# Will free electrons experience its own electric/magnetic field?

1. Feb 19, 2013

### iaMikaruK

Hi, everyone :)
Recently I've read a paper and found in that paper that the authors derived the wavefunction of moving free electrons from its own electric and magnetic field. It was quite a shock to me.
So, for free electrons without external electric and magnetic field, why additional "self-interaction" terms were added to the Hamiltonian? I don't remember any textbook has included such terms in Dirac equation for free electrons.
Thank you very much!

2. Feb 20, 2013

### Simon Bridge

Welcome to PF;
One way of looking at it is that there is always a probability that the electron has emitted a virtual photon (or how else does it interact with other electrons) which means there is a probability that it can interact with that photon. This means the electron is interacting with itself.

Have a look at the self-interaction bits concerning "renormalization".
http://en.wikipedia.org/wiki/Renormalization

Some care is needed - in QED you don't get an electron in a universe all by itself - that would mean there is nothing to measure it for eg. It has to come from some interaction and be going to another interaction.

3. Feb 20, 2013

### iaMikaruK

Anyhow, the interaction due to the virtual photon should not enter the Hamiltonian as $e\boldsymbol{\sigma}\cdot\mathbf{A}$,$\boldsymbol{\mu}_e\cdot\mathbf{B}$ and $-eV$? Sorry I know little of QED.
In that paper, the authors simply added $e\boldsymbol{\sigma}\cdot\mathbf{A}$ and $-eV$ to the Hamiltonian, where $\mathbf{A}$ and $V$ are fields created by the moving free electrons themselves in the laboratory frame, as a correction. I don't understand why they include the fields created by the free electrons as interactions.

4. Feb 20, 2013

### Simon Bridge

The interaction does not enter in at the Hamiltonian level as such - but in the perturbation theory. Did you read the link?

QED = Quantum Electrodynamics ... the field theory of electrons and photons. Nobel prize for Feynman and some people less famous.

To be able to address your specific case, though, I need the reference.
If the term is added "as a correction" they should tell you what they are correcting.

If there is a charge density, then the electrons are not "free" electrons - they experience each other's fields.

5. Feb 20, 2013

### iaMikaruK

Sorry, I cannot open the wiki page because of some reasons. But it works for me now.
I've sent the reference link to you by private message. Thank you.

6. Feb 20, 2013

### andrien

can you give a link to this paper.

7. Feb 20, 2013

### iaMikaruK

I've sent you the link in private message.

8. Feb 20, 2013

### Simon Bridge

Please don't do that - if you want public replies, you should give public information.
S. M. Lloyd, M. Babiker, J. Yuan, and C. Kerr-Edwards
Electromagnetic Vortex Fields, Spin, and Spin-Orbit Interactions in Electron Vortices
Electron vortices are shown to possess electric and magnetic fields by virtue of their quantized orbital angular momentum and their charge and current density sources. The spatial distributions of these fields are determined for a Bessel electron vortex. It is shown how these fields lead naturally to interactions involving coupling to the spin magnetic moment and spin-orbit interactions which are absent for ordinary electron beams. The orders of magnitude of the effects are estimated here for ȧngström scale electron vortices generated within a typical electron microscope.
Phys. Rev. Lett. 109, 254801 (2012) [5 pages]

The article does not seem to deal with free electrons at all.

9. Feb 20, 2013

### iaMikaruK

Well, it does deal with free electrons. The electric field [Eq.(9)] and magnetic field [Eq.(10)] of the vortex beam are all evaluated from the solution of Schrodinger equation of free electrons [Eq.(1)]. Then ,the authors claim that "To determine how electric and magnetic fields interact
with the electron vortex, we start from the Dirac equation in the presence of electromagnetic fields, with vector and scalar potentials A and $\Phi$. These potentials can be external, or they could be those corresponding to the vortex fields derived above."
What puzzling me is that why these electric and magnetic fields derived from a vortex beam interact with the vortex beam itself?

Last edited: Feb 20, 2013
10. Feb 20, 2013

### Simon Bridge

The beam is made up of individual electrons which individually interact with the fields of all the other electrons. Ergo - the beam interacts with it's own field.

11. Feb 20, 2013

### iaMikaruK

This argument is reasonable but I don't think it's applicable here.
I just made a simple calculation. For a 200 kV electron, its speed is about 0.7c. Assuming that the cross-section of the beam is 1x1 angstrom^2 and the current density about 1 nA taken from the reference, then we can calculate the density of electrons. The calculation result is about 1 nA*1s / (0.7c*1 angstrom*1 angstrom*1s)=3x10^(-9) electrons/angstrom^3. So I think it is of very low probability for two electrons to interact with each other.

12. Feb 20, 2013

### ZapperZ

Staff Emeritus
You should learn a little bit about beam physics for particle accelerators. Here, the charge per bunch, and the size of each bunch can be of significant importance due to space-charge effects. Such an effect is, by definition, the bunch's self-interaction. In free-electron lasers, this interaction causes an increase in the beam emittance, which is something we don't want.

Zz.

13. Feb 20, 2013

### iaMikaruK

I agree that for a contiunous emittion of electrons, the electrons will interact with each other. I tried to calculate the space-charge effect in electron microscopy but failed to get a reasonable value.
But I still have the question that the Hamiltonian of single electron, if we have taken the space-charge effect into consideration, will still have the form as described in the reference?
Thanks very much.

14. Feb 20, 2013

### iaMikaruK

I reread the paper and found that the authors claimed that "Note that the fields are due to the charge and current arising from the flow of electrons associated with the vortex and we assume that electron-electron interactions are negligible, thus ignoring the Boersch effect" Any idea?

15. Feb 21, 2013

### Simon Bridge

the fields are due to the charge and current arising from the flow of electrons associated with the vortex ... I would read that as neglecting direct, individual, e-e interactions, but the vortex comes from someplace.

It's a bit like pointing out that the e-e B-field interactions for a current is negligible when dealing with the effect of the B-field due to the current.

It's still not individual electrons here ... if you started your model as individual free electrons, you'd need to correct for the fact that there are other things going on.
That's how it works - you start with a simple model that has easy math and include corrections as more different things get taken into account.

But also - bear in mind what ZapperZ wrote.

16. Feb 21, 2013

### iaMikaruK

A single electron can form a vortex itself. So the vortex needs not come from someplace.
So you are suggesting that two vortex beams are interacting with each other via electric field? But as I recalled the Boersch effect, it was treated completely as electric field interaction between electrons, for example: J. Vac. Sci. Technol. 16, 1676 (1979). And this effect was also neglected by the authors. So what I understand is that the authors have neglected both the electric and magnetic field interactions between individual electrons. Is my understanding correct?

Then what I have come to conclude:
(1) I wrote down a Schrodinger/Dirac equation for a free electron and find the solution;
(2) I evaluated the electric and magnetic field from the solution;
(3) I should add the electric and magnetic field self-interaction back to the Hamiltonian although there is no external field or electron-electron interaction?

17. Feb 21, 2013

### Simon Bridge

When modelling a beam, there will be correction terms to account for the real circumstances of the beam. You need to look deeper into the nature of the beam being used in the experiment to understand more what the authors are describing.

If you still don't believe the answers you have been getting - I suggest writing to the authors and asking them what they are talking about.

18. Feb 21, 2013

### iaMikaruK

Here I listed three references. The first two use Dirac equation and the last one mass-corrected Schrodinger equation. But they all do not taken the electric and magnetic field self-interaction into Hamiltonian. The experimental set-up are the same as in PRL 109, 254801 (2012).
[1] PRL 99, 190404 (2007).
[2] PRL 107, 174802 (2011).
[3] Ultramicroscopy 111, 1461-1468 (2011).

I've written a mail to the author but got no response by now.

19. Feb 21, 2013

### andrien

20. Feb 21, 2013

### Jano L.

No, there is no need for that. You have to decide which situation you want to describe. If you have just one electron, free or in potential, there is no need to introduce self-action, because there is no experimental evidence for it, and it is also very difficult to make it exact and consistent with other things.

But if you have many electrons that interact, you can describe them effectively as one object, and then this composite object will always experience "self-interaction", due to mutual interaction of different electrons. For example, the current in the antenna feels radiation resistance, "self-force", and this can be explained as being due to mutual interaction between distinct electrons.