1
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.