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Divergence of integral over vacuum energies (Free field)

  1. Aug 24, 2015 #1

    The Hamiltonian for the free scalar field, expressed in terms of the creation/annihilation operators, is

    [itex] H = \int d^{3}p [\omega_p a^{\dagger}_p a_p + \frac{1}{2}\omega_p \delta^{3}(0)] \hspace{3mm} [/itex]

    I thought: [itex] \omega_p [/itex] is a function of p as [itex] \omega^{2}_p = |p|^{2} + m^2 [/itex] and so the dirac delta will sift out the value of [itex] \omega_p [/itex] at [itex] p = 0 [/itex]. Could someone tell me why this statement is incorrect? I think I've made some significant conceptual error. Is the first term divergent for infinite p as well?

    P.S. wherever p appears above, it is to be taken as the 3-momentum
    Last edited: Aug 24, 2015
  2. jcsd
  3. Aug 24, 2015 #2


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    What you expect would happen if the Dirac delta was ## \delta^3(p) ##. But ## \delta^3(0) ## is simply equal to infinity, or more precisely, the volume of whole space, which is a constant.
  4. Aug 24, 2015 #3
    Ah, of course. Thanks, that was silly of me.
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