical input in dimensional analysis come from? In other words, how do we know that the dependence of a quantity on various parameters obtained from purely dimensional considerations is physically true?

## 2 Answers

I assume that you refer to a form of the field given by:

$$E(z,t) = \left(\frac{2\omega^2}{V\epsilon_0}\right)^{1/2} q(t)\sin (kz).$$

The solution from Maxwell's eqs. is

$$E(z,t) = A q(t)\sin (kz),$$ where $q$ satisfies the eq. $$\frac{d^2q}{dt^2} + c²k^2q = 0.$$

Now, you must be able to get the magnetic field $B$, and use the fact that the total energy of the field is

$$H = \frac{1}{2}\int dV\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right)$$

If the integration is performed on a box of sides' length $L_x, L_y, L_z$, then, if you choose $A$ as shown in the first equation of this pots, you end up with

$$H = \frac{1}{2}(\dot{q}^2 + \omega^2q^2),$$

which resembles the energy of a harmonic oscillator of unit mass. The amplitude $A$ is picked like that because of this reason I believe.

I think a great answer to your question can be found in this lecture from 16:49 to 19:45. To reproduce Dr. Lukin's derivation, we want the amplitude associated with a single cavity photon to be equal to the energy of that photon, which we know to be $\hbar \omega$. The time-averaged energy $dU$ in a differential volume $dV$ due to an electromagnetic wave with amplitude $E_0$ is given by

$$dU = \epsilon_0 E_0^2 dV$$

Where the usual factor of $1/2$ that comes from taking the time average of the electric field is cancelled by the (identical) energy contribution of the magnetic field. So the energy of a photon with amplitude $E_0$ in a cavity with effective volume $V$ is

$$\epsilon_0 E_0^2 V = \hbar \omega$$ which gives for the amplitude associated with single photon $$E_0 = \sqrt{\frac{\hbar\omega}{\epsilon_0 V}}$$ I suppose as a result you would say that if you have $N$ photons in a cavity mode the field amplitude is $E_0N^{1/2}$ to give a final energy of $N\hbar\omega$.

So although this is not a rigorous quantum field theoretic derivation, I think the thought process is a little more careful than pure dimensional analysis.

Regarding your second question, dimensional analysis provides a sense of the natural orders of magnitude for quantities involved in a problem, I don't think anyone would say it does more than that.