How do individual photons make up an EM wave?

Question

I'm trying to understand the connection between the wave model and the particle model for light.

It's understood that the energy of a photon is given by E=hf, but from my understanding of fourier analysis, the only kind of wave that has a precise frequency is a plane wave. The plane wave is an idealization, since no real wave permeates all of space and time.

So imagining a more realistic EM pulse, the frequency spectrum will have some kind of spread depending on the shape of the pulse. Is the pulse a single photon? Or is it a collection of photons, each with different frequencies?

In the photoelectric effect, it's usually described as a single photon with sufficient energy being absorbed, kicking the electron out of its orbit. Let's imagine the pulse is symmetrically centred around the frequency with energy exactly equal to the metal's work function. What exactly happens to such a pulse? Will the whole pulse be absorbed, since its average frequency has energy of the work function? Or will the half of the pulse that has the higher frequency be absorbed, leaving the rest to reflect or what-not?

Answer 1

So imagining a more realistic EM pulse, the frequency spectrum will have some kind of spread depending on the shape of the pulse. Is the pulse a single photon? Or is it a collection of photons, each with different frequencies?

A weak EM pulse could be a single photon (with a fuzzy frequency). A stronger pulse would consist of a larger number of photons. A classical pulse would be a coherent superposition of photons, each with a fuzzy frequency, and with the number of particles not well defined (i.e., the pulse isn't an eigenstate of particle number).

Or will the half of the pulse that has the higher frequency be absorbed, leaving the rest to reflect or what-not?

This is the right idea, but the details won't actually work. A metal can absorb photons that are below the energy corresponding to the work function. The work function is not perfectly well defined, since the surface is never going to be perfectly clean and uniform. An electron that picks up an energy only infinitesimally greater than the work function will not actually escape, because of energy loss on the way out through the metal while traveling toward the surface.

But anyway, the spread in the photon's energy is passed on to the electron. It's possible for the electron to be in a superposition of two states, one in which it is ejected and one in which it isn't.

The basic idea here is that quantum mechanics (1) is linear, and (2) conserves energy exactly (not just on a statistical basis). Therefore an initial state that is a superposition of energies will continue forever to have the same superposition of energies.