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Helical Electron Polarisation?

  1. Oct 11, 2004 #1
    In what manner can an electron be said to be helically polarised?
    I have heard of spin-polarisation, but that should be strictly up or down shouldn't it?
  2. jcsd
  3. Oct 19, 2004 #2
    helical polarization???

    You sure you aren't referring to chirality (the position of momentum p compared to the position of the spin-vector). In QFT the four-fermion interaction of the weak force is described by the V - A-current. Basically this means that the fermionic current to which the massive gauge-bosons of the weak interaction couples, can be written as a vector and an axial (A) vector. This V-A structure arises when you apply the left-chirality-operator onto the fermionic current, calculating the left handed components of the invloved matter fields.

    Keep in mind that only left handed chiralities "feel" (this means : are subject to) the weak interaction. Experiments show us that only fermions of left handed chirality will undergo weak (eg decay-processes) interactions. In other words : the massive gauge bosons of the weak interaction only couple to left handed fermionic currents (ie the matter particles with left handed chirality)...

  4. Oct 19, 2004 #3
    I suspect Jacko means that the spin of the electron would be coaxial to its direction of movement.
  5. Oct 19, 2004 #4
    Because of this difference in behavior between left and right handed chiralities and the fact that a mass-term in the Lagrangian mixes up these chiralities, all elementary particles are massless. Once the breakdown of symmetry has occurred (this means : the original symmetry of the Lagrangian is no longer respected by the vacuum or groundstate) mass is "generated" through interaction with the Higgs-field. Photons are always massless because they do not interact with the Higgs field. The reason for this is that after this symmetry-loss of the vacuum-state the "universal" U(1)-symmetry of elektromagnetism is still respected by both the vacuum and the Lagrangian. So photons do not "feel" this loss of symmetry because their U(1)-symmetry is still there...

    You gotta think of this Higgs-interaction as if a particle were moving through some kind of background that is originally uniform. When the particle appears in this Higgs-field the uniformity will be disturbed, yielding a higher density of Higgs-particles around the elementary particle. This higher density is the mass that is generated around the particle. This socalled Higgs-particle is also referred to as the Higgs-boson. Many Higgs-bosons together constitute a Higgs-field just like many photons constitute an EM-field. This Higgs-boson has not yet been observed.

  6. Oct 19, 2004 #5
    So he is talking about chirality...

  7. Oct 19, 2004 #6
    I'm new to this (the reasoning leading from electron spin to the Higgs particle), but it all seems quite hypothetical. How are people looking for this Higgs particle?

    Could you clarify how a left-hand electron (which doesn't decay at all as far as I know) would behave differently than a right handed one?
  8. Oct 19, 2004 #7
    Indeed it doesn't. Yet electrons are invloved into decay processes, just look at the electron in beta-decay. The difference in behaviour is exactly the fact that only left handed fermions feel the weak force.

    About the search for the Higgs-particle : here is a nice site :

    Last edited: Oct 19, 2004
  9. Oct 19, 2004 #8
    The Higgs-boson is indeed speculative to some extent yet the Higgs-mechanism is the best thing we have up till now for explaining the mass of elementary particles. The discussion of mass mixing up the two chiralities is inherent to the theory of QFT and experiments have shown us that only left handed chiralities couple to the weak interaction. So all elementary particles must have ecquired mass after the spontaneous breakdown of symmetry. This is all proven and shown by experiments. Only the Higgs-particle is a well-considered "guess"... So this is not just some speculation...

  10. Oct 20, 2004 #9
    Just a little precision : we're talking here about charged current, because right particles can couple to the Z boson (neutral current).

    These different comportments can be "seen" when you derive the QED lagrangian. You will find different couplings between fermionic, Z and W fields.

    As Marlon said, you will find that only the left handed fermionic fields are coupled to the W, but both right and left handed fields are coupled to the Z and the photon, with different amplitudes (which explain the different comportments even for the neutral current).
  11. Oct 20, 2004 #10
    Welcome Major !

    I disagree with your statements. You and Marlon seem to imply that there is a theoretical reason for the fundamental particles to be left-handed doublets and right-handed singlets. I wish I understand that, but I think it is wrong. Would right-handed neutrinos occur in Nature, they would be in the standard model as well. I fear that there is no well-justified deep theoretical reason for this bearkdown of chiral symmetry.

    Chiral symmetry breaking in QED for weak coupling
    J C Huang and T C Shen 1991 J. Phys. G: Nucl. Part. Phys. 17 573-581

    Gauge bosons in an [tex]SU(2)_{right} \otimes SU(2)_{left} \otimes G_{leptonique}[/tex] electroweak model
  12. Oct 20, 2004 #11
    You're right, there is no theoretical reason to that. I have just wanted to say that the different comportments between the two chiralities can be directly seen in the lagrangian.
  13. Oct 20, 2004 #12
    So is it correct to say that all beta-decayed electrons are left-handed?
  14. Oct 21, 2004 #13
    Like stated before, only left handed fermions are considered when interaction with the W+ and W- vector bosons is regarded. I am not talking about the Z-vector bosons...

    Last edited: Oct 21, 2004
  15. Oct 21, 2004 #14
    yes, electrons involved in weak decays are left handed. The neutrino is either left or right handed so the corresponding neutrino current cannot be zero. this is the famous two-component theory of the neutrino created by Salam, Yang, Lee and Landau in 1957...
    The reason for this is that given the left handed electrons invloved, the neutrino current must be written in a "certain" way invloving Dirac-matrices and so on. In this model you can "choose" certain variables to have certain values. Once chosen, the chirality of the neutrino is determined by anticommutation-relations that give you info whether the neutrino current is zero or not. So basically you "chose" the neutrino-chirality in such a way that the current is not zero, otherwise beta-decay would not be possible.

    I know this sounds vague but this is quite advanced QFT-matter and the best way to study this is looking this up in books like Weinberg or Siegel...

    http://arxiv.org/PS_cache/hep-th/pdf/9912/9912205.pdf (this is a difficult book, so you would need the necessary QFT-background)

  16. Oct 21, 2004 #15
    Euuh, give the above site some time to load...

  17. Oct 21, 2004 #16
    Yes, you should give this time, because it will be invaluably rewarded :approve:
    Siegel's "Fields" is incredible. Some people have expansive books of their own, which do not cover even half of this marvelous masterpiece.
  18. Oct 22, 2004 #17
    Hi Humanino,

    like i said, i went to Paris for opto2004. Marvellous technology, very interesting. You know, since i am studying applied physics right now i am totally into photonics and nanotechnology. I was really blown away by the state of the art technology that was presented there. Combine this with the many beautiful women in Paris and you know what i am talking about. We should really meet someday in Paris so you can show me the coolest places to visit...(i am not talking about the Eiffel tower here...)

  19. Oct 23, 2004 #18
    Thanks for all your replys. Just to clarify, the reference to helical electron polarisation: produced by circularly polarised laser light on a gallium-arsenic photocathode.
  20. Oct 24, 2004 #19
    From what I understood from marlon, electrons produced this way would differ according to whether the light is right-circular polarized or left-circular polarized. In one case, they could experience the nuclear weak force, but not in the other.

    It then seems that whether they are weak-force enabled or not also depends on if they are shot out of the sample on the front side (where light is incident) or the backside (should be possible for a very thin sample).

    Confirmation of this by someone would be appreciiated.
  21. Oct 25, 2004 #20
    From what I understand, both types of electrons interact via the weak force (maybe only via neutral current) and leave in the same direction. I still dont quite understand how these eletrons are 'helically polarised'. Is it something to do with their intrinsic spin?
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