# Electron recoil due to photon emission

1. Mar 25, 2015

### Danyon

Consider a single accelerating electron, this electron emits a single photon wave which radiates out spherically in a superposition, What direction does the electron recoil if there is no defined direction for the photon?

Last edited: Mar 25, 2015
2. Mar 25, 2015

### Staff: Mentor

That situation requires QED which doesn't analyse the situation in such terms - eg a single accelerating electron emitting a photon is pretty much meaningless in that formalism - not entirely so - but its part of something a lot different called Feynman diagrams.

Anything you read about things like that outside a Quantum Field Theory (QFT) textbooks is likely wrong - and unfortunately QFT is rather difficult - but doable if you have done a basic QM course and are willing to put in the effort:

Thanks
Bill

Last edited by a moderator: May 7, 2017
3. Mar 26, 2015

### Staff: Mentor

For a complete picture, you have to include whatever is accelerating the electron. As you may already know a free electron that is not interacting with anything else cannot simply emit a (real) photon, because it's not possible to conserve both energy and momentum in such a process.

After the photon is emitted, the energies and momenta of all three objects (the electron, the photon, and the object that is causing the acceleration) are indefinite and must be described using probability distributions, with the constraint that the total momentum and total energy must be conserved (i.e. be the same as before the emission).

4. Mar 26, 2015

### neilparker62

I'm sorry I can't answer your question but I wondered about this situation in the context of an electron in transit from one orbital to another at lower energy. I considered the electron to be 'transiently free' and thus able to absorb recoil energy independent of the nucleus. And I wrote:

$$\Delta\rho_e=\rho_p=\frac{hf}{c}\Rightarrow\Delta v=\frac{\Delta\rho_e}{m_e}=\frac{hf}{m_e c}\Rightarrow \Delta E_e=\frac{(hf)^2}{2m_e c^2}$$

My idea was that the total transition energy would be the sum of photon energy and electron recoil energy as determined above ie:

$$E_T=hf+\frac{(hf)^2}{2m_e c^2}$$

So there would be a very small difference (Stokes shift) in the frequencies of emission and absorption spectral lines presuming that recoil applies only to emission and not absorption.

Just simple-minded me - I do hear what is written above about needing to know QED for a better appreciation of such situations.

5. Mar 26, 2015

### Staff: Mentor

There's no such thing as "in transit" and hence no such thing as "transiently free". The electron is never between one orbital and another, there are no "in-between" states.

6. Mar 27, 2015

### neilparker62

Thanks for the response - could you perhaps help a QED ignoramus conceptualize (at least to some extent!) what happens when an electron 'de excites' (I hesitate to say moves down!) from higher energy to lower and emits a photon. Is there a place for consideration of electron recoil energy at all ? How is it that in Compton scattering electrons within atoms are considered to absorb recoil energy after collison with photons (which thereby emerge at lower energy/frequency) but this is apparently not possible in a photon emission event?

Apologies to original poster - perhaps the last 3 posts here including this one should be moved to a different thread (?)

7. Mar 27, 2015

### vanhees71

This can be misleading, because the emission of a photon is a continuous process, described by unitary time evolution. However, in relativistic QFT it is not possible to give the transient state a proper meaning in terms of a particle interpretation. The only observable quantities are S-matrix elements, describing the transition probabilities from one asymptically free state to another.

Bremsstrahlung, as discussed here, is however a subtle issue, because there are IR and collinear divergences, which are, however, an artificial problem, originating from the fact that in theories, where massless particles are involved (here the photon), the usually used plane waves are not the true asymptotic free states. These are coherent states, which can be obtained in perturbation theory by appropriate soft-photon resummations.

E.g., in leading order of scattering of an electron on an external (Coulomb) potential, emitting one additional real photon, you need to take into account the next-to-leading order for elastic scattering, because you cannot distinguish physically the elastic scattering of the electron on the potential and the scattering plus the emission of the photon with an energy smaller than the energy resolution of your detectors. Look for the keywords "Bloch and Nordsieck" in any good book on QED. The scattering of an electron on a external Coulomb field is nicely discussed in Itzykson, Zuber, Quantum Field Theory.

8. Apr 7, 2015

### neilparker62

Hmm - can I again pose the question: is electron recoil possible or not possible in the case of photon emission ? If it is possible, then I conjecture it would manifest in there being a very small frequency difference (Stokes shift) between absorption frequency and emission frequency which I estimate at about 12GHz in the case of the H 1s-2s transition. I do not know whether it is possible to separately measure (for eg in the case of H 1s-2s transition) absorption and emission frequencies or is it just taken for granted that both are the same ?

Apologies once again to the original poster for perhaps 'hijacking' this thread and taking it in a direction different to what may have been originally intended.

9. Apr 7, 2015

### Staff: Mentor

In QFT emission is not described by that sort of model. Rather a quantum field is a superposition of 0 particle states, one particle states, two particle states etc in a so called Fock space:
.http://en.wikipedia.org/wiki/Fock_space

Thanks
Bill

10. Apr 7, 2015

### neilparker62

Copenhagen interpretation: "rhubarb, rhubarb, rhubarb ..."
Many Worlds interpretation: "wara, wara, wara ..."
Shut up and calculate: 12 GHz
Now may I suggest shut up and measure (if possible ??)

11. Apr 7, 2015

### Staff: Mentor

Measure what? Something the most accurately verified theory of all time says doesn't exist?

Thanks
Bill

12. Apr 7, 2015

### neilparker62

Appreciate your continuing answers to my probably very ignorant questions - many thanks.

Help me a little here - when the Hydrogen 1s-2s transition frequency measurement is reported down to however many places (+- 10Hz or so I think), is that an emission frequency or an absorption frequency they are measuring? I presume it's an emission frequency and what I'm asking is if the absorption frequency could be separately measured. Just in case there's an unexpected difference.

13. Apr 9, 2015

### neilparker62

14. Apr 9, 2015

### Staff: Mentor

Neils Bohr said all sorts of things, some of them quite dubious - of which the above is an example. That of course in noway changes that he was a great physicist, its just he was a well known mumbler who often spoke in terms that were not necessarily models of clarity.

The reason I didn't answer your previous query is I simply don't know.

Thanks
Bill

15. Apr 10, 2015

### neilparker62

Thanks again - would any other 'resident expert' care to comment on the above query re measurement of the H 1s-2s transition frequency ? Have tried sending an email to Prof. T.W. Hänsch at ULM. Probably a little precocious of me to hope for an answer from a Nobel Prize winner!

16. Apr 22, 2015

### neilparker62

If I may ask a related question, how does one interpret the data one can find at the following solar monitoring station:

http://bass2000.obspm.fr/solar_spect.php

I put in 1215, 1 to view the Lyman H-alpha line and what you find are twin peaks with (I assume) an absorption dip in the middle. On the wavelength scale I measured the bottom of the dip to be at 1215.6304 A. That is somewhat short of the accurately measured emission wavelength at 1215.6845 A. In frequency terms the difference amounts to a few tens of GHz.

I am presuming that theoretically the Lyman H alpha emission/absorption lines are at the same frequency.