# Interpretation of the field operator

1. Apr 17, 2009

### martin_blckrs

Hi,

I've just finished a 2 semester course on Quantum Mechanics and I am now eager to learn some Quantum Field Theory. I started to read some books on my own, but I have some problems understanding a few concepts.

My main problem is this:
In the Canonical Quantization procedure, we take some classical field and make the (infinite number of) positions and momenta into operators and postulate the commutation relations between them. Now if we, for example, take the real scalar field, we obtain the field operators $$\Phi(x)$$. Since the field is real, these operators are all Hermitian and thus should correspond to an observable (by QM). The questions are: How should I interpret this observable? What does the expactation value of this operator tell me? Can I view the operator as a creation operator of a particle with a given position (in space-time)?

2. Apr 17, 2009

### meopemuk

I will tell you my opinion, which is not widely shared by others (unfortunately): there is no any physical interpretation for fields operators, they are just formal mathematical objects.

The main purpose of quantum field theory is to provide description of physical processes, in which the number of particles can change. So, in contrast to ordinary quantum mechanics, QFT should have interaction Hamiltonians, which do not commute with particle number operators. In principle, you can construct such Hamiltonians without using quantum fields, e.g., as abstract matrices acting in the Fock space. Quantum fields and creation/annihilation operators simply provide a convenient mathematical tool for writing such interaction operators in a compact form. Quantum fields are particular linear combinations of creation/annihilation operators, which have simple Lorentz transformation properties. As discussed in Weinberg's textbook, these properties guarantee that you can build relativistically invariant Hamiltonians simply as products of few field operators. This is the only reason why quantum fields are useful. After you constructed the interacting Hamiltonian of your theory (as a product of fields) and proved its relativistic invariance, you don't need quantum fields anymore. You can do quantum mechanics with this Hamiltonian in a usual way and obtain, for example, amplitudes for transitions between asymptotic states with different number of particles (the S-matrix elements), or energies of bound states.

In conclusion, quantum fields are just formal mathematical symbols that allow one to simplify the notation considerably. Do not try to find physical counterparts to quantum fields. The more you try, the more confused you'll be.

3. Apr 18, 2009

### martin_blckrs

OK, but then another question: Why do we interpret the vacuum expactation value as the amplitude of the scattering process?

4. Apr 18, 2009

### ExactlySolved

In single particle quantum mechanics the amplitude of a scattering process is the kernel of the time evolution operator i.e. the Green's function for the operator exp(-i t H). To see the connection between Green's functions and expectation values you might want to read about the general concept of a n-point correlation function, which are ultimately the solutions that field theorists would like to compute.

This article does not seem too well written, but it might get you started:

http://en.wikipedia.org/wiki/Green's_function_(many-body_theory [Broken])

Last edited by a moderator: May 4, 2017
5. Apr 18, 2009

### xepma

Probably the best thing to keep in mind while reading on QFT is that in the end, it's all about the correlation functions. In the end, everything is aimed at determining these fellas.