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In this answer, Lubos explains that in quantum field theory there are linear hermitian operators representing observables.

Quantum field theories are a subset of quantum mechanical theories. So they obey all postulates of quantum mechanics, they have Hilbert space, linear Hermitian operators i.e. observables, obey the superposition principles, calculate probabilities from squared absolute values of complex amplitudes, and so on.

  1. In QM, all dynamical variables position, momentum, angular momentum etc are also hermitian operators and observables.

But in QFT, all dynamical variables are not hermitian. While quantizing a classical field theory, it is not clear which variables are promoted to hermitian operators and which are not. For example, for complex scalar field, the field operator $\hat\phi(\textbf{x},t)$ is not hermitian but the Hamiltonian is. The number operator is hermitian. What is the general rule? Which objects are promoted to hermitian operator?

  1. What are observables in QFT? Should we say the Hamiltonian or the Number operator etc are the observables or the scattering amplitudes, decay rates etc as the observables? If the latter, then they are not associated with operators of any kind.

  2. In quantum mechanics, there are various hermitian operators which forms basis in the Hilbert space. But in QFT, the number operators seems to be the only operator whose eigenstates from a basis (that too in free theory). Is that correct?

  3. Can the field momentum considered to be hermitian (and observable) and therefore, its eigenstates forming a continuous basis like in QM?

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  • $\begingroup$ Note: There is no such thing as a "classical complex scalar field". Conceiving of two real scalar fields as one complex field is a reformulation that makes the theroetical treatment easier, but it does not change the fact that the "fundamental" dynamical variables are real-valued fields. This subtlety in the quantization of complex scalar fields is also discussed here. $\endgroup$ – ACuriousMind Jan 4 '17 at 18:00
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What is the general rule? Which objects are promoted to hermitian operator?

The general rule is that if the classical variable is real, the quantum operator is hermitian.

What are observables in QFT? Should we say the Hamiltonian or the Number operator etc are the observables or the scattering amplitudes, decay rates etc as the observables?

If by observable you mean "something that can be observed", then the latter are observables while the former are not. But in general, when one speaks of observable in the context of QM, then one usually means "an hermitian operator", in which case the former are observables and the latter are not.

If the latter, then they are not associated with operators of any kind.

Scattering amplitudes (of which the decay rate is just an example) are associated to the $S$ matrix operator, which is not hermitian but unitary.

In quantum mechanics, there are various hermitian operators which forms basis in the Hilbert space. But in QFT, the number operators seems to be the only operator whose eigenstates from a basis (that too in free theory). Is that correct?

No, it is not correct. In general the complete set of commuting operators is formed by the generators of translations $P^\mu$, the Casimir operators of the Poincaré Algebra, the generators of internal symmetries (e.g., the charge operator, which is just the number operator with a funny normalisation), etc. That this set is finite is one of the working hypothesis of any axiomatic approach to QFT.

Can the field momentum considered to be hermitian (and observable) and therefore, its eigenstates forming a continuous basis like in QM?

If by field momentum you mean $P^\mu$, then yes (as in the previous item). If you mean $\pi(x)$, then in general no (see Physical interpretation of canonical conjugate momenta in field theory) for more details).

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    $\begingroup$ Your second-to-last paragraph strikes me as odd (because I don't understand what "there are various hermitian operators which forms basis in the Hilbert space" is supposed to mean to begin with). Are you trying to say that the operators you list form a complete set of commuting observables, i.e. their joint eigenstates yield an eigenbasis with no degeneracy? $\endgroup$ – ACuriousMind Jan 4 '17 at 18:02
  • $\begingroup$ @ACuriousMind yes, that's exactly what I meant. $\endgroup$ – AccidentalFourierTransform Jan 4 '17 at 18:03
  • $\begingroup$ @AccidentalFourierTransform The reason why operators are hermitian in QM is that their eigenvalues are real. Any measurable quantity should be real. But in QFT, we see something which is not measurable, is also hermitian (for some vague reason). $\endgroup$ – SRS Jan 4 '17 at 18:18

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