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When electron is diffracted after the slit it might follow different direction, than before the slit. That means, that going through the slit it gains some acceleration. And accelerated charge emits photons.

enter image description here

Thus - does the diffracted electron radiate photons?

Paweł Tokarz
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  • Of course, but for low (few eV) electrons it's just so minute that it doesn't matter for our purposes. You couldn't even measure it in comparison to the momentum exchange with the slit. In an x-ray source, however, the effect is huge. – CuriousOne Dec 30 '14 at 14:54
  • @CuriousOne - I don't care if it matters or not in practice, but whether it is true or not ;) You say that the efect is visible in X-ray source. But in X-ray lamp electrons are not diffracted on slit, but on an electrode, which is not the same - there is electromagnetic interaction between the electron and the electrode - not pure diffraction. – Paweł Tokarz Dec 30 '14 at 15:14
  • The electron doesn't know the difference between a slit and an atomic lattice... they are both made of the same stuff: atoms. There is nothing else we can make them of. – CuriousOne Dec 30 '14 at 15:16
  • Consider the Mössbauer effect, where you have momentum exchange not with a single particle but with an entire lattice; I suspect the answer is "usually, but not necessarily" but I'm not comfortable expanding into an answer. – rob Dec 30 '14 at 15:57
  • @rob: The Moessbauer effect requires absorption on a resonant nuclear state, so that the "interaction time" is of the order of magnitude of the acoustic propagation time of the phonons trough a significant volume of the crystal. A simple elastic scattering event won't even be noticed by the lattice until the electron is long gone. – CuriousOne Dec 30 '14 at 17:13
  • The electron interact with the electrons on the edge of the slit. This takes place with photon exchange and the loss of energy. The system radiates. And more - my permanent insistence - the diffraction pattern behind every edge (not only slits or multiple slits) is the result of the quantized filed between this electrons. This fact is included in @CuriosOne 's dialouge with anna v. "Even if the electron were to form a Rydberg atom with the slit it would still be in an excited state and radiate." – HolgerFiedler Dec 31 '14 at 07:09

3 Answers3

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I believe the answers given so far are off the mark and do not directly answer the OP's question: do diffracted electrons emit bremsstrahlung radiation/soft photons?

So I give the answer here: The answer is a resounding yes!

However since the diffracted electrons are scattered in the far forward direction (tiny scattering angles), the momentum/energy transfer is minute and therefore the emitted photons have extremely long wavelengths. Therefore, in any practical diffractive experiment, the radiation is undetected. That the radiation is not detected is very important in order to maintain coherence of the system, and therefore leave the diffraction pattern unspoiled.

(I think CuriousOne was alluding to this kind of answer in the comments section of others' answers)

An amusing thought just crossed my mind: it may be possible to observe the radiation for components of the electron trajectories deflected at sufficiently large scattering angles, in which case it is possible to deduce, in principle, the direction of the scattered electron, leading to the loss of the interference pattern at larger angles!

QuantumDot
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  • what about the double slit interference pattern of single electrons? It is there http://en.wikipedia.org/wiki/File:Double-slit_experiment_results_Tanamura_2.jpg with large angles. – anna v Dec 31 '14 at 09:31
  • @annav Bremsstrahlung radiation occurs for the double-slit case, too. Also recall that in these experiments that the detection screen is placed far beyond the position of the double slits, so that even the outer-most parts of the observed interference patten result from small-angle deflection. Also remember that the amplitude for truly large-angle deflections (where significant bremsstrahlung occurs) is much too low to readily observe the interference pattern out there. A high-intensity beam would be needed for observation of pattern at such angles. – QuantumDot Dec 31 '14 at 17:44
  • @annav : I agree with you, your answers are always very illuminating. Just on one thing, you say that the effect of diffraction is analogous with the situation in the atom. But in the atom there is a Coulombian potential, and in the slit there isn't. I understand now what you wanted to say (in the beginning I didn't). Anyway, I gave you an upvote to correct a -1, placed by I don't know whom, and I see another -1. Can you introduce a few more words in your answer? That would allow me to add one more upvote. If you do, please be kind and notify me. – Sofia Dec 31 '14 at 17:49
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    @Sofia you can only upvote a post once. (And as I mentioned in another comment elsewhere, if you're going to upvote a post to cancel out a downvote, please at least make sure it's the kind of post that actually deserves an upvote - in other words your upvote should be because the post is good, not really because someone else downvoted.) – David Z Dec 31 '14 at 18:52
  • @DavidZ : don't you think that I am a grown up person? Please, I don't deserve being told such things. Did you read the answer before writing me these words? Because I read it deeply, and the debate around it. It is not that "at least I made sure", it is that it is an illuminated answer. Otherwise, I wouldn't mix in all this issue. Now, that I can vote only once, yes, this I didn't know, and I learn from you now. And thank you for telling me. I respect rules. I only strongly regret the -1 that I find unjustified. – Sofia Dec 31 '14 at 22:58
  • @Sofia I really don't care whether you're a grown up person. I only care about your behavior on the site; and the fact is, there are people who upvote posts that they wouldn't otherwise upvote, simply to cancel out a downvote. If you don't do that, great. But I had no way to know that from your comments, and I wanted to make sure you know that upvoting merely to cancel out other people's downvotes is bad for the health of the site. I do the same every time I see someone talking about canceling out votes. (That being said, if you really want to do it, nobody can or will stop you.) – David Z Jan 01 '15 at 00:38
  • @DavidZ: I apologize, please accept my apology. There is a problem here. It happens many times that really it is impossible to post an answer. It is a deep problem, and I consider to bring it to the attention of the moderators, one of them being you. Maybe we can discuss the issue sometime when you are in the room. I would be very glad. Again, please accept my apology. – Sofia Jan 01 '15 at 00:55
  • @Sofia no problem, and do feel free to bring your concerns up in chat sometime or ask about them on meta. – David Z Jan 01 '15 at 13:03
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Thus - does the diffracted electron radiate photons?

In your question, you use terms acceleration from classical mechanics and photon from quantum theory. Since these theories are not mutually compatible, the question is badly stated. To get a useful and clear answer, you have to state which theory you are asking.

If you ask "does the electron radiate EM waves when it passes through the slit?" it is a question in the realm of classical theories and the answer is yes based on Maxwell's equations, because electron accelerates in the vicinity of the slit and any accelerated charged particle radiates EM waves.

On the other hand, if you ask "does the electron radiate photons when the double slit experiment is done?" the answer is not so clear, because it is a question in the realm of theory with photons, e.g. quantum field theory. The answer depends also on other assumptions such as initial conditions for the electron and EM field. The corresponding description of diffraction in time would be quite complicated, but no doubt the quantity representing EM field will be non-trivially time-dependent, especially near places electron is present during the diffraction. So I would say there is radiation even in quantum field theoretical description.

I would not insist this means photons are produced, because the field may not even be in state that could be described by the concept of photon number. When people talk about photon transition, it is usually an expression of the idea that some measurement suddenly changed the state of the EM field into state with definite photon number. That is not necessary to calculate solutions of the equations, and describe the double-slit experiment.

Ján Lalinský
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As in my reply to the question Diffracted electron - where does it gain additional momentum? it depends on the size of the slit and the energy of the electron. If the energy of the electron is large enough and the slit large enough, the scattering on the field of the atoms at the sides of the slit will radiate a photon and change the direction of the electron.

If the size of the slit and the energy of the electron are of the order of h_bar where the wave nature of the diffracted electron will become evident, the answer to this is the same as the answer of "why the electron does not fall on the nucleus but stays about the atom without radiating photons". It led to the Bohr model of the atom to start with and then to the full Schrodinger equation and solutions of the hydrogen atom. These solutions give probability distributions for locating the electron.

Thus the answer is that the electron is a quantum mechanical entity, and the solution of "electron of vector momentum p + slit" has as a solution a probability distribution that allows the electron various paths, without radiating any energy. This is related to the Heisenberg uncertainty principle.

Here is the double slit experiment with accumulation of single electrons:

double slit

>Electron buildup over time

One observes that even though a single electron can be anyplace on the screen, the probability distribution built up during the experiment displays an interference pattern. It is a quantum mechanical effect dependent on the geometry of the slits but it shows that the electron is not radiating away random photons, as there would be no interference pattern. The question is about a single slit, but the physics is the same, just the boundary conditions are different for the quantum mechanical problem.

Edit after comments, as the answer has gotten a number of negative votes, to clarify radiation or not by the electron

Take the electron at the maximum off center location in the top picture. It is evident that it has gotten a momentum in the x direction. Treated classically, how could it get there with all its energy? Classically a neutral ball can scatter elastically, momentum being conserved by the whole solid where the ball would hit/graze. Classically though a charged particle changing direction drastically should radiate and it would lose energy, more energy the higher the angle. This random loss of energy would destroy an interference pattern, as the questioner guesses. As interference is observed and we know that the framework is quantum mechanical at the level of atoms and electrons, a wavefunction must exist giving the probability distribution of finding an electron given the boundary conditions of two slits with no energy loss , similar to the energy levels in atoms where the electron might be in any position in the orbitals without radiating.

Community
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anna v
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  • Why it's down voted? I find this quite nice answer, but waiting for others. – Paweł Tokarz Dec 30 '14 at 15:10
  • I think my grammar/syntax was wrong. It read "but stays about the atom by radiating photons" . I corrected it – anna v Dec 30 '14 at 16:54
  • As much as I like your usual answers, this one is simply false. The only no-radiation state in an atom is the ground state. All higher states radiate (even though some may be long lived because of selection rules). The uncertainty principle has absolutely nothing to do with the exchange of photons in electromagnetic scattering processes, since it doesn't include the charge. An uncharged particle, after all, does not radiate like a charged electron, even though it is subject to the same quantum mechanical rules. – CuriousOne Dec 30 '14 at 17:17
  • @CuriousOne Atoms are in their ground state unless they have been excited. I know no solutions of "uncharged particles" in a potential well. If there exists a potential there exists some type of "charge", colored or what not. – anna v Dec 30 '14 at 18:12
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    The electron in this example is not in a ground state. It is in an excited state (even though the spectrum is continuous) with non-zero angular momentum in the observer system. As a result it can radiate and, like all accelerated charges that can change their energy and angular momentum it does radiate. Like I said, a 1eV electron just doesn't radiate much. Make it a 15-25keV electron and you got yourself a mammography machine, make it 120keV or so and it's a regular x-ray machine. Make it a GeV and it's a free electron laser... – CuriousOne Dec 30 '14 at 19:03
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    @CuriousOne You are seeing it as a scattering , I am answering as if the slit is of the size to show difraction. In the two slit experiment for example, the electron is not radiating when it has changed direction at the hit on the screen. I will clarify – anna v Dec 30 '14 at 19:21
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    It is scattering. I am curious, though, why you think of a lower energy (1eV) scattering event as fundamentally different from a low energy event (1keV). I know that you know that 1keV electrons will create soft x-rays. So what makes you think that dialing down the energy by three orders of magnitude will fundamentally prevent the emission of low energy photons? Even if the electron were to form a Rydberg atom with the slit it would still be in an excited state and radiate. – CuriousOne Dec 30 '14 at 19:27
  • @CuriousOne I am sure that if there were photons accompanying electrons in the two slit experiment they would have been detected by now. It is not the energy, it is the HUP combination really that allows changes in direction without loss of energy, and the HUP holds for specific measurements and variable couples. – anna v Dec 30 '14 at 19:49
  • As a high energy physicist you should know that the total radiated power for Bremsstrahlung is proportional to the square of the acceleration. The resulting radiated power (and energy loss) for a low energy electron is so small that nobody cares for the purpose of this experiment. – CuriousOne Dec 30 '14 at 20:11
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    @CuriousOne whether you like it or not there exists a quantum mechanical regime. Brehmstrahlung is classical and is what prevents the existence of atoms and their spectra to be modeled classically. – anna v Dec 30 '14 at 20:14
  • The "quantum mechanical regime" doesn't help because it clearly says that the electron is not in a ground state with respect to the crystal lattice, hence it will radiate and generate phonons until it has come to rest. Phonons have extremely low energies and photons don't have a lower energy, at all, the free electron can make lots and lots of them without violating any selection rules. – CuriousOne Dec 30 '14 at 20:24
  • @CuriousOne you are talking of a different boundary condition problem. Note: the answer is about electron+slit not, electron+crystal . – anna v Dec 31 '14 at 04:24
  • Could you please forward me that address of the scientific supply company that can sell a slit to me that is not made out of atoms? Having said that, you still haven't told me what makes you think that real electrons running trough an imaginary perfect slit wouldn't radiate in theory? Alternatively , shall we just suspend all disbelief and run non-radiating imaginary charges trough an imaginary perfect slit? That would solve the problem, too. Like I said, I like your other posts, but in this case you need to go back to square one. – CuriousOne Dec 31 '14 at 04:39
  • @CuriousOne have a look at WetSava... answer here http://physics.stackexchange.com/questions/95143/can-we-observe-diffraction-even-if-the-slit-size-is-tending-to-zero/95177#95177 . The mathematics would be similar for electron probability waves , no energy losses, just diffraction. Please note that radiation means energy losses. – anna v Dec 31 '14 at 04:50
  • What's that got to do with the simple fact that an accelerated charge will experience bremsstrahlung? A real physical effect doesn't go away just because you don't enter its existence into a theoretical calculation. All that happens is that your calculation is not giving you the correct result but something idealized, which in this case is perfectly fine. – CuriousOne Dec 31 '14 at 04:55
  • @PawełTokarz : it is no more downvoted. The answer is good. I have no idea who downvoted, but I repaired that, by upvoting. – Sofia Dec 31 '14 at 16:36
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    @Sofia in general: just make sure that if you do that, it's the sort of answer that you think actually does deserve an upvote (not because someone else downvoted it). – David Z Dec 31 '14 at 18:50