0
$\begingroup$

Why do the complex scalar fields defined using two sets of creation and annihilation operators (one for particle and one for antiparticle) describe spin-0? Is it because that they originated from scalar QED in which $e^-$'s spin is neglected?

$\endgroup$
4
  • $\begingroup$ Are you asking why a field with 2 sets of creation/annihilation operators describes a spin-0 particle as opposed to a higher-spin particle? Note that spin-0 particles are adequately described by real scalar fields; complex scalar fields describe charged spin-0 particles. $\endgroup$
    – bapowell
    Mar 20 '18 at 15:47
  • $\begingroup$ I am wondering why it describes spin particles at all. $\endgroup$
    – John
    Mar 20 '18 at 15:48
  • $\begingroup$ And to your comment, why are spin 0 particles described by real scalar fields? And the complex scalar fields is the real scalar fields plus photons? $\endgroup$
    – John
    Mar 20 '18 at 15:49
  • $\begingroup$ First off, by "scalar" what is really meant is "Lorentz scalar"; that is, a field that is invariant with respect to Lorentz transformations. Rotations are a subset of Lorentz transformation: if a field is unchanged by a rotation, it cannot have spin (which has a direction). Therefore, a field is spin-0 if and only if it is a Lorentz scalar. A complex scalar field still only describes spin-0 particles (photons are spin-1, or vectors under Lorentz). The additional imaginary degrees of freedom of the complex field are manifested as charge. $\endgroup$
    – bapowell
    Mar 20 '18 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.