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B Dirac spin matrices

  1. Jan 1, 2017 #1
    So, we can break down the Dirac equation into 4 "component" equations for the wave function.

    I was going to post a question here a few days ago asking if a fermion (electron) could possess a "spin" even if it were at rest, I.e., p=0.

    I did an internet scan, though, and found out that, indeed, you can have zero momentum and still be in a spin up or down state.

    Why is that?

    What's the purpose of the Pauli "spin" matrices if you don't need them to imbue a particle with spin? What's their purpose? From what I gather from Viascience, if you're in a rest state, you can be in "pure" up or down spin state, but once you start moving, you confound that pure state when you start moving and add momentum to the equation.



    But, my central question remains. The Dirac equation is a 4 component coupled equation with 4 solutions. The first two are positive energy solutions and the next two are negative energy solutions. However, if you set the momentum to zero, there doesn't seem to be anything in the math that would suggest a "spin." Having a spin would seem to be a 3-D property. Once you set the momentum to zero, it seems as if the contribution of the spin matrices are irrelevant.

    The only thing I can think of is that perhaps there is something intrinsic to the two (say positive energy) solutions that imbue a spin up or spin down character just by virtue of the equation itself?
     
    Last edited: Jan 1, 2017
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  3. Jan 1, 2017 #2

    PeterDonis

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    I'm not sure why this is a problem, physically speaking. What's wrong with having a spin 1/2 particle with zero momentum?

    It might be that the non-relativistic formulation of the Dirac equation is confusing you. Bear in mind that that is only an approximation; the more exact underlying theory is quantum field theory, which for spin 1/2 particles means the relativistic Dirac equation:

    $$
    \left[ i \gamma^\mu \left( \partial_\mu + i e A_\mu \right) - m \right] \psi = 0
    $$

    Here ##\gamma^\mu## are a set of four 4 x 4 matrices, the Dirac matrices; they are the relativistic generalization of the Pauli spin matrices. Since they include a "time component" ##\gamma^0## as well as the "space components" ##\gamma^1##, ##\gamma^2##, ##\gamma^3##, the formalism is exactly the same for the case of zero momentum as for the case of nonzero momentum. In a relativistic theory these cases aren't fundamentally different anyway, it's just a choice of reference frame with no difference physically.
     
  4. Jan 1, 2017 #3
    Don't you need three (space) dimensions (or at least two) to have angular momentum? I guess that's my dilemma. Spin is about angular momentum and I don't see how you get that when you set p=0.
     
  5. Jan 1, 2017 #4

    bhobba

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    See the following - things will be a LOT clearer:
    https://www.amazon.com/Physics-Symmetry-Undergraduate-Lecture-Notes/dp/3319192000

    I am studying it right now and it has cleared up a lot for me eg the exact assumption being made in deriving Maxwell's equations from SU(1) symmetry.

    Actually spin is about symmetry where its defined as j1+j2 from the generators of the Poincare group (see page 85 of the above book).

    Thanks
    Bill
     
    Last edited by a moderator: May 8, 2017
  6. Jan 1, 2017 #5

    PeterDonis

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    There are three. What's the problem?

    Quantum angular momentum does not mean some little billiard ball is actually spinning about an axis. Nor does the linear momentum p being zero have anything to do with the number of space dimensions. So I still don't understand what the problem is.
     
  7. Jan 2, 2017 #6

    vanhees71

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    For a massive particle spin is defined by the representation of the rotation group (or more precisely its covering group SU(2)) for the particle modes at ##\vec{p}=0##. This induces a complete representation of the orthochronous special Poincare group via Wigner's construction of the momentum eigenstates for ##\vec{p} \neq 0## via boosts. See Weinberg, Quantum Theory of Fields, vol. 1 or appendix B in my QFT script:

    http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
     
  8. Jan 2, 2017 #7

    stevendaryl

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    In quantum mechanics, angular momentum comes in two types: Intrinsic angular momentum (spin), and "orbital" angular momentum.

    Orbital angular momentum is indeed defined in terms of momentum: [itex]\vec{L} = \vec{r} \times \vec{p}[/itex]. So if [itex]\vec{p} = 0[/itex], then [itex]\vec{L} = 0[/itex]. Intrinsic spin is independent of momentum.
     
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