If Klein Gordon equation is for spin-0 particles, I write massless fields as $\square A=0$, how can I say $A_\mu=\epsilon^\mu e^{-ikx}$ as a wave function of polarized photon (spin-1) ?
[Physics] Klein Gordon for spin-1 particle photon
klein-gordon-equationquantum-field-theoryquantum-spin
Best Answer
As in the comments, fulfilment of the Klein Gordon equation is only a necessary condition for a field and it is fulfilled by all fields. For example, the Dirac equation for an electron implies the Klein Gordon equation, but not conversely. If you've never seen this, try working out the following. Begin with the Dirac equation $(i{\partial\!\!\!\big /} - m) \psi = 0$ and then impart the operator on the left hand side again to both sides of the equation: you'll find $(i{\partial\!\!\!\big /} - m)^2 \psi = 0$ is equivalent to the Klein Gordon equation for all four components of the Dirac spinor independently. $(i{\partial\!\!\!\big /} - m) \psi = 0$ is therefore a strictly tighter constraint than $(i{\partial\!\!\!\big /} - m)^2 \psi = 0$. Likewise for a massless spin one field: you of course get D'Alembert's equation. But you also need a Lorenz gauge condition, leading to the tighter constraint.