Since field-operators are not always hermitian (for example in case of a complex scalar field, or the dirac-field), they don't (in the quantum-mechanical sense) correspond to observables.

Does that mean that field-operators are just a mathematical tool, that works out nicely, but there isn't any interpretation on what their eigenvalues (if available) or their time-evolution means?

I am especially asking because I don't have a problem imagining for example a classical complex-valued scalar-field. Asigning a complex value to any point in space seems pretty doable for me, and in classical field-theory, the fields (even if complex) where always the objects in question.

The process of quantization is often said to replace classical values with operators, whose eigenstates correspond to this values via the eigenvalues they have. For non-hermitean field-operators, this isn't the case.

Similarly, quantum mechanics is often described as a qft on 1 time-dimension, instead of 1 time-dimension and 3 space-dimensions, and the position-operator being an analogue of the field-operator. For non-hermitean field-operators, this analogy equally breaks down.

Can I somehow "safe" the two analogies given above? Is there another way in which the field-operators, even when being non-hermitean, correspond to a classical field? Or is it really just a mathematical tool? In that case, what meaning does it even have that a field-operator undergoes time-evolution?

EDIT: I think I know how to Narrow down my question, so it is answerable: Quantum-Fields in general have been introduced to me as operator-valued distributions: I was under the impression that means that at every point in space, a quantum field can be seen as a superposition of classical field values. This or this answers kind of support this idea.

Now that I know that quantum field operators are not neccessarily hermitian, this seems to contradict the two answers I cited above, and the notion of a quantum field as superposition of field values, because states are not necesarily decomposable into eigenstates of the field-operator anymore. How can I clear this contradiction?

Since I can see any operator as a sum of a hermitian and an anti-hermitian operator, I could argue that (for the example of the scalar field) a complex field has a real part and an imaginary part, and there simply isn't a base of the hilbert-space where both components are sharp. Is that a way one can look at quantum-fields?

  • 1
    $\begingroup$ Regarding your example: A non-hermitian scalar field operator may be an observable (I mean, the hermitian and anti-hermitian parts may be considered observables), but not if its coupled to a gauge field in the usual way (because observables should be invariant under local gauge transformations). More generally, non-observable field operators are useful for constructing observables, partly because field operators tend to satisfy simpler equations of motion. (The example of a non-hermitian scalar field coupled to a gauge field illustrates this.) Do any of these thoughts address the question? $\endgroup$ Oct 17, 2019 at 13:18
  • $\begingroup$ @ChiralAnomaly: Treating the hermitian and anti-hermitian parts as observable addresses my question (although that would mean that I couldn't measure the "real" and the "imaginary-part" of the corresponding classical field at the same time in general), because I still try to see field operators as a mathematical object that encodes probabilities for classical field configurations. $\endgroup$ Oct 17, 2019 at 16:44
  • $\begingroup$ You might consider looking in to the wave functional formalism that uses the Schödinger equation, etc. An example of a question using this approach: physics.stackexchange.com/questions/142180/… $\endgroup$ Oct 18, 2019 at 11:29


This site is temporarily in read-only mode and not accepting new answers.