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Units of Fourier expanded field

  1. Feb 17, 2015 #1
    If I write the basic scalar field as $$\phi(x)=\int\frac{d^3k}{(2\pi)^3}\frac1{\sqrt{2E}}\left(ae^{-ik\cdot x}+a^\dagger e^{ik\cdot x}\right),$$ this would seem to imply that the creation and annihilation operators carry mass dimension -3/2. That's the only way I can get the total field ##\phi## to have mass dimension 1, since the ##d^3k## carries three dimensions of mass and the ##1/\sqrt{2E}## factor carries dimension -1/2. This seems contradictory to the fact that ##a^\dagger a## is usually defined as the number operator, which should be dimensionless. Am I missing something here? (My question at first was just how to get the dimensions of the Fourier-expanded field to come out right.)
  2. jcsd
  3. Feb 18, 2015 #2


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    It's all about the normalization of the creation and annihilation operators. Using this convention, your normalization is such that the annihilation and creation operators fulfill
    $$[a(\vec{p}),a^{\dagger}(\vec{q})]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{q}).$$
    Thus ##N(\vec{p})=a^{\dagger}(\vec{p}) a(\vec{p})## are momentum-space number densities, i.e., the dimension for your annihilation and creation operators is, of course, correct since ##N(\vec{p})## must be of mass dimension -3.

    Then you directly have for the total energy, e.g.,
    $$H=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \sqrt{m^2+\vec{p}^2} N(\vec{p}).$$

    Note that there are many different conventions in the literature concerning the norm of the annihilation and creation operators. The unanimous commutation relation is the equal-time commutation relation of the field operators,
    $$[\phi(t,\vec{x}),\dot{\phi}(t,\vec{x})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{p}).$$
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