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Regarding intuition of QFT

  1. Aug 27, 2013 #1
    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:
    [tex]\phi \mid \psi \rangle = \psi(x) \mid \psi \rangle[/tex]
    and the following according to the second:
    [tex]\phi \mid \psi \rangle = \kappa \mid \phi, \psi \rangle.[/tex]

    Is the first interpretation wrong or can they be fused togheter in some way?
  2. jcsd
  3. Aug 27, 2013 #2


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    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:

  4. Aug 27, 2013 #3
    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
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