The electron is not point-like?

[birdhouse]

TL;DR Summary

My understanding is that we take electrons to be point-like from experimental results where electrons interact with photons. But photons don't exist at only one point. Perhaps it is only the interaction between electrons and other particles which is point-like, and electrons are larger than we have observed. Please tell me why I'm wrong.

Let me start with my understanding of a photon. A source emits a single photon, which can be described as an excitation of the EM field. This excitation radiates outward, producing isochrons which in pure vacuum would be spherical. Then at some point the photon is absorbed by some atom. By principle of least action, the photon must have taken a direct line path from the source to the sink. Until the photon was absorbed, it was "everywhere". My interpretation of this is that a photon is a probability distribution in the shape of a "sphere" of radius c*dt where dt is the elapsed time since emission, with the highest probability being along the path of least action.

Now, when we perform scattering experiments to determine the shape and size of an electron, we observe point-like interactions and thus assume the electron is point-like, whizzing around in its orbital. But in this conceptualization of the spatially-distributed nature of a photon, it is entirely feasible for the photon to interact with the electron at multiple locations, conceivably even at every location the electron could possibly be, so long as those interactions are sufficiently isochronous. Then our observation that the electron was interacted with at a discrete point and thus our assumption that the electron is point-like may be called into question.

Am I fundamentally misunderstanding something? Is anything I said correct? Thanks in advance.

 

[PeroK]

Quantum Theory is not locally realistic. In a way that can be made precise. In particular, the quantized electromagnetic and electron fields cannot be fully described in classical terms.

That's a general issue in terms of explaining quantum theory. Moreover, trying to reconcile quantum theory with your ad hoc assumptions and assertions is not possible.

 

[.Scott]

By principle of least action, the photon must have taken a direct line path from the source to the sink.

In many ways it is as if it the photon traveled in a direct line, but in many ways it is not.
If you want to think of it in terms of "trajectories", then during the photon transmission, it is capable of interfering with counterfactual versions of itself. In other words, the paths that it does not take have a statistical affect on its final trajectory.

As PeroK said, QM, at its essence, is non-local. Even a point-like particle is not really located at a point.

 

[gentzen]

Taking "electrons to be point-like" just means that they have no inner structure. Taking a proton or neutron to be point-like would just mean to ignore their known inner structures. As long as the involved energies are sufficiently small, nothing is wrong with taking protons and neutrons as point-like too. For electrons, we simply have no evidence that taking them to be point-like would no longer be fine for extremely high energies.

 

[anuttarasammyak]

Let me start with my understanding of a photon. A source emits a single photon, which can be described as an excitation of the EM field. This excitation radiates outward, producing isochrons which in pure vacuum would be spherical. Then at some point the photon is absorbed by some atom. By principle of least action, the photon must have taken a direct line path from the source to the sink. Until the photon was absorbed, it was "everywhere". My interpretation of this is that a photon is a probability distribution in the shape of a "sphere" of radius c*dt where dt is the elapsed time since emission, with the highest probability being along the path of least action.

This configuration reminds me of the discussion at a Solvey conference raised by Einstein. He suggested the imcompleteness of QM pointing out the sudden contraction of wavefunction. However, I do not think point-like nature of particles matters in this scenario.

[EDIT] @birdhouse if spreading cloud like substance as is wavefunction is a counterpart of your point-likeness, this configuration shows electron is observed point-like.

 

[Nugatory]

He suggested the imcompleteness of QM pointing out the sudden contraction of wavefunction.

Although it was far from obvious at the time, this was not so much a problem with QM as with the Copenhagen interpretation of QM.

 

[PeterDonis]

My interpretation of this is that a photon is a probability distribution in the shape of a "sphere" of radius c*dt where dt is the elapsed time since emission, with the highest probability being along the path of least action.

This is not correct. If you have an isotropic light source, which is what you are implicitly assuming with your "sphere" description, then the probability of detecting a photon is the same in all directions. There is no "path of least action".

If the photon is absorbed by a particular atom at a particular location, then after the fact it might seem tempting to attribute to the photon a "path of least action" from the source to the atom. But this temptation should be resisted. The photon does not have a well-defined path where it is not measured. The reason it was absorbed at a particular location is that that was the location of the atom that absorbed it. It could just as likely have been absorbed by another atom in a different location, if other atoms were present. In the idealized case where atoms are equally distributed in all directions from the light source, it is equally likely that the photon would be absorbed by any one of them.

It is also worth noting that most light sources do not emit "photons"--that is, the light they emit is not in an eigenstate of photon number. It is actually quite difficult technically to construct a photon source that actually does emit single photons one at a time. With an ordinary light source, or even a sophisticated one like a laser, it is perfectly possible for multiple "photons" to be detected by different atoms even if the source is at very low intensity.

 

[vanhees71]

In many ways it is as if it the photon traveled in a direct line, but in many ways it is not.
If you want to think of it in terms of "trajectories", then during the photon transmission, it is capable of interfering with counterfactual versions of itself. In other words, the paths that it does not take have a statistical affect on its final trajectory.

As PeroK said, QM, at its essence, is non-local. Even a point-like particle is not really located at a point.

Photons are not localized point-particle-like objects. They do not even admit a position observable in the usual sense. One has to forget about false pictures of 1905 (in the best of all world these false pictures wouldn't have been taught to students to begin with).

Also photons cannot be understood with QM. It's a massless quantum (even with spin 1) and thus can only bee adequately described by relativistic quantum field theory. There's no way to describe it with non-relativistic QM!

 

[PeterDonis]

photons cannot be understood with QM. It's a massless quantum (even with spin 1) and thus can only bee adequately described by relativistic quantum field theory. There's no way to describe it with non-relativistic QM!

I think this is too strong. In contexts like, for example, quantum computing, or experiments using photons to test for Bell inequality violations, the relevant photon degrees of freedom being measured are polarizations, not positions. In those contexts, a non-relativistic approximation in which photons are treated as non-relativistic spin-1 particles and only the spin degrees of freedom are involved in measurements works fine. Of course this is only a limited subset of all experiments involving quantum aspects of the electromagnetic field; but these experiments are important and do not require a full QFT analysis to be understood.

 

[.Scott]

Also photons cannot be understood with QM. It's a massless quantum (even with spin 1) and thus can only bee adequately described by relativistic quantum field theory. There's no way to describe it with non-relativistic QM!

I believe you are attempting to presume the most comprehensive context of "understanding". If your task was to learn everything known about the photon and discover even more, that would be an appropriate presumption.

But if you think that only a complete and accurate model of the photon is acceptable, then you you may be out of luck. Mankind may not have completed a full and accurate description of the photon.

@PeterDonis gave examples where simpler models will suffice. If ones interest is only in measuring the speed of light, then they wouldn't need to deal with photons at all.

 

[vanhees71]

One should, however, use the best theory we have available to talk about photons and not perpetuate some predecessor theories, which are only of historical interest, and Einstein's naive "light-quanta picture" of 1905 is definitely of this kind. Einstein himself was most critical about it, by the way. The idea that photons were localized point-particle-like objects is completely wrong and misleading.

Indeed for the photoelectric effect you do not even need the quantized electromagnetic field. It is sufficient to consider quantum-mechanically described electrons bound in the metal interacting with a classical electromagnetic field to understand the photoelectric effect on the level we discuss here. Of course, to understand this you need quantum mechanics, the notion of bound states (energy eigenvalues in the discrete spectrum of the Hamilton operator) and time-dependent perturbation theory.

One can also use a modified semiquantitative argument like Einstein. You just have to take the classical electromagnetic field and assume that the metal can absorb electromagnetic field energy only in portions of size $\hbar \omega$, and that this can lead to the emission of a photon. Since energy is conserved in these "elementary quantum process", the electron can only be kicked out if this "photon's energy" is greater than the binding energy of the electron. This picture is much closer to the full theory than the naive picture of point-particle-like photons.

 

[PeterDonis]

One should, however, use the best theory we have available to talk about photons

If it's necessary, yes. But if it's not, that's just wasted effort. Physicists use approximate models all the time when the extra effort involved in using a better model is not worth it. If we always used the best theory available, we would not use Newtonian physics, or non-relativistic QM, or special relativity. We would always force ourselves to do full GR or QFT computations, every time. But nobody does that.

 

[PeterDonis]

for the photoelectric effect you do not even need the quantized electromagnetic field.

One can also use a modified semiquantitative argument like Einstein.

You're contradicting yourself. First you say we must always use the best theory available; then you say we can use approximate models that aren't the best theory available. If you insist on using the best theory available, where's your full QFT analysis of the photoelectric effect?

 

[birdhouse]

Thanks everyone, I've already learned from this thread. Can anyone recommend a book or other resources that I can learn from in order to have a better grasp of the concepts underlying photons, matter, and quantum fields? For instance, how could it be that a source emits anything but a whole number of photons?

It is also worth noting that most light sources do not emit "photons"--that is, the light they emit is not in an eigenstate of photon number. It is actually quite difficult technically to construct a photon source that actually does emit single photons one at a time. With an ordinary light source, or even a sophisticated one like a laser, it is perfectly possible for multiple "photons" to be detected by different atoms even if the source is at very low intensity.

Thanks again!

 

[PeterDonis]

how could it be that a source emits anything but a whole number of photons?

Look up coherent states. This Wikipedia article is a decent start and has some good references:

https://en.wikipedia.org/wiki/Coherent_state

The short answer is that most states of the quantum electromagnetic field are not eigenstates of photon number (coherent states are just the most commonly encountered such states), so they do not have "a whole number of photons". "Photons" are not what the quantum EM field is "made of"; rather, they are just a name for particular kinds of states of the quantum EM field, that turn out to be very difficult to actually produce, however useful they are theoretically.

 

[PeroK]

Thanks everyone, I've already learned from this thread. Can anyone recommend a book or other resources that I can learn from in order to have a better grasp of the concepts underlying photons, matter, and quantum fields? For instance, how could it be that a source emits anything but a whole number of photons?

Thanks again!

How did you find PF?: Google search

I finished an undergraduate degree in physics (conc: computational) last winter and I'm looking into grad schools and trying to work in field while I take a little time off from school.

If you are talking about physics grad school, then you should be looking the appropriate graduate texts.

 

[A. Neumaier]

There's no way to describe it with non-relativistic QM!

One can describe it perfectly within conventional quantum mechanics by taking the Silberstein vector as the wave function.

 

[vanhees71]

That's still relativistic physics, i.e., using the representation of the Lorentz group as $\text{SO}(3,\mathbb{C})$ for a massless vector field, $\vec{E}+\mathrm{i} \vec{B}$, which is of course equivalent to the transformation of an antisymmetric tensor, $F_{\mu \nu}$.

 

[A. Neumaier]

That's still relativistic physics, i.e., using the representation of the Lorentz group as $\text{SO}(3,\mathbb{C})$ for a massless vector field, $\vec{E}+\mathrm{i} \vec{B}$, which is of course equivalent to the transformation of an antisymmetric tensor, $F_{\mu \nu}$.

Relativistic physics yes, but represented in the formalism of ordinary quantum mechanics (not field theory).

 

[vanhees71]

This is classical field theory. The photon cannot be described with a wave function, because it can be annihilated and created. For massive particles there's a non-relativistic limit and a finite energy gap for creation/annihilation, i.e., in this non-relativistic limit you can neglect such creation/annihilation processes and work with the "1st-quantization formalism" for a fixed number of particles, i.e., with a wave function.

 

[A. Neumaier]

This is classical field theory. The photon cannot be described with a wave function, because it can be annihilated and created. For massive particles there's a non-relativistic limit and a finite energy gap for creation/annihilation, i.e., in this non-relativistic limit you can neglect such creation/annihilation processes and work with the "1st-quantization formalism" for a fixed number of particles, i.e., with a wave function.

The photon can be described as a 1-particle relativistic field theory with the Silberstein vectors as wave functions. It gives precisely the 1-photon sector, which is conserved in QED in the absence of matter.

It works in the same sense as the positive energy solutions of the Dirac equation give 1-particle electrons, although in QED these can generate arbitrary number of photon pairs and electron-proton pairs.

 

[vanhees71]

Maybe that makes sense for free photons. Rather than to describe it with some 1st-quantization formalism you can as well use the 2nd-quantization formalism which is the more adequate description, particularly when it comes to interactions with matter, which you need to describe photon detection etc.

For free particles/photons all you need are indeed the single-particle modes. Then you write annihilation and creation operators instead of the classical amplitudes in the Fourier integrals. A 1st-quantization formalism is never needed. You can formulate QED in this somewhat clumsy formalism. This leads to Dirac's "hole-theoretical formulation" of QED, but I've never seen this approach used for bosons, because there the Dirac sea makes even less sense than for fermions ;-).