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Does QFT respect conservation of angular momentum?

  1. Apr 2, 2013 #1
    Hi there

    Iive been reading the intro chapter to Peskin and Schroder's 'An intro to QFT'
    I have a question regarding the conservation of angular momentum during particle collisions/scatterings

    As an example they talk about e+ e- --> μ+ μ-

    The take the Centre of Mass (CM) frame.

    Consider the case where
    e+ is right handed and moving in + z-direction
    e- is left handed and moving in negative z-direction.
    This gives us a total of + one unit of angular momentum in the z direction

    These then couple to a photon which has polarisation vector (0,1, i,0)
    - this is circularly polarised in the z direction
    - is has angular momentum of plus one in z direction

    Then we look at the muons
    these may have a polarization (0, cosθ, i, -sinθ) where θ is the scattering angle
    - so the muons have a total angular momentum of +1 in the direction that is rotated by θ from the z direction


    So im confused how this conserved angular moment
    - at the start we had +1 in the z-direction
    - now we have +1 in the direction at angle θ to the z direction

    Can anyone explain this to me please. Is it to do with the reference frame somehow?

    Thanks
    Soph
     
  2. jcsd
  3. Apr 2, 2013 #2
    Is conservation of angular momentum defined to be conserved in the direction of motion? Maybe this is a possible answer to my question?
     
  4. Apr 2, 2013 #3

    tom.stoer

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    Angular momentum as a generator of the Poincare group is always conserved. The generators are {Pμ=H,Pi; Li; Ki} with 4-momentum P, angular momentum L and boosts K.

    Momentum and angular momentum conservation means that [H,Pi] = [H,Li] = 0. This eq. holds strongly as operator equation (and w/o quantization anomaly, hopefully ;-). In rel. QFT the operators H, P, L, K are defined in terms of field operators ψ etc., i.e. H[ψ], P[ψ], ... The commutation relations can be calculated using the commutation relations of the fundamental fields ψ. The algebra of the operators H, P, L, K and the algebra of the classical 4*4 matrices generating the 4-dim. rep. of the Poincare algebra are identical.
     
    Last edited: Apr 2, 2013
  5. Apr 2, 2013 #4

    Vanadium 50

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    A virtual photon is described with a propagator, not a polarization vector.
     
  6. Apr 2, 2013 #5

    vanhees71

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    A virtual photon is not a photon ;-).
     
  7. Apr 2, 2013 #6

    Bill_K

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    With all the favorable comments heard here about the Peskin/Schroeder book, I'm a bit disappointed at how they handle this example. (The relevant pages are available for inspection on Google Books.)

    Angular momentum is certainly conserved, and the obvious way to see it is to quantize everything along the same axis. For reasons unknown, P/S quantize the electrons along one axis and the muons along the other. :frown:

    Well you can, of course, represent a spin-up state in terms of spin states along some other axis, provided you use a coherent superposition, and that's what they're doing. Fig 1.3 says "One possible set of spin orientations". They subsequently calculate M for the other orientations and add them all together (as coherent amplitudes).
     
  8. Apr 14, 2013 #7
    I'm missing something here. I thought they had used the same axis for both muon and electron. The muons are scattering at an angle theta, so be rotating by theta they have lined the muon axis up with the electron? i.e both defined along the z axis in figure 1.3

    So are you saying that individually each spin orientation expression doesn't conserve angular momentum but when you have added them all for the total amplitude, this does?

    Sorry if i am thinking of this too simply. I have done the full calculations for compton scattering from chapter 5 so my level of understanding is slightly better than someone who's just started QFT but I am still finding certain things hard to grasp.

    In the compton scattering chapter (5.5) it doesn't really talk about things in the same way... so chapter 5 made sense to me. Looking at the expression for the amplitude calculated for compton scattering the polarization vectors are contracted with gamma matrices, so I guess somehow this part of the algebra is what conserves angular momentum?

    Going back to the bit in chapter one that is confusing me -
    I think that the main thing thats bothering me is the sentence 'Since H_I should conserve angular momentum, the photon to which these particles couple must have the correct polarization vector to give it this same angular momentum.' Then then seem to align the photon with the electron angular momentum, but not the muon. They make a point of making sure the photo polarisation vector matched the electron, but they don't do this with the muon and i can't understand why.


    Thankyou for your comments :)
     
  9. Apr 14, 2013 #8
    I understand this and have done many calculations of amplitudes using propagators, but I am simply repeating and trying to understand what is written in chapter one of Peskin :)
     
  10. Apr 14, 2013 #9

    Bill_K

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    [This was almost two weeks ago!]
    No, they never rotate anything. They are treating both electrons and muons throughout as right- and left-handed, that is, quantizing the spin of each particle along its direction of motion. In addition to showing this in Fig 1.3, they say this repeatedly. They calculate four amplitudes, M(RL→RL), M(RL→LR), M(LR→RL) and M(LR→LR) and add them together. But they need to calculate the dot product between the weak current for the electrons and the weak current for the muons (Eq 1.3). The electron current points along the z-axis (Eq 1.4) while the muon current points at an angle θ (Eq 1.5).

    Given a state which is spin up along the z axis, you can write it as a superposition of states with definite spin projection along the x axis. Individually, these components seem to have different spins. Does not disconserve angular momentum, because when you add them together, the state still has spin up.
     
  11. Apr 14, 2013 #10
    Sorry for my slow reply originally, been busy with family stuff over Easter and such like!
    Anyway, thanks for your reply, it makes sense, I think it's starting to click in my head now :)
     
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