Interpretation of the photoelectric effect

[DrDu]

I'm not arguing that the conduction electrons are in a bound state to the bulk metal. I'm arguing that they are not in a discrete energy state, which is the premise in the original post.
Zz.

In this effect the k vector of the EM wave may be set to 0. Hence in the transition the k vector of the electronic Bloch states is conserved. But the Bloch hamiltonian with fixed k has only discrete states.

 

[ZapperZ]

In this effect the k vector of the EM wave may be set to 0. Hence in the transition the k vector of the electronic Bloch states is conserved. But the Bloch hamiltonian with fixed k has only discrete states.

I'm not sure I understand this.

1. setting k=0 means that one is emitting from only electrons at the Γ point. This is only true if you have a material that has the Fermi level right at this point. This is obviously not true for metals which has an occupied band all the way to the Fermi energy. If this is a toy-model, then having the Fermi level right at k=0 and having an emission just from that point means that this state is not only discrete, but it is also singular (as in single, not as in a mathematical pole). This toy model does not represent a typical metal.

2. the total k vector is not conserved in a photoemission process. The in-plane k-vector, i.e. parallel to the surface of emission, is conserved, but the perpendicular, or out-of-plane k-vector is not. We can already see this from reflection photoemission, where perpendicular momentum depends on whatever's left of the photon energy after a photoelectron leaves the bulk material. This applies even for emission from the Γ-point. (the photon momentum doesn't count in photoemission since it is typically minuscule when compared to the electronic momentum).

Zz.

 

[DrDu]

I'm not sure I understand this.

1. setting k=0 means that one is emitting from only electrons at the Γ point. This is only true if you have a material that has the Fermi level right at this point. This is obviously not true for metals which has an occupied band all the way to the Fermi energy. If this is a toy-model, then having the Fermi level right at k=0 and having an emission just from that point means that this state is not only discrete, but it is also singular (as in single, not as in a mathematical pole). This toy model does not represent a typical metal.

2. the total k vector is not conserved in a photoemission process. The in-plane k-vector, i.e. parallel to the surface of emission, is conserved, but the perpendicular, or out-of-plane k-vector is not. We can already see this from reflection photoemission, where perpendicular momentum depends on whatever's left of the photon energy after a photoelectron leaves the bulk material. This applies even for emission from the Γ-point. (the photon momentum doesn't count in photoemission since it is typically minuscule when compared to the electronic momentum).

Zz.

ad 1. I was talking about setting the k vector of the EM field to 0, not that of the electron. This is what you also assume in your last sentence of 2. However, I avoided to speak of the photon momentum, as we want to show that the photoelectric effect occurs also without quantizing the EM field.
ad. 2. The k vector I was talking about is the crystal momentum of the electron, not its total momentum. Obviously crystal momentum is only defined for an electron inside the crystal.

 

[ZapperZ]

ad 1. I was talking about setting the k vector of the EM field to 0, not that of the electron. This is what you also assume in your last sentence of 2. However, I avoided to speak of the photon momentum, as we want to show that the photoelectric effect occurs also without quantizing the EM field.

But is this realistic? Having k=0 for EM wave means that... er... there's no wave?

In the last sentence, I did not assume that it is zero. I emphasized that the photon's momentum doesn't participate in the momentum of the photoelectrons. This has been rigorously worked out, and I can try and dig a reference for you if this is needed.

ad. 2. The k vector I was talking about is the crystal momentum of the electron, not its total momentum. Obviously crystal momentum is only defined for an electron inside the crystal.

So what is this process that you claim that momentum is conserved? I assumed that since we're talking about photoemission, that this is the process. The in-plane momentum of the electrons in the crystal is conserved upon emission. The out-of-plane momentum of the electrons does not. It is why we can map the band structure of the electrons in the crystal using ARPES.

Zz.

 

[DrDu]

Ok, let's try to x this out: A Bloch wave can be written as $\psi_{nk}(r)=u_n(r) \exp(ikr)$ where $u_n(r)$ has the periodicity of the lattice.
The coupling to the field is $W=e/m\;\{A,p\}$, where $A=A_0 \exp(iqr)$ is the vector potential of the EM field, e electric charge of an electron and p the electronic momentum operator $p=-i\hbar \nabla$.
As q is much smaller then the inverse lattice spacing, the transition amplitude is
$\langle n'k'| W |n k\rangle \approx -\frac{eA_0}{m} \left(\int_\mathrm{EC} d^3r (u^*_{n'k'} i\hbar\nabla u_{nk}-u_{nk} i\hbar\nabla u^*_{n'k'}) \right) \sum_\mathrm{R} \exp(i (k+q-k')R)$, where the integration is over the elementary cell and the sum over all lattice vectors. The latter gives a Kronecker delta $\delta_{k', k+q}$. Now q is usually much smaller than k so that we can simply set $k'\approx k$, which is what I meant.
Hence the transition amplitude can be calculated from the periodic functions $u_{nk}(r)$ with fixed k. The u functions are eigenvalues of the hamiltonian
$H_k=-\frac{1}{2m} (p-i\hbar k)^2 +V(r)$ which only has a discrecte spectrum for fixed k whose eigenvalues are labeled by n.

 

[ZapperZ]

Ok, let's try to x this out: A Bloch wave can be written as $\psi_{nk}(r)=u_n(r) \exp(ikr)$ where $u_n(r)$ has the periodicity of the lattice.
The coupling to the field is $W=e/m\;\{A,p\}$, where $A=A_0 \exp(iqr)$ is the vector potential of the EM field, e electric charge of an electron and p the electronic momentum operator $p=-i\hbar \nabla$.
As q is much smaller then the inverse lattice spacing, the transition amplitude is
$\langle n'k'| W |n k\rangle \approx -\frac{eA_0}{m} \left(\int_\mathrm{EC} d^3r (u^*_{n'k'} i\hbar\nabla u_{nk}-u_{nk} i\hbar\nabla u^*_{n'k'}) \right) \sum_\mathrm{R} \exp(i (k+q-k')R)$, where the integration is over the elementary cell and the sum over all lattice vectors. The latter gives a Kronecker delta $\delta_{k', k+q}$. Now q is usually much smaller than k so that we can simply set $k'\approx k$, which is what I meant.
Hence the transition amplitude can be calculated from the periodic functions $u_{nk}(r)$ with fixed k. The u functions are eigenvalues of the hamiltonian
$H_k=-\frac{1}{2m} (p-i\hbar k)^2 +V(r)$ which only has a discrecte spectrum for fixed k whose eigenvalues are labeled by n.

A discrete spectrum at a fixed k, yes, within this formulation. But k isn't fixed at only a single value, and I argued that k has a continuous range of values from k=0 all the way to k=k_F in the conduction band.

This is what I don't understand from the original argument.

Zz.

 

[DrDu]

When you square the

A discrete spectrum at a fixed k, yes, within this formulation. But k isn't fixed at only a single value, and I argued that k has a continuous range of values from k=0 all the way to k=k_F in the conduction band.

This is what I don't understand from the original argument.

Zz.

Yes, any electron with any k value in the filled bands has a chance to be ionized if it is energetically possible.

 

[David Neves]

The way it is normally taught in school, they imply that Planck was motivated by solving the ultraviolet catastrophe that results from applying the equipartition theorem to the electromagnetic theory of light. That is not true. In reality, Planck found the idea of energy discreteness in Boltzmann's 1877 paper on statistical mechanics, where he introduced it as a purely mathematical device to count configurations in a model of a molecular gas.

 

[vanhees71]

The history of Planck's radiation law is very interesting. The key advantage for Planck was that at this time in Berlin at the Physikalisch-technische Reichsanstalt (at this time the equivalent of today's NIST in the US) they wanted to create an objective reproducible standard for the brightness of light sources (perhaps because electric light bulbs started to substitute for the gas lights around this time). It was of course well known that thermal radiation is a natural standard, i.e., it's independent of the material of the emitter. So they started to measure the thermal spectrum of electromagnetic waves using a Platin cavity brought accurately to a stable temperature to have a (nearly) exact black-body emitter at a well-defined temperature and then measured the spectrum over a very large range of frequencies (from the IR to the far UV, if I remember right).

From the very accurate results Planck in 1899 first guessed the right law by interpolating between then known low- and high-frequency limits. Then he thought about, how to derive the law, using Maxwell electromagnetics and (phenomenological!) thermodynamics, which failed. Then, very reluctantly, he started to use Boltzmann's methods of Statistical Mechanics. Planck was not in favor of Statistical Mechanics before, but then he used Boltzmann's methods after his trials before failed. Indeed, he used Boltzmann's method to count the number of microstates of a given macrostate to get the entropy (writhing down $S=-k_{\text{B}} \ln \Omega$ for the first time!) and from that the black-body spectrum. The remarkable thing was, how he counted wave modes with discretized energies, i.e., the idea that the energy exchange between the em. field of frequency $f$ ($\omega$) and the material (theorized as harmonic oscillators, which is the most simple model, but it doesn't matter anyway for the universal black-body radiation how you model the cavity walls) is in discrete quanta of the size $E=h \nu=\hbar \omega$. The first motivation was to make $h \rightarrow 0$ at the end of the calculation, but Planck realized that this is precisely what he shouldn't do, because with the discretized energy quanta he got precisely the spectrum he had guessed and which he knew to describe the measured spectrum very accurately. Then for the rest of his life he tried to get rid of this conclusion again and to find an argument, how to derive the black-body spectrum from classical physics, but of course in vain.

What's very interesting about Planck's derivation of the black-body spectrum is the combinatorics he used. From our modern understanding it's easy to understand, because intuitively he used the correct counting for bosons, but at his time this was not so straight forward. Maybe he was inspired from Boltzmann's solution of the Gibbs paradox, where he introduced ad hoc the famous factor of $1/N!$ for a gas consisting of $N$ "indistinguishable particles", although from the point of view of classical mechanics particles are always distinguishable through their initial state in phase space, which implies the unique state in phase space at all times, i.e., one can follow the trajectory of each particle in phase space (in principle) and thus each particle is identifiable. Anyway, I could not understand his counting method without using the argument from modern quantum theory to count the microstates for bosons. Also Planck indeed had no idea to think of his energy quanta in terms of particles, and he fought always against the idea of "light particles", and as we know now, he was not completely wrong with that since indeed the modern photon, defined through relativistic QFT is far from being particle like in a classical sense of a localizable entity.

 

[kith]

This is getting off-topic but if you are interested in the accurate historical developments of QM, there's a marvelous book called "Quantum Concepts in Physics: An Alternative Approach to the Understanding of Quantum Mechanics" by Malcom Longair.

What @vanhees71 has sketched in his previous post is explained in the book in detail. Certainly, there's also an in depth chapter on how exactly Einstein reasoned about the photoelectric effect but I don't have the book handy.

 

[vanhees71]

Another very detailed and comprehensive source is the multi-volume work by Mehra and Rechenberg.

 

[kith]

Another very detailed and comprehensive source is the multi-volume work by Mehra and Rechenberg.

The main difference between the two books is that Mehra and Rechenberg is huge (just out of curiosity: how much of the thousands of pages of Mehra and Rechnberg have you actually read?) because it is a book about the history of science and wants to include every detail and every exchange of letters between important scientists.

Longair on the other tries to explain how the early quantum pioneers arrived at their ideas to modern phycisists. He uses historical details as a tool and not as an end in itself. You can also see this of you look at the ratio of equations to text. It is much higher in Longair's book.

 

[maline]

Maybe he was inspired from Boltzmann's solution of the Gibbs paradox, where he introduced ad hoc the famous factor of 1/N!1/N!1/N! for a gas consisting of NNN "indistinguishable particles", although from the point of view of classical mechanics particles are always distinguishable through their initial state in phase space, which implies the unique state in phase space at all times, i.e., one can follow the trajectory of each particle in phase space (in principle) and thus each particle is identifiable. Anyway, I could not understand his counting method without using the argument from modern quantum theory to count the microstates for bosons.

Take a look at this paper by Hjalmar Peters, who shared it with me here on PF:

http://www.springerlink.com/openurl.asp?genre=article&id=doi:10.1007/s10955-010-0077-7

It turns out the 1/N! factor can be recovered for the case of distinguishable but identical classical particles, using a straightforward combinatorical argument.
I also came up with a very basic form of this argument, in the OP of that PF thread.

 

[vanhees71]

Well, of course, it's a straight-forward combinatorical argument. However, the physics argument, why one should apply this argument, is imho not within classical mechanics. For me the lesson taught by the discovery of QT is that there's no consistent classical theory of matter within classical physics, but one needs QT. That's also why on a foundational level quantum statistical physics is much easier than classical statistical physics. That's also why usually in the university theory course statistical physics and thermodynamics is placed after the QM1 lecture.

 

[maline]

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.

 

[DrDu]

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 amount of information entropy.

 

[vanhees71]

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").

 

[vanhees71]

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 amount 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.

 

[maline]

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?"

 

[edguy99]

.. 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?

 

[ZapperZ]

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.

 

[maline]

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?

 

[ZapperZ]

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.

 

[vanhees71]

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.

 

[maline]

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?

 

[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.