Can one electron decay into an electron plus a phonon?

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In a metal, can one electron decay into one lower-energy electron plus one phonon? (i.e., can the attached Feynman diagram occur?)

If we replace phonons by photons and consider the process in a vacuum, I guess this is prohibited because you can always boost to a frame where the incoming and outgoing electron velocities are the same. Thus, the electron has no ability to transfer energy to the photon because there are no photons with finite q but zero energy.

However, in a metal, Galilean and Lorentz invariance are broken by the crystal lattice, so it seems to me that the process ought to be allowed. Is my thinking here correct?
 

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  • #2
Simon Bridge
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That is not, strictly speaking, a decay.
It corresponds to an electron losing energy - presumably it gained energy someplace.
I'd consider the diagram incomplete.
 
  • #3
ZapperZ
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In a metal, can one electron decay into one lower-energy electron plus one phonon? (i.e., can the attached Feynman diagram occur?)

If we replace phonons by photons and consider the process in a vacuum, I guess this is prohibited because you can always boost to a frame where the incoming and outgoing electron velocities are the same. Thus, the electron has no ability to transfer energy to the photon because there are no photons with finite q but zero energy.

However, in a metal, Galilean and Lorentz invariance are broken by the crystal lattice, so it seems to me that the process ought to be allowed. Is my thinking here correct?

How is this different than the typical electron-phonon interaction?

For example, a transition in an indirect band-gap semiconductor would require both energy and momentum change, the latter being an interaction with phonons. The electron-electron bound state in Cooper pairing for conventional superconductors requires electron-phonon coupling (your Feynman diagram is one-half of the Feynman diagram for Cooper pairing).

Now, whether one can call this as a "decay", that's a different matter and, to me, seems rather odd.

Zz.
 
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That is not, strictly speaking, a decay.
It corresponds to an electron losing energy - presumably it gained energy someplace.
I'd consider the diagram incomplete.

Can you elaborate on what you mean when you say that the Feynman diagram is incomplete?

Nomenclature aside, what I am trying to ask is, in a metal or semiconductor, can one electron in a state at energy [itex]\omega[/itex] spontaneously scatter into one electron at a lower energy state [itex]\omega-\Omega[/itex], plus one phonon at energy [itex]\Omega[/itex].

If not, then the deeper question is, what are the lowest order Feynman diagrams that can explain the fact that the electrons in any real metal have a finite lifetime?
 
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How is this different than the typical electron-phonon interaction?

For example, a transition in an indirect band-gap semiconductor would require both energy and momentum change, the latter being an interaction with phonons. The electron-electron bound state in Cooper pairing for conventional superconductors requires electron-phonon coupling (your Feynman diagram is one-half of the Feynman diagram for Cooper pairing).

Now, whether one can call this as a "decay", that's a different matter and, to me, seems rather odd.

Zz.

Thanks, and yes, the typical electron-phonon interaction is exactly what I am trying to describe. I am interested to know if Feynman diagrams would be an appropriate way to describe the electron-phonon interaction in a "cartoon" fashion. If so, what do the lowest-order diagrams look like?
 
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  • #7
DrDu
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Electron phonon scattering is in deed one of the principal mechanisms behind electronic resistivity, however, only at finite temperatures. The resistivity should increase with temperature like ##T^5##. This is called Bloch's ##T^5## law.
At very low temperatures the probability for an electron near the Fermi surface to emit a photon goes to zero due to phase space arguments. This is usually discussed within the framework of Landau's Fermi liquid theory.
 
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So, suppose you have a metal that is not superconducting at any temperature, like copper or silver. It also is a perfect crystal with no impurities or defects of any kind.

What would be the allowed mechanisms for the remaining resistance at low temperature? Does copper at 0,1 K actually and observably have resistance 100 000 times less than the same copper at 1,0 K?
 
  • #9
DrDu
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Finally the low temperature resistance of very pure metal samples is limited only by scattering from crystal surfaces.
 
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As ZapperZ said, this is precisely what the electron-phonon coupling diagrams look like:
https://www.google.com/search?tbm=isch&q=electron+phonon+diagram

Feynman diagrams are indeed an appropriate way to describe this interaction (in a very rigorous way!), see e.g. Mahan's Many Particle Physics book.

Great! I am declaring my original post to be entirely correct. Thanks for everyone’s comments.

I'd consider the diagram incomplete.

This statement that the diagram is incomplete is not right, as far as I can tell. It would only be accurate for the example with electrons and photons in a vacuum, and the reason is that the diagram I drew would violate simultaneous energy and momentum conservation. The situation is different for the electron-phonon interaction in a crystal or semiconductor.

Electron phonon scattering is in deed one of the principal mechanisms behind electronic resistivity, however, only at finite temperatures.

I should note that I am an ARPES guy, so the perspective I bring is slightly different from that of transport. If you imagine a system at zero temperature except for one isolated electron at finite energy [itex]\omega[/itex], then phase space restrictions likely do not prevent the electron from scattering to a lower-energy state. Thus, there is a finite lifetime associated with this quasiparticle, even in the absence of any phonons, in the absence of any other electrons, and in the absence of impurities.

Nevertheless, the comments I see above regarding transport in the limit of zero temperature are very interesting. Snorkack, in practice I think that impurity scattering is often impossible to eliminate, so I would guess that the resistance of copper in the low-temperature limit approaches a finite value.

Actually, however, if you were able to fashion an absolutely perfect crystal and eliminate all sources of scattering (from other electrons, as well as phonons and impurities), then Ohm’s law would break down, and the electrons would oscillate back and forth in time under the influence of an applied electric field. The resistance of the material would increase toward infinity rather than dropping to zero. Such Bloch oscillations have actually been observed in certain very clean systems.
 
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  • #13
ZapperZ
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Most , if not all, resistivity measurement on metals will show a residual resistivity approaching 0K. I remember reading a couple of physical Review papers on this that showed the T^2 dependence at very low temp due to electron-electron scattering as expected from Fermi Liquid Theory. Wish I can find those papers. They all showed a finite, non-zero resistivity when you extrapolate that to T=0K.

Zz.
 

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