# Units of Fourier expanded field

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.)

Related Quantum Physics News on Phys.org
vanhees71
Gold Member
2019 Award
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,
$$[\phi(t,\vec{x}),\dot{\phi}(t,\vec{x})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{p}).$$