1
$\begingroup$

I'm studying the Källén-Lehmann representation for Dirac spinors on the book Quantum Field Theory: Lectures of Sidney Coleman (chapter 23, around page 490). I'm having trouble to analyse the multi-particle terms: the author implies that to the sum/integral \begin{equation} \sum_n\langle 0\rvert\psi'(0)\lvert n\rangle\langle n\rvert\overline{\psi}'(0)\lvert 0\rangle, \end{equation} where $n$ runs over all multi-particle states of the theory, and $\psi'$ is the renormalised field, only states of spin $\frac12$ contribute.

I think that one could prove that $\overline{\psi}'(0)$ creates from the vacuum a state with spin $\frac12$ only: once this is done, then the state $\lvert n\rangle$ would be orthogonal to the previous one and vanish from the sum unless it has spin $\frac12$ itself. However I can't find a way to prove the statement in bold (above), about which I'm not really sure anyway, in a mathematical fashion. Where could I start? If $\psi'$ were the free field, and $\lvert 0\rangle$ the "bare vacuum", then it would be obvious, but here we have the "physical vacuum", and the renormalised field, so I'm not so sure that the deduction is the same.

$\endgroup$

1 Answer 1

2
$\begingroup$

As the spin is an observable, it is an Hermitian operator. Therefore, two eigenstates of the spin operator corresponding to different eigenvalues (that is - states which have different spins) must be orthogonal to each other. If $|0\rangle$ is a state with spin $0$ (that is - an eigenstate of total the spin operator with eigenvalue 0), and $\bar{\psi}'(0)$ is an operator that creates a particle with spin-$1/2$, then $\bar{\psi}'(0)|0\rangle$ will be an eigenstate of the total spin operator with egienvalue $1/2$, and only states $|n\rangle$ with the same value of total spin will can be not-orthogonal to it.

$\endgroup$
2
  • $\begingroup$ Sorry, maybe I should have expressed myself more clearly in the question, the part about the orthogonality of the states is no problem for me, I worried about proving that $\psi'$ creates a spin-$\frac12$ state from the vacuum. $\endgroup$
    – yellon
    Jan 21, 2020 at 22:30
  • $\begingroup$ oh. I see. While I don't have access to the book, the renormalization of the field usually will not change the fundamental properties such as spin content. It depends on the theory, of course, but basically you require that $\{ \psi'(x), \bar{\psi}'(y)\} = \delta(x-y)$ which means that they create normalized particles from the vacuum, and then assuming that the interaction preserves spin, the renormalization process will also maintain the spin content of the operator. A way to see it will be to examine the commutation relations of $\bar{\psi}'$ with the (renormalized) spin operator. $\endgroup$
    – user245141
    Jan 22, 2020 at 9:43

Your Answer

By clicking “Post 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.