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Will a boost change spin angular momentum? Or what's spin angular momentum in QFT?

  1. Jan 7, 2012 #1
    By "spin" we sometimes refer to spin angular momentum, sometimes we refer to a specific representation of Lorentz group, in the following I'll refer to the former, otherwise I'll write "spinor representation"

    Say an spin up(z direction) electron at rest, state vector [itex]|m,+\rangle[/itex] being an eigenstate of [itex]J_z[/itex] , after boosting(x direction) it to the state [itex]|p,+\rangle[/itex], is no longer an eigenstate of [itex]J_z[/itex], because of the nonzero commutation relation [itex][J_z,K_x][/itex]. This fact is not surprising per se, because [itex]J_z[/itex] is total angular momentum, and we can't expect [itex]|p,+\rangle[/itex], which is a plane wave, to be its eingenstate. However an interesting question arises: How should we divide the total angular momentum into spin part and orbital part?
    In nonrelativistic QM it's pretty straightforward, orbital part [itex]L_z=(r\times p)_z[/itex], spin part [itex]S_z=\frac{1}{2}\sigma_z[/itex](setting [itex]\hbar=1[/itex]). And an spin up eigenstate of [itex]S_z[/itex] will stay an spin up eigenstate after any boost.
    However, in relativistic case the division is not so straightforward, for a Dirac field, by Peskin page 60, eqn(3.111)
    [tex]J_z=\int{d^{3}x\psi^{\dagger}[(x\times(-i\nabla))_z+\frac{1}{2}\Sigma_z]\psi}[/tex]
    It's very tempting to define the first term as the orbital part and second term as the spin part, but it might not be quite right, because if we apply the second term to [itex]|p,+\rangle[/itex] we don't see an eigenstate(as a reminder [itex]|p,+\rangle[/itex] is related to [itex]|m,+\rangle[/itex] by a pure boost), i.e. the good old property we had in nonrel QM is lost, now there are 3 possibilities of this problem:
    (1)Indeed [itex]|p,+\rangle[/itex] is not an eignestate of [itex]S_z[/itex] in relativistic QM
    (2)We should redefine [itex]S_z[/itex] so that [itex]|p,+\rangle[/itex] is an eigenstate.
    (3)We should redefine [itex]S_z[/itex] by some other reason, and [itex]|p,+\rangle[/itex] is not necessarily an eigenstate.
    So which one is correct? And what should we take as the starting point of defining [itex]S_z[/itex]?

    PS: In case of confusion, I'm not asking about Wigner rotation, and I'm just asking if an electron is spin up in its rest frame, then after a pure boost is it still spin up? And in what sense?
     
    Last edited: Jan 7, 2012
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  3. Jan 7, 2012 #2

    Bill_K

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    Re: Will a boost change spin angular momentum? Or what's spin angular momentum in QFT

    The Wikipedia article on the Pauli-Lubanski vector answers your questions.
     
  4. Jan 7, 2012 #3

    strangerep

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    Re: Will a boost change spin angular momentum? Or what's spin angular momentum in QFT

    IIRC, unlike the nonrelativistic case, there's no longer an invariant distinction between spin and orbital angular momenta in the relativistic case.

    (Sorry, don't have time right now to dig out a reference. Maybe later if no one else gives one.)
     
  5. Jan 9, 2012 #4

    Bill_K

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    Re: Will a boost change spin angular momentum? Or what's spin angular momentum in QFT

    In either case, the invariant distinction between spin and orbital angular momentum is that spin is defined to be the angular momentum in the rest frame.
     
  6. Jan 9, 2012 #5

    strangerep

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    Re: Will a boost change spin angular momentum? Or what's spin angular momentum in QFT

    Doesn't it need to be bit more general than that? (The above doesn't work in the massless case.)
     
    Last edited: Jan 10, 2012
  7. Jan 10, 2012 #6
    Re: Will a boost change spin angular momentum? Or what's spin angular momentum in QFT

    Thanks for the replies, but I'm a bit busy recently, I'll check out Pauli-lubanski later in detail.
     
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