The short answer is, No, because the fields don’t commute with the Hamiltonian. Only operators that commute with the Hamiltonian have definite values in the eigenstates of the Hamiltonian. This, in fact, holds in general.

Thank you Stam Nicolis .

I can follow the following line of reasoning. An equation, which has a commutator containing the Hamiltonian, for the time derivative of the field-amplitude operator results (at least for the kinds of fields I have in mind) in this time derivative being related to the momentum density operator. This together with a commutator rule between momentum density and field amplitude implies that the field amplitude operator does not commute with its time derivative when both are evaluated at the same time point. In other words, the field amplitude and its time derivative are not simultaneously knowable. From this conclusion it can be argued that if the field state has a definite value of field amplitude at one point in time, it will not have a definite value of field amplitude at other times.

I get the above conclusion, that if the field state has a definite value of field amplitude at one point in time, it will not have a definite value of field amplitude at other times. However, I don’t see how this conclusion rules out the following conjecture. I don’t claim that this conjecture is correct. In fact, I am asking if it is correct. But so far it still looks to me to be possible. The conjecture is that the field-amplitude operator has eigenstates that have definite values of field amplitude. If the field is prepared in one of these eigenstates, it will not remain in an eigenstate but it is still true that eigenstates have definite values of field amplitude. Is this conjecture correct?

Yes, that's what eigenstates of operators are, by definition. The definite value is the eigenvalue of the operator for that eigenstate. But that doesn't have anything to do with physics, it's a mathematical statement. The physics enters the picture by realizing that there's a particular operator of interest, namely the Hamiltonian and its eigenstates are of particular significance.

Thank you again Stam Nicolis . I think I can figure out how to show that in a finite-dimensional space (in the sense that operators can be represented by finite matrices), every operator (Hermitian or not) has at least one eigenvalue. However, for an infinite-dimensional space (an infinite number of basis vectors spanning the set of state vectors), it is not clear to me that the field-amplitude operator necessarily has any eigenvectors. If it does, just knowing that it does is profound information. If such eigenstates exist, then there exists states having definite values of field amplitude. So another way to ask the original question of this thread is to ask if such states exist. Your conclusion, "Yes, that's what eigenstates of operators are, by definition." assumes existence. But existence is not obvious to me. So let me ask the question this way. Does there necessarily exist eigenstates of the field-amplitude operator?

In that case the relevant distinction is between the finite and infinite dimensional case, not whether one is discussing the field amplitude operator. The same question can and has been posed for the Hamiltonian. In fact the same issues arise, already, in quantum mechanics. So it's misleading to focus on the field operators, if the question is about the mathematics of operators that act on Hilbert spaces. A lot is known about them, e.g. http://web.math.ku.dk/~durhuus/MatFys/MatFys4.pdf and the ``canonical'' reference is the textbook of Reed and Simon, http://www.astrosen.unam.mx/~aceves/Metodos/ebooks/reed_simon1.pdf

For quantum field theory, more particularly, the ``canonical'' reference would be the textbook of Glimm and Jaffe, ``Quantum Physics: A Functional Integral point of view'', this summary: https://www.arthurjaffe.com/Assets/pdf/CQFT.pdf might be useful.

The course, https://www.ias.edu/math/qft hopefully can, also, help.

These references are particularly useful for understanding what are the appropriate questions to actually ask-not all questions one might think about make sense.

Thanks again Stam Nicolis . Maybe I should not want to know the answer to the question that started this thread but I still want to know and I don't believe that an answer has been given yet. However, you provided a number of links where I believe I probably will find the answers. There is a lot to study and it will take me a long time to study it all, but thank you for the links.

Why the field amplitude is not an observable in QFT? Is the particle position an observable in ordinary QM?

A question asked by Jose Gaite is:

"Why the field amplitude is not an observable in QFT? Is the particle position an observable in ordinary QM?"

That sounds to me like a pretty good question and I asked the same question in another thread. The answer that I understand is as follows. We are considering the case in which the field amplitude is a complex number and not a real number (in contrast to the position coordinates of a particle which are real numbers). For this to be an eigenvalue of the field-amplitude operator, the operator cannot be Hermitian (because Hermitian operators have only real eigenvalues). Also, an assertion made in a post in another thread is that if an operator is not Hermitian then the quantity is not observable. This assertion is not obvious to me. (Can't two commuting observables be assembled as the real and imaginary parts of an observable that is not Hermitian?) But given this assertion, we conclude that the field amplitude, which was already found to be non-Hermitian, is not observable.

If anyone has anything more to say (in agreement or disagreement) about the above answer, I am interested because I am trying to understand this stuff myself.

Thus, we agree that a real field amplitude is an observable, presumably. I think this can be extended to a complex field, which just consists of a couple of real fields, assembled in a complex number to better represent the O(2) symmetry.

What you say makes sense to me. I wish some QFT expert would get into this conversation and confirm for sure (instead of "I think" like you and I have to say) whether or not a field amplitude is a couple of real and observable fields assembled into a complex number.

We can not say QFT operator's has a definite value of field amplitude even when the field amplitude is not observable and the eigenvalue is not a real number. Because mathematically if such values exist then they can not be expressed or plotted as per field.From my point of view mathematically you are talking about the roots of a complex number and we can proceed only practically further if something is present to be considered, other wise mathematically we can not.

In KTP - no.

In the IEA - yes.

MEA is a method of energy information analysis.

MEA replaces the wave function with a deterministic probability function (DPF).

FDP when differentiating in time or in space is - a quantum with one period, which has the same frequencies: Planck (amplitude), Prony (attenuation), Fourier (harmonics).

MEA - makes quantum computing closer to natural computing.

I asked for inputs from experts without realizing that we already had that from Stam Nicolis. I was not able to recognize the expertise of his input until I studied some more literature. I think I get it now. I was obsessed with the question (that started this thread) of whether the field-amplitude operator has eigenstates of definite values of field amplitude because I have a personal dislike for discussing any entity before it has been defined. Literature that I read on QFT begins with statements like “The field amplitude is promoted to an operator.” I ask: What does that mean? What does this operator operate on? What are the mapping rules for this operator? In other words, how is this operator defined? If it was true (as I hoped) that this operator has the property that there is a set of basis vectors, eigenvectors of this operator, having definite values of field amplitude (even when a complex number and not a real number), that would provide a definition of the operator, and I would have been satisfied that the operator is defined. But, as Stam pointed out, I have been asking the wrong question. There are other ways to define operators. A textbook that I was just looking at shows that the field-amplitude operator is a superposition of operators that are individually defined by specifying their mapping rules (and are recognized as creation or annihilation operators). This superposition serves as a definition of the field-amplitude operator. So there is a definition that I can understand and I am happy now.

To answer my own question, it may or may not be true that the field-amplitude operator has eigenstates having definite values of field amplitude, but (now that I have a better understanding of what Stam said) I don't care about that anymore.

An eigenvector is a non zero vector that changes at most by a scalar factor when characteristics of linear transformation is applied to it. Geometrically an eigenvector corresponding to a real non zero eigenvalue points in a direction in which it is streched by the transformation and the eigenvalue is the factor by which it is streched. If now field amplitude operator is not Hermitian or real then every eigenvalue should also be non real. So if we still assuming some definite eigenvalue for non real operator can not provide sense in the way how much amplitude is streched in particular direction by particular eigenvector for an assumed eigenvalue.

I'm not sure what your point is Anil Kumar Jain. It is possible for an operator to have a complex eigenvalue. For example, consider the operator

exp(-i t H/h-bar)

where H is a Hermitian operator and therefore necessarily has eigenvalues (that are all real). If E (a real number) is any eigenvalue of H, then the complex number

exp(-i t E/h-bar)

is an eigenvalue of the above operator. But I'm not sure if you are disputing this because I'm not sure what your point is.

Anyway, I think I figured out the answer to the question that started this thread. (Unless I am making a mistake. Am I?) There are example Lagrangians for which the implied (by choice of the Lagrangian) properties of the amplitude operator include the property that it does not commute with its Hermitian conjugate. The focus here is on these kinds of Lagrangians. That the amplitude operator does not commute with its Hermitian conjugate implies that there are no field states having definite values of both, the field amplitude (a complex number) and its complex conjugate. This implies that there are no field states having definite values of both, the real part of the field amplitude and the imaginary part of the field amplitude. This implies that there are no field states that have a definite value of the (complex number) field amplitude. The conclusion, and answer to the question in this thread (if I am not making a mistake), is that the field amplitude operator has no eigenstates for the kinds of Lagrangians considered here.

By the way, the above conclusion is consistent with what Stam Nicolis said a long time ago. The square of the modulus of the field amplitude is observable (which can be shown to be eigenvalues of a Hermitian operator) but the amplitude is not, and it is not very productive to search for eigenstates of this operator. I'm not very smart and it takes me a long time to properly appreciate his inputs.