Interpretation of the photoelectric effect

[vanhees71]

Yes, without quantizing the em. field all the energy eigenvectors of the atom are stationary, i.e., they don't change in time. Quantizing the em. field, there's spontaneous emission, i.e., if the atom is not in a ground state, there's a certain probability that a photon is spontaneously emitted and the atom goes into a lower state.

 

[maline]

But "without quantizing the EM field" means "pretending that charges do not affect the field"! You don't "need photons" to be able to see that there is a coupling between the electron and EM fields. You do need some kind of formalism of QED, which will end up involving photons, but I don't see that the photon idea is helping to explain why emission occurs.

 

[vanhees71]

Of course, you don't need quantizing the em. field to have interaction between charges and the em. field, but if you don't quantize the em. field, you don't get spontaneous but only induced emission (and absorption).

 

[maline]

Of course. What I'm trying to point out is that "not quantizing the EM field" means treating the field as externally given and ignoring the fact that the electrons are sources for the field. Even with induced emission we do not actually describe the light emitted at all- we just find the rate for the electron to go down in energy in response to the incoming field, and then say "well, by conservation of energy, that energy must have been emitted into the field".

As soon as we note that the field is affected by the charges, it is clear that we have no reason to assume that electron eigenstates will really remain stable. They are oscillating charges, and if they can emit energy they will, just like in the classical case. I don't see any obvious way to say that spontaneous emission "occurs because the field is composed of photons". The photons are needed by us, to enable us to work out the details using the perturbation expansions.

Suppose, for argument's sake, that we had never noticed the existence of the Fock Space basis, and instead would express QED purely in terms of field strength operators. We could write down the QED Hamiltonian and the Dirac field equations and attempt, rather hopelessly, to solve for the time development of the field in a given state. I think we would be able to predict, at least in a rough, general way, that spontaneous emission should occur. I am asking, what phenomena would we be likely to miss completely? What is it that makes sense to us using the photon construction, that would seem very surprising otherwise?

 

[vanhees71]

Particles in an energy eigenstate are not oscillating but at rest!

Indeed, in QED with the em. field quantized you get automatically spontaneous emission (aka Bose enhancement for photons).

 

[PeterDonis]

Particles in an energy eigenstate are not oscillating but at rest!

This is not correct. A particle in an energy eigenstate will (I think) also be in a momentum eigenstate, but there is no requirement that it be a momentum eigenstate with eigenvalue zero, which is what "at rest" implies.

 

[vanhees71]

In a bound state of an atom, the electron is not in a momentum eigenstate. Perhaps I should have said that the energy eigenstate is, of course, a stationary state, and thus all expecation values of all observables are constant. In this sense, the electrons in an atomic energy eigenstate are "at rest".

 

[PeterDonis]

In a bound state of an atom, the electron is not in a momentum eigenstate.

In which case the term "at rest" is even more inappropriate.

all expecation values of all observables are constant. In this sense, the electrons in an atomic energy eigenstate are "at rest".

I don't think "at rest" is a very good shorthand for "all expectation values of all observables are constant".

 

[ZapperZ]

Maybe "at rest" means "stationary solution", as used in calculus of variation.

Zz.

 

[vanhees71]

You are right: The correct statement is that the energy eigenstates are the stationary (pure) states of the system.

 

[edguy99]

Remember, we were restricting it to within the UV range. In this range of photon energy, there is a significantly higher probability of emission from predominantly the conduction band. You can see this if you look at ARPES data using UV light sources. The photoemission spectrum shows no such "bound" states to the atoms .

In contract, x-ray photoemission spectroscopy (XPS) will cause emission from the "core level", which is not from the conduction band. But this is not the typical photoelectric effect that is in question in this thread.

So what I say is based on not just the theory, but also from experimental observations that I've performed.

Zz.

My google lookup of ARPES pulls up this pdf as the first item.

In an effort to understand this better, I assume the hv jump from the bonded electrons (EB) are distinct and pretty clear. I also assume the jump from EF to a broader area is the conduction band you are talking about.

I am unclear on this:
1/ the electrons in the conduction band when hit with a specific wavelength will have somewhat diffuse energy because of their kinetic motion in the conduction band and that is why we see a broad area?
or 2/ the electron will be ejected with the same energy as the wavelength that hit it subject to the work function? (ie. for each specific wavelength used, you see only a very narrow energy line)

Thanks

 

[Dadface]

In terms of the practical applications of photo cathodes and the like are there any advantages in using the more modern theories? Are the theories good, for example, at showing how the rate of electron emission depends on the intensity of the incident radiation and what factors affect the efficiency of the cathode? Thank you.

 

[ZapperZ]

My google lookup of ARPES pulls up this pdf as the first item.
View attachment 143118
In an effort to understand this better, I assume the hv jump from the bonded electrons (EB) are distinct and pretty clear. I also assume the jump from EF to a broader area is the conduction band you are talking about.

I am unclear on this:
1/ the electrons in the conduction band when hit with a specific wavelength will have somewhat diffuse energy because of their kinetic motion in the conduction band and that is why we see a broad area?
or 2/ the electron will be ejected with the same energy as the wavelength that hit it subject to the work function? (ie. for each specific wavelength used, you see only a very narrow energy line)

Thanks

What you have cited is a basic photoemission spectroscopy that measures the energy spectrum. ARPES technique goes beyond that by also measuring the momentum of the photoelectrons. That picture also does not indicate the energy scale, i.e. what are the photon energies that will cause a noticeable emission from the core level?

Remember what I pointed out to be the issue with your post - that you were not aware that in a standard photoelectric effect experiment, using as high as UV photons, that the overwhelming signal will come from photoelectrons emitted from the conduction band, not from electrons bounded or localized at the various atoms of the metal. That figure already shows that because the energy band that crosses the Fermi energy is just from the conduction band.

To answer your specific question:

1. I don't what you mean by "diffuse energy". One can easily get very sharp "peaks" in the energy spectrum in ARPES experiments. BTW, here's an example of an ARPES spectra in which both energy and momentum are captured (i.e. the band structure).

You'll notice the very bright and well-defined dispersion curves, especially near the Fermi energy, which will be the dominant emission in this case. This is an example of a conduction band of a metal.

2. Er... electrons do not get "... ejected with the same energy as the wavelength that hit it subject to the work function..." Only the ones from the Fermi energy do. Electrons get ejected with a RANGE OF ENERGY, depending from which part of the band it came from. I stated the word "binding energy" already in this thread. This is the energy state of the electrons in the solid below the Fermi energy. In fact, this can be clearly seen in your own diagram. Those from well before the Fermi energy will have to overcome not just the work function, but the binding energy of the solid.

Zz.

 

[edguy99]

... One can easily get very sharp "peaks" in the energy spectrum in ARPES experiments. BTW, here's an example of an ARPES spectra in which both energy and momentum are captured (i.e. the band structure).

View attachment 145109

You'll notice the very bright and well-defined dispersion curves, especially near the Fermi energy, which will be the dominant emission in this case. This is an example of a conduction band of a metal...

Thank you for the post. I have read a lot about ARPES experiments in the last 2 days and hope to learn more. What do the 3 labeled dispersion curves/bandlines say to you?

 

[ZapperZ]

Thank you for the post. I have read a lot about ARPES experiments in the last 2 days and hope to learn more. What do the 3 labeled dispersion curves/bandlines say to you?

They are just different bands.

The band structure of a material isn't just a simple parabolic curve. Many of us in this field often call it the spaghetti lines. For example, look at the calculated band structure of gold:

ARPES experiments can directly measure such bands, especially close to the Fermi energy. It can also (with reconstruction) map out the Fermi surfaces.

But we're now going very far off the topic here.

Zz.