Energy of an EM wave compared to energy of a photon

Question

Several posts on this forum ask the question about the role of amplitude in calculating the energy of an EM wave. This struck me as odd since I learned that $E=h\nu$. There is no amplitude in the Planck relation. But I see why this might arise as a question since every other description of waves includes the amplitude in energy calculation.

This posts on this forum highlight the questions:

The energy of an electromagnetic wave

Energy in an EM wave should depend on frequency

Where is the amplitude of electromagnetic waves in the equation of energy of e/m waves?

In one post I read that:

The macroscopic electric field of a wave consisting of photons does have an amplitude which is statistically built up from the individual photons.

My interpretation of the answers in these posts is that the photon is a fundamental unit with a fixed amplitude. In the same way that the speed of light is a fixed velocity, it appears that the amplitude of a photon is also fixed. So this leads to the question of what is the amplitude of a photon.

Amplitude of an electromagnetic wave containing a single photon

I cant seem to determine whether the question was actually answered in that post. But why is it not valid to just use the fact that in a wave, energy is proportional to amplitude squared.

Why is energy in a wave proportional to amplitude squared?

$$ E=h\nu $$

$$ E \propto A^2 $$

so the amplitude of a photon should be:

$$ a\propto \sqrt{h\nu} $$

Answer 1

There is a sense wherein your last statement is right: the photon's amplitude (in the sense you talk about above) is proportional to $\sqrt{h\,\nu}$. One models the measurement of physical quantities with "observables", and for a quantised electromagnetic field, the electric field observable is the following operator (see the Wikipedia page for Quantization of the Electromagnetic Field:

$\hat{\mathbf{E}}(\mathbf{r}) = i\sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar\omega}{2 V\epsilon_0}} \left(\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} - \bar{\mathbf{e}}^{(\mu)} {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right)$

For the moment don't worry too much about it's meaning if you don't altogether get it, but note that it is indeed proportional to $\sqrt{\hbar\omega}$.

The photon itself does not have an amplitude in the sense that you're talking about above. The photon, or, more precisely, our mathematical model of it, is simply a state vector in Hilbert space whose components represent "probability amplitudes" for the entity to be in a certain eigenstate (read here: unit basis vector of the Hilbert space); I talk about this further here. The concept of the amplitude, in the sense you understand it above, arises when one makes a measurement. The measurement, in quantum measurement theory, is modelled by an "observable" $\hat{E}$, which is a linear operator (for now think of it as a square matrix) on that Hilbert space together with the following recipe for how to apply the operator and interpret its results:

1. After the measurement, the state vector $\psi$ is in one of the observable's eigenstates and the measurement outcome is the corresponding real eigenvalue two that eigenstate;
2. The $m^{th}$ moment of the probability distribution $p(\lambda)$ for the measurement $\lambda$ is $\psi^\dagger \hat{E}^m \psi$ in matrix notation (or in the physicists bra-ket notation $\left<\psi |\hat{E}^m | \psi\right>$).

One can do any unitary transformation on the Hilbert space one likes and still keep all the information about the problem (the observables undergo corresponding transformations too of course). So it is convenient, when talking about a particular measurement, to transform the Hilbert space so that the measurement's observable becomes a diagonal matrix. In these coordinates, the probability that the state is a particular eigenvector $\psi_0$ and thus the probability to observe a measurement equal to the corresponding eigenvalue $\lambda_0$ is particularly simple, to wit $\left<\psi_0 | \psi_0\right>$ (if the measurement is a discrete variable, then you get a probability distribution; if it's continuous variable you get a probability density function).