Will free electrons experience its own electric/magnetic field?by iaMikaruK Tags: electric or magnetic, electrons, experience, field, free 

#1
Feb1913, 11:51 PM

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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 "selfinteraction" 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
Feb2013, 12:15 AM

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PF Gold
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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 selfinteraction 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
Feb2013, 12:40 AM

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Anyhow, the interaction due to the virtual photon should not enter the Hamiltonian as [itex]e\boldsymbol{\sigma}\cdot\mathbf{A}[/itex],[itex]\boldsymbol{\mu}_e\cdot\mathbf{B}[/itex] and [itex]eV[/itex]? Sorry I know little of QED. In that paper, the authors simply added [itex]e\boldsymbol{\sigma}\cdot\mathbf{A}[/itex] and [itex]eV[/itex] to the Hamiltonian, where [itex]\mathbf{A}[/itex] and [itex]V[/itex] 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
Feb2013, 12:59 AM

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PF Gold
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Will free electrons experience its own electric/magnetic field?
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
Feb2013, 03:29 AM

P: 11

I've sent the reference link to you by private message. Thank you. 



#6
Feb2013, 05:43 AM

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#7
Feb2013, 05:55 AM

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#8
Feb2013, 06:15 AM

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PF Gold
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Here's the link I got: S. M. Lloyd, M. Babiker, J. Yuan, and C. KerrEdwards Electromagnetic Vortex Fields, Spin, and SpinOrbit 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 spinorbit 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
Feb2013, 06:26 AM

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with the electron vortex, we start from the Dirac equation in the presence of electromagnetic fields, with vector and scalar potentials A and [itex]\Phi[/itex]. 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? 



#10
Feb2013, 06:52 AM

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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
Feb2013, 07:22 AM

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I just made a simple calculation. For a 200 kV electron, its speed is about 0.7c. Assuming that the crosssection 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
Feb2013, 07:37 AM

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Zz. 



#13
Feb2013, 09:15 AM

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But I still have the question that the Hamiltonian of single electron, if we have taken the spacecharge effect into consideration, will still have the form as described in the reference? Thanks very much. 



#14
Feb2013, 08:17 PM

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#15
Feb2113, 01:34 AM

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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, ee interactions, but the vortex comes from someplace.
It's a bit like pointing out that the ee Bfield interactions for a current is negligible when dealing with the effect of the Bfield 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
Feb2113, 02:33 AM

P: 11

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 selfinteraction back to the Hamiltonian although there is no external field or electronelectron interaction? 



#17
Feb2113, 03:41 AM

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Self interactions occur for free particles in the perturbation theory. You need to learn more about selfinteractions and renormalizations to understand how this works.
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
Feb2113, 04:00 AM

P: 11

[1] PRL 99, 190404 (2007). [2] PRL 107, 174802 (2011). [3] Ultramicroscopy 111, 14611468 (2011). I've written a mail to the author but got no response by now. 


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