# Regarding intuition of QFT

1. Aug 27, 2013

### Kontilera

Hello!
I'm finally starting to get a grip around quantum field theory. The last hang up is the following:
I've been told that since we are quantizing a field, the field strength is the observable. Now analogous to QM we then define a field of hermitian operators, $\phi(x)$, which give a hermitian operator in every point in spacetime (i.e. for every observable). However in most books the field $\phi(x)$ is introduces as a excitation/creation field. These two interpretations seems to be contradicitve.
If say $\phi$ is nonzero only at a point x (lets forget about smoothness conditions), and we act on a state with it.. it seems to me that we should expect the following result according to the first definition:
$$\phi \mid \psi \rangle = \psi(x) \mid \psi \rangle$$
and the following according to the second:
$$\phi \mid \psi \rangle = \kappa \mid \phi, \psi \rangle.$$

Is the first interpretation wrong or can they be fused togheter in some way?

2. Aug 27, 2013

### vanhees71

You just choose different bases. With the annihilation and creation operators wrt. the single-particle momentum eigenstates you create the Fock basis (occupation-number basis), i.e., states with determined particle number. The other basis is overcomplete and the eigenstates of the fields. The latter you use to establish the path-integral. For further explanations, see my QFT manuscript:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

3. Aug 27, 2013

### The_Duck

It's just like the harmonic oscillator in regular QM. There the main observable is $x$, but we can write $x$ as a sum of creation and annihilation operators: schematically $x = a + a^\dagger$, ignoring normalizations.

In the QM of the harmonic oscillator, there are two useful bases for the Hilbert space. There's the position basis, in which $x$ is diagonal. Then there's the Fock basis, in which the Hamiltonian is diagonal and $a$ and $a^\dagger$ are seen to be lowering and raising operators.

Similarly in QFT you can consider two bases. In one the field operator $\phi(x)$ is diagonal. This basis is not too useful, though. The more useful basis is the Fock basis, consisting of states of definite numbers of particles each with definite momentum.

Last edited: Aug 27, 2013