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I Right and left handed electrons

  1. Jun 7, 2016 #1
    I am an independent learner studying particle physics and am using lecture notes and the like, obtained from open courseware etc. My problem is, understanding why there is a difference between the parameters that describe electrons with right handed and left handed spin. Specifically, why doesn't the e - right have isospin and yet has twice the hypercharge value of the e - left? Are these differences experimentally measured values or theoretically plugged in?

    I can follow the matrix algebra (just!) which gives rise to right and left handed spinors; that they Lorentz transform differently but this is just theory. Surely an electron is an electron. I understood them to have spin up or down = ± ħ / 2 from my undergraduate QM studies but this concept of differences in isospin and other quantum numbers has thrown me.

    Can anyone shed some light on the reason behind these differences in a relatively non-mathematical way?
     
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  3. Jun 7, 2016 #2

    Orodruin

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    The left and right handed fields are independent. It is only upon the acquisition of a mass term that they mix.

    The left handed electron field is part of an SU(2) doublet, while the right handed is a singlet. The experimental evidence for this is the V-A structure of charged current interactions.
     
  4. Jun 7, 2016 #3

    ChrisVer

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    I guess a very easy reason to see why the left and rights are not equivalent is taken from Parity Violation...
    Only interactions that can possibly violate parity see a difference between the left/right-handed particles.
    Also it can be confusing to use the term "isospin" for obvious reasons (because of this https://en.wikipedia.org/wiki/Isospin).
     
  5. Jun 7, 2016 #4

    PeterDonis

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    Not in an "A" thread. In fact, it's borderline even in an "I" thread, which is what I have reclassified this thread as. An "A" level thread assumes graduate level understanding of the subject matter; there's no way to get that in "a relatively non-mathematical way". An "I" thread assumes undergraduate level understanding of the subject matter; even that is problematic if you want it in "a relatively non-mathematical way".

    Bottom line, the subject you are trying to address here is an advanced one, and you need to be prepared to deal with advanced concepts in order to understand it.
     
  6. Jun 7, 2016 #5

    PeterDonis

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    We can't directly measure weak isospin and hypercharge because electroweak symmetry is broken in our current universe. But we can measure them indirectly by inducing various interactions in high energy experiments, such as in the LHC, and making measurements of them. Also we can relate them to other quantum numbers that we can directly measure: for example, electric charge ##Q## is related to them by the formula

    $$
    Q = I_3 + \frac{1}{2} Y
    $$

    So, for example, the left-handed electron, with ##Q = -1## and ##I_3 = - 1/2##, must have ##Y = -1##, while the right-handed electron, with ##Q = -1## and ##I_3 = 0##, must have ##Y = -2##.

    Not according to the Standard Model; according to the SM there are two electron fields, the left-handed one and the right-handed one. What you would normally think of as an "electron" is a combination of both of those fields (linked, as Orodruin pointed out, by the mass term in the Lagrangian).
     
  7. Jun 8, 2016 #6
    Thanks, these comments have helped, I get the point about there being two electron fields prior to acquiring mass now. Hence, I see how the two spinors arise from the zero mass case Dirac equation. I am familiar with the equation for Q above but am still unsure as to why the quantum numbers I3 and Y are different for the two types of electron. However, if this comes from more advanced theory I suppose I will just have to accept it for now.

    As for the level of the thread, I was uncertain - it is my first posting in this section of the forums. I didn't think it was suitable for the homework sections. I take your point about being prepared to study higher level theory; that is what I am trying to do - see my initial comments. However, the non-mathematical answers provided by yourselves has helped to elucidate the matter. Thank you.
     
  8. Jun 8, 2016 #7

    Orodruin

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    In a very condensed form, it has to do with how the fermions acquire masses through the Higgs mechanism and how the fields interact through weak interactions. In order to preserve the gauge symmetry in the Higgs interaction term that becomes the mass term when the Higgs takes a vev, the left-handed electron needs to be part of an SU(2) doublet together with the left-handed neutrino. This is where the quantum numbers come from. After symmetry breaking, it is just a particular combination of I3 and Y which remains a symmetry generator, corresponding to the charge.
     
  9. Jun 8, 2016 #8

    ChrisVer

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    that's because the R electrons are in an SU(2) singlet, while the L electrons are in an SU(2) doublet...
    That means that [almost by definition] they have different I3s... so in order to have the same charge their Ys need to be different as well...

    Again as to why R an L electrons belong to different representations, you'd better recheck the parity violation of the weak interactions... in particular you need to have something different between L and R... I guess that's also the reason why you never created a Right electron doublet as well (with a R neutrino)...and the SM's EW symmetry is not an SU(2)L+R x U(1)
     
  10. Jun 8, 2016 #9

    PeterDonis

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    You can actually add a right-handed electron neutrino to the Standard Model, but it has to be an SU(2) singlet, like the right-handed electron (because, as you say, there has to be a difference between L and R to account for parity non-conservation in weak interactions). Doing this is one way of accounting for neutrino masses.
     
  11. Jun 9, 2016 #10
    Thank you for these comments. They are helping me to get a better understanding of the subject. You have provided me with enough background material for me to look into this in more detail.
     
  12. Jun 10, 2016 #11
    I am assuming an SU(2) singlet transforms as exp(-iσα/2) whereas an SU(2) doublet transforms as exp(-iσα), where σ generators are Pauli matrices, or something similar, am I right?
     
  13. Jun 10, 2016 #12

    PeterDonis

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    No. An SU(2) singlet doesn't transform at all under SU(2); an SU(2) transform acts on it as the identity operator.
     
  14. Jun 13, 2016 #13
    Right. Thanks once again; I shall look into this in more detail now.
     
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