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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?

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

1 Answer 1

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I know that the creation and annihilation operator need to have dimension 1, since they related with one-particle state.

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 $-3$.

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