# Dimensional analysis of quantized Klein-Gordon Field

For the free Klein-Gordon Lagrangian density: $$\mathcal{L}=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu} \phi-m^2\phi^2 .$$ Since we need the dimension of Lagrangian density equal to 4 (in this case action dimension $$[S]=0$$ in 4D spacetime).

And the kinetic term usually have a large contribution, we infer from $$\partial^{\mu}\phi\partial_{\mu}\phi$$ that the dimension of $$\phi$$ is 1.

If we quantize $$\phi$$, (as in Peskin and Schroder's book) $$\phi(x)=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\mathbf{p}}}}\left(a_{\mathbf{p}} e^{-i p \cdot x}+a_{\mathbf{p}}^{\dagger} e^{i p \cdot x}\right).\tag{2.25/2.47}$$ I know that the creation and annihilation operator need to have dimension 1, since they related with one-particle state. This means that the dimension of $$\left[\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\mathbf{p}}}}\right]=0$$ From non-trivial thinking $$E_{\mathbf{p}}$$ should have dimension 1, so where is my problem?

• What do you mean by "Since we need the dimension of Lagrangian equal to 4"? Lagrangian is a scalar function Sep 19, 2022 at 6:19
• @basics as in the dimension of the Lagrangian is $m^4$. Sep 19, 2022 at 6:38
• I know that the creation and annihilation operator need to have dimension 1. Not so fast. Sep 19, 2022 at 6:41

I don't know what rule you think you're using, but this is wrong. Each ladder operator has dimension $$-\frac32$$ because$$[a_\mathbf{p},\,a^\dagger_\mathbf{q}]=(2\pi)^3\delta^3(\mathbf{p}-\mathbf{q})$$is of dimension $$-3$$.