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Just wondering where the normalisation factor $\frac{1}{\sqrt{2 \omega_{\mathrm{p}}}}$ comes from in this field operator expression?

\begin{equation} \phi(\mathbf{x})=\int \frac{d^{3} p}{\left(2 \pi^{3}\right)} \frac{1}{\sqrt{2 \omega_{\mathrm{p}}}}\left(a_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x}}+a_{\mathbf{p}}^{\dagger} e^{-i \mathbf{p} \cdot \mathbf{x}}\right) \end{equation}

Jolly Penguins
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1 Answers1

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This also bothered me when I was learning QFT. My current understanding is that this $1\over\sqrt{2\omega_p}$ is not important. The reasons why books are written in this I guess are as following.

  1. People used to work with harmonic oscillators, where they set $$x=\frac{1}{\sqrt{2\omega}}(a+a^\dagger).$$ Remember that QFT is merely many harmonic oscillators with different $\omega$ summed together. So they chose this historical convention.

  2. This factor can simplify the commutator of $a$ and $a^\dagger$ $$[a_{p},a_{p'}^\dagger]=(2\pi)^3\delta^3(p-p').$$ This makes $a$ and $a^\dagger$ more suitable for the name creation/annihilation operators.

You need not stick to this convention. Some books, for example, string theory textbooks, do not adopt this convention. And you will get a slightly different version of commutators.

Youran
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