Consider the Dirac spinor for a positron at rest, and the spinor for an electron with non-zero 3-momentum. In the Dirac basis it is clear that these are not orthogonal, as I would expect. Does this have a physical interpretation?

Of course the full states are orthogonal, but I would expect the spinors to be orthogonal also.

When I refer to spinors in this question, I mean the spacetime-independent object.


1 Answer 1


Why would you expect the spinors to be orthogonal?

Recall, for a vanishing electromagnetic field, the Pauli equation collapses to $$(\hat p ^2 /2m -i\hbar \partial_t )|\psi\rangle=0,$$ solved by
$$|πœ“(𝑝)\rangle = \exp(βˆ’π‘–π‘^2t/2π‘šβ„)|πœ“(0)\rangle . $$ This is a pure tensor product state of a momentum space (unnormalizable) wave function βŠ— a spinor, which does not change at all, so it dots to itself, to yield 1 , if normalized. The dotting of your two wavefunctions dots each of its tensor factors in the respective space. Dotting 1 to an exponential, however, will readily yield zero for $p\neq 0$.

What's weird about that?

  • $\begingroup$ Sure, the whole states are orthogonal, as I said. But it seems worth thinking about that the boosted electron spinor has a nonzero projection onto the positron rest spinor, while the two particles at rest have orthogonal spinors. If you have a physical interpretation of this, I'd like to hear it. $\endgroup$
    – ToKalon
    Commented Dec 19, 2020 at 17:57
  • $\begingroup$ I've no idea what puzzles you. You take a spin up electron at rest, and boost it in the z direction to something that's still spin up and has a nonzero momentum. Without spin, you may rotate (analogously to a boost) a state to an orthogonal state. $\endgroup$ Commented Dec 19, 2020 at 19:21

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