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Momentum-free spinors and the Dirac equation

  1. Jan 16, 2016 #1
    We can create a Dirac equation with no potential energy and zero momentum and still get spin? Is this correct? How do the Pauli spin matrices apply here? On the surface, the Dirac equation seems fairly straightforward, but when you dig even a little deeper, it's starts to become unwieldy

    I've got a million questions, but my first one is what does the zero potential, zero momentum Dirac equation tell us about spin? I don't get it.
  2. jcsd
  3. Jan 16, 2016 #2
    Ok, if you solve this equation, what you get is psi equals the exponential raised to the power of the mass, Ψ=e^i(M)t

    But you have 4 equations, with 4 eigenvalues, two of which are for the positive energy solution and two of which are for the negative energy solution. We can split each into two spin states but that tells (at least me) nothing as to how these states vary from one another. In my explorations of what "spin" means and is what it is telling us my inclination is to look at the Pauli spin matrices and try to decipher what the superpositions of the left versus the right moving wave functions are telling us, along with how each of the 4 equations mix up these directional wave dynamics. What I am not seeing is how when you halt the momentum and eschew the contribution of the Pauli matrices, how do you still get spin?
  4. Jan 16, 2016 #3


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    Hmm. It's probably best to take a step back and consider the theory of (nonrelativistic) quantum angular momentum in isolation. But I don't know how much you've studied that. E.g., do you have a copy of Ballentine? The crucial information here is in sect 7.1 (iirc). Quantized spin emerges from essentially nothing more than a requirement that elements of the rotation group ##SO(3)## be represented as unitary operators on a Hilbert space -- independent of any considerations about energy or linear momentum. From there, one can progress to the Pauli equation (nonrelativistic analog of the Dirac eqn), and so on.
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