# Interpretation of 4-vector quantum field operator

Peskin and Schroeder, on page 24, quotes the following expression for a generic (scalar) field operator: $$\phi(\mathbf{x})|0\rangle = \int \frac{d^3\,p}{(2\pi)^2}\frac{1}{2E_{\mathbf{p}}}e^{-i\mathbf{p}\cdot\mathbf{x}}|\mathbf{p}\rangle,$$ which, apart from the relativistic $1/2E_{\mathbf{p}}$ at the from, is the usual expression for $|\mathbf{x}\rangle$.

So, they say that the operator $\phi(\mathbf{x})$, acting on the vacuum, creates a particle at position $\mathbf{x}$.

What is a similar interpretatio fo 4-vector field such as $A^{\mu}$? Surely it cannot be creating a photon localised at $\mathbf{x}$?

• "Surely it cannot be creating a photon localised at x?" why not? would you mind elaborating on this statement, please? Dec 9 '15 at 18:48
• Photons move at the speed of light, I assumed they can't be localised? Dec 9 '15 at 19:26
• Dec 9 '15 at 19:32

The argument used by Peskin and Schroeder is that computing the scalar product between the above and a general one-particle state $|p'\rangle$ one has: $$\langle 0|\phi(x)|p'\rangle = e^{ipx}$$ therefore reminiscent of the momentum representation on the position states of a single particle in quantum mechanics. Nevertheless, the reasoning is flawed and it cannot be applied because the definition of particle in QFT is much more subtle and takes into the irreducible representations of some gauge groups under which the equations of motion must be invariant. Such representations can, in some cases, be classified and expressed in terms of some parameters $(m,s)$ that one reads as the mass and the spin of the particle; the correct argument is more complex, however the aforementioned is in a nutshell how the particles spectrum is introduced in QFT.
This said, given the expansion of the electromagnetic field $A^{\mu}$ as Fourier series, the interpretation is along the same lines as for the rest, that is each of the $\mu$ component of the vector potential is an independent field that creates and annihilates an infinite number of particles. The $A^{\mu}|0\rangle$ must not be interpreted as particles states; instead one has to compute the propagators and the scattering amplitudes vacuum-vacuum (or given any other configuration): once doing so, the vector indexes contract together after being summed upon and one is left with a real number that in principle expresses the probability of going from the initial configuration to the final one in presence of some electromagnetic scattering in the between.