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