- #1
jfy4
- 649
- 3
Hi,
I suppose I'm a little late to start here, but I just got hung up on the following: The field quanta in E&M is the photon and it comes from the gauge potential in QED [itex]A(x)[/itex]
[tex]
A(x)=\int \frac{d^3 p}{(2\pi)^3 \sqrt{2\omega_p}}\sum_{\lambda=1,2}\left[ \epsilon(p,\lambda)a_{p,\lambda}e^{-ipx}+\epsilon^{\ast}(p,\lambda)a^{\dagger}_{p. \lambda }e^{ipx} \right]
[/tex]
which is an operator on the fock space that creates a particle with helicity 1, momentum k, energy [itex]|k|[/itex], and no mass, the photon. But classically the potential [itex]A[/itex] is not an observable, so how come the photon is an observable?
Thanks,
I suppose I'm a little late to start here, but I just got hung up on the following: The field quanta in E&M is the photon and it comes from the gauge potential in QED [itex]A(x)[/itex]
[tex]
A(x)=\int \frac{d^3 p}{(2\pi)^3 \sqrt{2\omega_p}}\sum_{\lambda=1,2}\left[ \epsilon(p,\lambda)a_{p,\lambda}e^{-ipx}+\epsilon^{\ast}(p,\lambda)a^{\dagger}_{p. \lambda }e^{ipx} \right]
[/tex]
which is an operator on the fock space that creates a particle with helicity 1, momentum k, energy [itex]|k|[/itex], and no mass, the photon. But classically the potential [itex]A[/itex] is not an observable, so how come the photon is an observable?
Thanks,