# Units of Fourier expanded field

1. Feb 17, 2015

### copernicus1

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. Feb 18, 2015

### vanhees71

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,
$$[\phi(t,\vec{x}),\dot{\phi}(t,\vec{x})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{p}).$$