Interpretation of the photoelectric effect

  • #51
DrDu
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Peters is saying that Gibbs' original paradox was based on inconsistent reasoning!
As you wrote, the assumption of "indistinguishability" makes no sense for classical particles. Yet it is thought to be necessary in Boltzmann statistical mechanics, implying that the concept of classical particles is fundamentally flawed. As Peters shows, this is incorrect. The classical assumption that particles are distinguishable leads to the same 1/N! factor as the opposite assumption. Statistical mechanics for classical particles is internally consistent, with no need for any assumption beyond the uniform distribution of probability.
I tend to second this statement. The point is that although classical particles are in principle distinguishable due to their paths, we never know the starting point of these paths whose knowledge would correspond to a tremendous ammount of information entropy.
 
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  • #52
vanhees71
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Ok, I'll try to study the paper soon to understand how this paradoxical sounding statement is to be understood: I'm a bit puzzled about the statement that I'm allowed to introduce the factor ##1/N!## for distinguishable particles, because if the microstate is not changed by interchanging particles (and that's the usual reasoning why one has to put thisfactor), I'd consider them as indistinguishable rather than dinstinguishable by definition.

Also if the particles are distinguishable also in the quantum sense, I should indeed get "mixing entropy", i.e., if I mix two gases of the same kind (i.e., consisting of the same sort of particles), there must be no increase in entropy, while if they are of different kind (i.e., consisting of different sorts of particles; e.g., hydrogen gas in one part and deuteron gas in the other is already enough) there should be an increase in entropy (the socalled "mixing entropy").
 
  • #53
vanhees71
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I tend to second this statement. The point is that although classical particles are in principle distinguishable due to their paths, we never know the starting point of these paths whose knowledge would correspond to a tremendous ammount of information entropy.
Ok, that's a valid statement. So you say that we have to introduce the ##1/N!## factor because we cannot know the positions and momenta of all the particles with arbitrary precision, it's not possible to dinstinguish any two particles as individuals since their trajectories in phase space are known only so unprecisely that even after picking two particles out of the gas, after a short time we cannot track them with sufficient precision to still distinguish them from the other particles.
 
  • #54
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I'm a bit puzzled about the statement that I'm allowed to introduce the factor 1/N!1/N!1/N! for distinguishable particles, because if the microstate is not changed by interchanging particles (and that's the usual reasoning why one has to put thisfactor), I'd consider them as indistinguishable rather than dinstinguishable by definition.
Yes, distinguishable means that the microstate is changed by interchanging particles. The main insight, if I understand correctly, is that this also makes it meaningful to ask, "Which, out of all possible sets of particles, are the ones in this system?"
 
  • #55
edguy99
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.. The photon energy that was used in this type of experiment usually goes up to as high as the UV range, and they are usually performed on metals. What this means is that the photoelectrons are emitted from the conduction band of the metal. ...
Why do you make this conclusion? Wouldn't higher energy photons tend to pop out the electrons that are bound in some manner rather then the "effectively free" conduction band electrons?
 
  • #56
ZapperZ
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Why do you make this conclusion? Wouldn't higher energy photons tend to pop out the electrons that are bound in some manner rather then the "effectively free" conduction band electrons?
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.
 
  • #57
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Several times in this thread, it has been mentioned that explaining spontaneous photoemission requires photons. Now of course, we cannot talk about photoemission without a theory of how the EM field is affected by quantum-particle charges, and that theory is in fact QED, in which calculations are usually carried out using Fock space i.e. photons. Is this all that is intended by the statement that "photons are required", or does the photon concept directly address a point about photoemission that would otherwise be problematic?
As far as I can tell, the answer to "why is light emitted in quanta of ħω", even in the case of spontaneous emission, is "because the light is generated by a passage between electron energy levels, which involves oscillation at frequency ΔE/ħ". The fact that ħω also happens to be the difference between energy eigenvalues for the free quantum EM field seems almost irrelevant! Except, of course, for the fact that it tremendously simplifies -or rather makes feasible- the calculations that we need to carry out.

Hard to even imagine calculating emission rates using only non-perturbative methods... how about a massive lattice simulation of the fields directly from the Hamiltonian... sounds horrendous, but it should be possible in principle, correct?
 
  • #58
ZapperZ
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Several times in this thread, it has been mentioned that explaining spontaneous photoemission requires photons. Now of course, we cannot talk about photoemission without a theory of how the EM field is affected by quantum-particle charges, and that theory is in fact QED, in which calculations are usually carried out using Fock space i.e. photons. Is this all that is intended by the statement that "photons are required", or does the photon concept directly address a point about photoemission that would otherwise be problematic?
As far as I can tell, the answer to "why is light emitted in quanta of ħω", even in the case of spontaneous emission, is "because the light is generated by a passage between electron energy levels, which involves oscillation at frequency ΔE/ħ". The fact that ħω also happens to be the difference between energy eigenvalues for the free quantum EM field seems almost irrelevant! Except, of course, for the fact that it tremendously simplifies -or rather makes feasible- the calculations that we need to carry out.

Hard to even imagine calculating emission rates using only non-perturbative methods... how about a massive lattice simulation of the fields directly from the Hamiltonian... sounds horrendous, but it should be possible in principle, correct?
This thread has gone way past the original topic.

There is a problem with your scenario because you have restricted light creation ONLY as an electronic energy level transition. If this is true, then all the synchrotron light sources in the world, and FELs, will not work.

Zz.
 
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  • #59
vanhees71
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Of course the emission of bremsstrahlung is also included in QED. It's in fact an interesting calculation, because you are forced to deal with the IR problems.
 
  • #60
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I guess my question wasn't clear enough. The OP pointed out that the photon concept is not needed to explain the photoelectric effect. What I want to know is, what phenomena in atom-photon interactions do need photons for the physical explanation of why they occur, rather than to enable calculations? Is this true of spontaneous emission?
 
  • #61
vanhees71
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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.
 
  • #62
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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.
 
  • #63
vanhees71
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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).
 
  • #64
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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?
 
  • #65
vanhees71
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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).
 
  • #66
PeterDonis
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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.
 
  • #67
vanhees71
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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".
 
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  • #68
PeterDonis
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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".
 
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  • #69
ZapperZ
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Maybe "at rest" means "stationary solution", as used in calculus of variation.

Zz.
 
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  • #70
vanhees71
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You are right: The correct statement is that the energy eigenstates are the stationary (pure) states of the system.
 
  • #71
edguy99
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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.
arpes.jpg

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
 
  • #72
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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.
 
  • #73
ZapperZ
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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).

atmospheric01.jpg


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.
 
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  • #74
edguy99
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... 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?
 
  • #75
ZapperZ
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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:

b717686b-f13.gif


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

fermiSurfaces.JPG


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

Zz.
 
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