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QFT of Klein Gordon fields

  1. Dec 18, 2014 #1
    I was reading about the Klein Gordon equation of scalar fields. I notice that the hamiltonian is not Hermitian:

    0(Φ,π)T = matrix((0,1),(-p2,0)) (Φ,π)T

    The Hamiltonian operator iH = matrix((0,1),(-p2,0)) is not a hermitian matrix.

    What does this mean? Does this mean Klein Gordon fields don't conserve particle number, or that this is not Unitary? Is a quantum field theory with just Klein Gordon fields not valid?
     
  2. jcsd
  3. Dec 18, 2014 #2

    vanhees71

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    The point is that in relativistic physics the idea of a single-particle wave function as in nonrelativistic wave mechanics does not lead to a sensible quantum theory of relativistic particles. The reason is that collisions of interacting particles at relativistic collision energies lead to production and destruction of other particles, i.e., the relativistic quantum theory should always be formulated as a many-body theory. The best class of models, very successful physics wise (although not understood completely in a strict mathematical sense), are local relaivistic quantum-field theories.

    The idea of "canonical quantization" is to make the fields to operators and use the Lagrange formalism to determine the expressions for the quantities as defined by Noether's theorem from Poincare invariance (translations in time and space give energy and momenum, rotations angular momentum, and boosts center-momentum motion).

    For the free real Klein-Gordon field the Lagrangian reads
    [tex]\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2.[/tex]
    Now you write (in this case self-adjoint) field operators for ##\phi##. The Hamilton density is
    [tex]\mathcal{H}=\Pi \dot{\phi}-\mathcal{L}, \quad \Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}\; \Rightarrow \; \mathcal{H}=\frac{\Pi^2}{2} + \frac{(\vec{\nabla}\Phi)^2}{2} + \frac{m^2}{2} \phi^2.[/tex]
    As you see this is (from a naive point of view) a positive semidefinite self-adjoint operator, and so is the Hamiltonian
    [tex]H=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathcal{H}.[/tex]
    So as a many-body theory the Klein-Gordon equation immediately makes sense.

    There's a subtle point here, however. Even in the free case, the Hamiltonian as written above is ill-defined, and a naive mode-decomposition reveals that it's actually divergent. The reason is that we multiply operator-valued distributions at one space-time point, which is not a well-defined procedure. To cure this one introduces normal ordering using the mode decomposition of the free field. Normal ordering means to bring all creation operators to the left of all annihilation operators in this mode decomposition.

    Now, the normal-ordering process depends on whether you try to quantize the field as bosons or as fermions. In the fermion case, when you exchange annihilation and creation operators to bring them in normal order, this gives an additional sign, while in the boson case it doesn't. So it makes a difference whether you quantize the KG field as bosons or fermions. It turns out that only the boson choice leads to a positive semidefinite Hamiltonian. That reflects a general theorem: Fields of half-integer spin have to quantized as bosons, those of integer spin as fermions. The KG field has spin 0 and thus must be quantized as bosons in order to have a positive semidefinite Hamiltonian as you want to have to guarantee that the ground state ("vacuum", i.e., no particles present at all) of lowest energy exists. The single-particle energy-momentum eigenstates states describe particles with the usual dispersion relation ##E=+\sqrt{m^2+\vec{p}^2}##. That's why this formalism describes spin-0 bosons.

    Sine we've chosen a real field to begin with, we have only one sort of particles here, no antiparticles. This is called a "strictly neutral" particle. The Lagrangian doesn't admit the definition of a conserved charge. In this sense it's "neutral".

    You can as well quantize the complex Klein-Gordon field. Then you get particles and antiparticles, and the Lagrangian admits the definition of a conserved. Again you need normal ordering, and you have to quantize the field as bosons. This again leads to a positive semidefinite Hamiltonian and to antiparticles that carry the opposite charge quantum number than particles.

    Examples for applications of this formalism are effective models for neutral and charged pions.
     
  4. Dec 18, 2014 #3
    I kind of get it.

    So [itex]e^{iH[Φ,π]}[/itex] would presumably act on some complex Fock space for many particles? Which is Unitary because H is real? But there's not equivalent of the shrodinger equation as such for a single particle. I suppose actually the term π∂0Φ represents in a Feynman diagram of a π particle changing into a Φ particle so the particle number is not conserved for the individual Φ and π fields? Is that right? Although I can't see how you'd ever end up with more than one π or Φ. Unless the set of states is just |>,|Φ> and |Ψ>????

    The final piece I'm unsure of is how you get from that to:

    [itex]G(x,y) = \int e^{i S[Φ]}Φ(x)Φ(y)[/itex]D[Φ]

    In other words how do you get from the fock/hilbert space representation to the Green's function representation? I suppose if it was that simple to explain Feynman wouldn't have got a Nobel prize for it. :w I read somewhere that Feynman never actually learned the creation/annihilation operator formalism because he couldn't get his head round how a particle could be created!

    On a second unrelated note. What's the big deal with describing everything as Unitary? exp(iS) where S is just a scalar belongs to the group U(1) but is also identical to O(2). Therefore couldn't we just as well call everything orthogonal? Just a thought.
     
    Last edited: Dec 18, 2014
  5. Dec 18, 2014 #4
    Because these are operators acting on quantum states which can have complex amplitudes. On a complex vector space, unitarity, not orthogonality, is required for things to stay normalized.
     
  6. Dec 18, 2014 #5
    True but I just meant that a 1 dimensional complex space is equivalent to a 2 dimensional real space. If all amplitudes are complex numbers, equivalently all amplitudes are pairs of real numbers. A 1 dimensional unitary operator iH is equivalent to a 2 dimensional orthogonal operator ((0,H),(-H,0)). Just a thought. Not really important.Just an equivalence between U(1) and O(2). Thus the unitary operator exp(iH) is equivalent to the orthogonal operator exp( ((0,H),(-H,0)) ). Hence the terminology "Unitary" should be equivalent to "Orthogonal over a 2D real space". It's just replacing complex amplitudes (A+iB) with pairs of real numbers (A,B).
     
  7. Dec 18, 2014 #6

    atyy

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    He might have learnt it later, but the first evidence for the equivalence of the operator picture and the path integral calculations were shown by Dyson. The rigourous equivalence for QFT came much later with things like the Osterwalder-Schrader conditions. (This is only really helpful for interacting fields. It shouldn't be necessary to make life so complicated for the free KG field that you asked about, but it should still work.)

    http://en.wikipedia.org/wiki/Dyson_series
    http://bolvan.ph.utexas.edu/~vadim/Classes/2011f/dyson.pdf
    http://www-tkm.physik.uni-karlsruhe.de/~carr/documents/MPAGS-QFT-II.pdf
     
  8. Dec 18, 2014 #7
    I'm not sure what you're trying to say, but this doesn't really sound correct, unless you mean this in some specific sense. For example, O(n,R) has two connected components, U(n) is connected.
     
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