Interpretation of 4-vector quantum field operator

Question

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 $\mathbf{x}$?

Answer 1

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.

Besides that, the mathematical objects and operators that appear in QFT are very different from the ones in QM, especially because of the definition of the Fock space as (infinite) direct sum of single particle Hilbert spaces (and all that derives from there) or of the usage of particular functional derivatives and measures in the path integral formulation, which by no means appear in QM (the path integral does appear, but the measures and the functional derivatives are different). For this reason, namely being the objects and the formalism different, interpretations by similarity ought to be discarded.

From the point of view of the physics, QFT is a theory of many particles whose objects are fields and an infinite number of particles being created and annihilated everywhere in the universe at any time, with any momentum and energy. One does not look at localisations, rather we calculate scattering amplitudes given some field configurations, which is what people may experimentally observe in laboratories and thus compare to the theoretical outcomes.

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.