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Is the delta in the commutation relations of QFT a dirac delta or a kronecker?

  1. Jun 18, 2012 #1
    If it's a dirac delta doesn't it mean it's infinite when x=y? Or is it a sort of kronecker where it's equal to one but the indices x and y are continuous? I'm confused.
  2. jcsd
  3. Jun 19, 2012 #2


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    Both the Dirac and Kronecker deltas can occur in QFT.

    Post a specific example of a CR you don't understand...
  4. Jun 19, 2012 #3


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    In QFT you can have commutation relations like

    [\phi^i (x), \pi_j (y)] = \delta^_i \delta^{(D)}(x-y)

    where phi and pi are conjugate fields, of which you can have, say, N, and D is the dimension of spacetime.

    The first delta states that these fields don't commutate if they are different components. That's a Kronecker delta. The second is a distribution, and is there because the fields are really distributions. These kind of relations only make sense if you integrate them with a test function. Otherwise you would naively say that the commutator blows up if x=y.

    You can compare it with the commutators in QM; those only make sense if you apply them to a wave function.
  5. Jun 20, 2012 #4
    Are all operators distributions then, and not just commutators? In QM, X and P are distributions. In QFT, creation/annihilation operators and fields are distributions?
  6. Jun 20, 2012 #5
    Could you elaborate on this?
  7. Jun 20, 2012 #6
    I was just asking haushofer a question. He said fields in QFT are distributions, and I was wondering why. He said you integrate them over test functions, so I was wondering if he meant that any linear operator is a distribution in QM, e.g.,

    <x|P|x'> takes ψ(x') and maps it to:

    <x|P|ψ>=-i d/dx[ψ(x)]

    I have no idea what he meant, so I just wanted to learn :)
  8. Jun 21, 2012 #7


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    In QFT fields are operator-valued distributions where (in position space) x is a "continuous index". The delta-distribution is the identity-operator.

    In QM the operators are X (and P, ...) - no fields - and there is only a discrete index (i=1,2,3, ... for dimensions etc.) and therefore you get a Kronecker delta as identity-operator. The delta-function in QM is not an operator but a matrix element.
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