In the elementary quantum mechanics (QM) of a single particle responding to a given environment, the state of the particle can be specified by specifying a set of commuting (i.e., simultaneously knowable) observables. Examples of observables include energy and angular momentum. Although not simultaneously knowable, other examples include the three rectangular spatial coordinates and the three components of linear momentum. Each observable in QM is a real number and is an eigenvalue of some Hermitian operator. Now consider quantum field theory (QFT) which considers a field instead of a particle. First consider the classical (before introducing QFT operators) description of the state of the field at a selected point in time. This is the field amplitude at every spatial location at the selected time point. For at least some kinds of fields, the field amplitude at a given space-time point is a complex number. Now consider the QFT corresponding to the selected classical example of a field. Is the field amplitude an observable even when it is not a real number? It is not an eigenvalue of any Hermitian operator when not real. So if the field amplitude is an observable, there is no Hermitian operator associated with this observable. My guess (and my question is whether this guess is correct) is that the real and imaginary parts of the field amplitude are simultaneously knowable observables, with a Hermitian operator (assigned to each space-time point) for each. This would at least explain how the field amplitude can be an observable but not real and not have any associated Hermitian operator. Is my guess correct?
Similar as in quantum mechanics, only now they're probabilities per unit time and unit volume. What were modes in quantum mechanics are particles in quantum field theory.
The field amplitude isn't observable, any more than the wavefunction is; it's the modulus squared of the field amplitude that describes the probability density of the number of particles, created by the field, at each point in space and time.
Stam Nicolis said "The field amplitude isn't observable, any more than the wavefunction is; it's the modulus squared of the field amplitude that describes the probability density of the number of particles, created by the field, at each point in space and time."
Thank you Stam. That answers my question.
Hi L.D. Edmonds: At the 1927 Solvay Conference, Bohr asked Dirac what he was working on, and Dirac said, “I’m trying to get a relativistic theory for the electron” And Bohr said, “But Klein has already solved that problem”. Everybody, except Bohr, had already noticed that there is no such thing as a negative probability. Feynman reinterpreted Klein’s negative energy states as positive energy particles moving backwards through time, and antiparticles were properly invented. Negative energy states are positive energy antiparticles. So, now we can answer your question. In QFT, operator valued fields must be Hermitian fields if they are to be observable.
George Soli said: "In QFT, operator valued fields must be Hermitian fields if they are to be observable."
Thank you George. That answers the question I was going to ask next. A question that you anticipated was coming next.
That brings me to the third question. I now see that the operator associated with the field amplitude at a given space-time point is not Hermitian, and the field amplitude is not observable. Can we still say anything about eigenstates (if they exist) and eigenvalues of this operator even though not observable or even real numbers? Specifically, can we say that an eigenstate has a definite value of field amplitude (equal to the eigenvalue) even though it is not an observable or even a real number? If not, what can we say about the eigenstates? Do eigenstates even exist? Or, maybe we don't want to say anything about the eigenstates because the primary significance of this operator is to operate on given states to construct new states. Is that true?
With the original question already answered (thank you Stam Nicolis), and the next question I otherwise would have asked also already answered (thank you George Soli), my attention now is on the question in my latest post. This is a little different than the original question that started this thread so I am starting a new question thread dedicated to the newest question. But I am still paying attention to the present thread because there might be additional interesting inputs from other people.
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