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Why does this integral go to zero?

  1. Apr 28, 2010 #1
    Why does this integral go to zero? Where alpha+ is a creation operator?

    [tex]\int \frac{d^3k}{w_k} k^j \left(\alpha^+ (\underline{k}) \alpha^+ (-\underline{k})e^{2iw_k x^0}\right)[/tex]
  2. jcsd
  3. Apr 29, 2010 #2
    odd integrand
  4. Apr 29, 2010 #3
    Sorry - stupid question, why is it an odd integrand? If it is an odd integrand, alpha(+k) and alpha(-k) are either odd and even (or even and odd) or both odd. An odd function has the property that f(-x)=-f(x) and an even function has the property that f(-x)=f(x) - why must the creation operators have these properties? As I understand it, they are just coefficients of the general expansion of the field, so they should be totally arbitrary..
  5. Apr 29, 2010 #4
    as far as I see, there is a k in front of the parenthesis..
  6. Apr 29, 2010 #5
    yes but that k is just a constant. We are integrating wrt vector k, I think.

    EDIT: In anycase, if the k was a variable in the integration, why would this make the integral zero? (sorry if this is a stupid question)..
  7. Apr 29, 2010 #6
    why isn't it the j-th component? j = 1,2,3

    d^3k = dk1 dk2 dk3

    so the integral is a 3-vector relation.

    where did you find this expression?
  8. Apr 29, 2010 #7
    Oh okay, that makes sense.

    The original question itself is this:


    [tex]\int d^3.\underline{x}T^{0j}(x)[/tex]

    in terms of creation and annihilation operators.

    Basically, when I expand it out (very long), I get two integrals like the one in the OP - which should go to zero according to the solutions I've been given... I just dont understand why.
  9. Apr 29, 2010 #8
    well the integral you posted first is odd, so it is zero

    and from

    \int d^3.\underline{x}T^{0j}(x)

    you that it is a j0 tensor which is what you have posted in your post. i.e it SHOULD depend on the j-th coordinate
    Last edited: Apr 29, 2010
  10. Apr 29, 2010 #9
    Yes it should depend on the jth coordinate.

    Sorry I'm sure this is a really stupid question, but would I be right in saying that the alpha functions must both be even (or one even one odd, or both odd), for the integral to be zero (as evenxevenxodd=odd (the standalone k is odd).

    Why must this be the case? The alpha functions are Fourier coefficients of a general field, why can we impose such a constraint to these functions?
    Last edited: Apr 29, 2010
  11. Apr 29, 2010 #10
    you see that the integral is zero since it is odd due to the k^j factor

    Also you have a(-x)a(x) which is even under x -> -x

    so it doesen't matter if a(x) is odd or even under x-> -x, since they come in this combination.
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