Number of photons in an arbitrary EM field

[Dale]

It is an extremely good approximation. (Probably a million or more times better than the approximation a 100W bulb puts out a constant amount of light.)

Thanks, that is really helpful. Do you think the same approximation would be valid for a dipole antenna?

 

[Vanadium 50]

Sure. Probably better since the power is higher (usually) and the energy per photon lower.

 

[vanhees71]

I don’t see him clearly identify the complexified field with the wavefunction. In your opinion, is that implied? I have seen it somewhere, but I don’t remember the source and therefore I could also be misremembering the statement that you could write $\vec \Psi = \vec E + i \vec B$.

No, not in these slides. There's a nice review here:

https://iopscience.iop.org/article/10.1088/1751-8113/46/5/053001

This actually is about counting photons. I would like to show, e.g. as you take a classical dipole wave and reduce the amplitude to the point that you have a small expected number of photons that it still is a dipole field. I should wind up with a wavefunction that matches the classical dipole fields and gives a very small probability of detecting a photon anywhere, but with no gaps.

There is no wave function, but of course a corresponding state. You get it by calculating the dipole solution for the field operators assuming a classical charge-current distribution. You can just copy the calculation from the classical theory because it's solving linear equations and there's not much formal difference between operator and "c-number"-valued fields. The state is then given by applying the positive-frequency part of the field-operators to the vacuum. The result is a coherent state, and the photon-number probabilities are given by a coherent state. The calculation for this "hemiclassical approximation" is discussed in detail in

C. Itzykson and J.-B. Zuber, Quantum Field Theory,
McGraw-Hill Book Company, New York (1980).

Of course a coherent state can have as low a "intensity" as you like. You can even make the average photon number as small as you want. It's just that in such cases the largest contribution in the decomposition in terms of Fock states is the vacuum contribution. Thus you find with the probabilities of a Poisson Distribution,
$$P(N)=\frac{\lambda}{N!} \exp(-\lambda N)$$
$N$ photons.

 

[Dale]

The result is a coherent state, and the photon-number probabilities are given by a coherent state.

Interesting, that explains why @Vanadium 50’s approach above works for many sources.

Many thanks everyone! I found this helpful.

 

[Paul Colby]

I might point out that $E+iB$ is a self dual field combination. Your field combination only describes one of the two possible polarizations. The other equally valid combination is $E-iB$.

Also, it's my understanding that $E$ and $B$ as operators in QM still obey Maxwell's equations. The natural thing to do would be to view them as such. Then the photon number question is as others have said, an operator expectation based on the given field state which you must first supply.

 

[f95toli]

I would suggest looking up some papers about testing of single photon detectors. This is frequently done using heavily attenuated lasers or other coherent sources and in some cases it is done in free space.
Nothing really unusual happens and this is mainly a "classical" calculation, you can estimate the mean number of photons/second by dividing by hf and when that number gets small enough your detector will measure a Poisson distribution; i.e.exactly what you would expect.