Are quantum fields observable? I have an elementary question. In quantum field theory observables are operators. But the quantum fields are operators too. This means the quantum fields are observable? I read somewhere that quantum fields are not observable.
 A: I am familiar with the quantum field theory of the standard model of physics, and my answer is within this frame.
The elementary particles in the axiomatic table of the standard model, the electrons, quarks, etc are mathematically represented as a field, an electron field, a quark field, etc, by the plane-wave wave functions of the corresponding quantum mechanical equation without a potential, Dirac for electron, quantized maxwell for photon etc. , at each point in space time. As is well known the  wave-function $Ψ$ is not measurable (only $Ψ^*Ψ$), and afaik, these fields are not operators.
Creation and annihilation operators operate on these fields, creating a particle and annihilating it, and that creation would be observable, if it were not the difficulty of plane waves that can not be  normalized to probability amplitudes. Real particles have to be represented by wavepackets  of the plane waves.
Fortunately for experiments in particle physics it is by the use of Feynman diagrams that the observable crossections and decays are calculated, and the predictions are validated continually in new experiments.
By successfully predicting using Feynman diagrams ( see paragraph 4) a specific decay, or a crossection, using creation and annihilation operators on the relevant fields, one can say that the existence of the fields is validated, and in that sense the fields are observable, because if they were not there the creation and annihilation operators would not work. So it is the combined  "creation and annihilation operators on the fields" that are really observable.
A: Observables in QFT are generally constructed as products of field operators such as $\hat{\phi}^- \hat{\phi}^+$. Note that such combination are Hermitian as they need to be to qualify as observables. Individual fields operator may not always be Hermitian. Even in those cases where the field operator is Hermitian (such as the photon), the single field operator would then to give zero. For instance $\langle n|\hat{E}|n\rangle = 0$, where $\hat{E}$ is the field operator for the electric field and $|n\rangle$ is a number state for the photon.
