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Perplexing commutator

  1. Apr 10, 2009 #1
    When simplifying this

    [tex] \int d^3x' [\pi(x), \frac{1}{2}\pi^2(x') + \frac{1}{2} \phi(x')( -\nabla^2 + m^2)\phi(x')] [/tex]

    we know that

    [tex] [\pi(x), \pi(x')] = 0 [/tex]

    [tex] [\phi(x), \pi(x')] = -i\delta(x-x') [/tex]

    how does that simplify to

    [tex] \int d^3x' \delta(x-x')( -\nabla^2 + m^2)\phi(x')[/tex]

    I know that

    [tex] [\pi(x), \pi^2(x')] = 0 [/tex]

    but not sure how does the laplacian gets factored out like that and one-half disappears?
    Last edited: Apr 10, 2009
  2. jcsd
  3. Apr 10, 2009 #2


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    Have you tried to see what part integration brings you ?
  4. Apr 11, 2009 #3
    By expanding the commutator we get,

    \frac{1}{2} \int d^3x' \pi(x)\phi(x')( -\nabla^2 + m^2)\phi(x') - \phi(x')( -\nabla^2 + m^2)\phi(x')\pi(x)

    the laplacian in the second term is sandwiched between two phis, integration by part doesn't seem to help to factor it out.
  5. Apr 12, 2009 #4
    I think you're better off using [tex][A,BC] = [A,B]C + B[A,C][/tex] and then integration by parts. Also, it's a good time in your life to realize that a commutator behaves very much like a functional derivative (recall QM 101).
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