# Electromagnetic Potential as an Observable

1. Aug 23, 2011

### jfy4

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 $A(x)$
$$A(x)=\int \frac{d^3 p}{(2\pi)^3 \sqrt{2\omega_p}}\sum_{\lambda=1,2}\left[ \epsilon(p,\lambda)a_{p,\lambda}e^{-ipx}+\epsilon^{\ast}(p,\lambda)a^{\dagger}_{p. \lambda }e^{ipx} \right]$$
which is an operator on the fock space that creates a particle with helicity 1, momentum k, energy $|k|$, and no mass, the photon. But classically the potential $A$ is not an observable, so how come the photon is an observable?

Thanks,

2. Aug 23, 2011

### Ben Niehoff

But A is observable classically! Specifically, the combinations

$$-\nabla \phi + \partial_t \vec A, \qquad \nabla \times \vec A$$

are observable. A has extra gauge degrees of freedom. But the true degrees of freedom, after removing the gauge redundancy, are observables. You'll notice that after removing the gauge redundancy, A (a four-vector) has two remaining degrees of freedom; these are exactly the two observable polarizations of light.

3. Aug 23, 2011

### jfy4

Thanks Ben,

That's what I thought, that A is observable after preforming those operations on it. I specifically recall that the potential field cannot be directly observed, only the field strength tensor, which is what you have above.

Thanks,