# Completeness relation

I have dug several resources from the internet, but none happen to explain the following formula:

##1 = \int \frac{dp}{(2\pi)^{3}} |\vec{p}><\vec{p}|##

I have done basic quantum mechanics, so I know that this is the completeness relation. Also, I understand that an integral is being taken over all the momentum states. That's all fine by me.

What's tripping me up is the factor of ##(2\pi)^{3}##.

Can someone explain where they come from. I've thinking it has to do with Fourier analysis or something because the same funny factor appears when Fourier transform into momentum space.

## Answers and Replies

Orodruin
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Indeed it is related to the Fourier transform. The state ##|p\rangle## is the state normalised such that ##\langle x|p\rangle = e^{-ip\cdot x}##. Using ##\langle x|x'\rangle = \delta(x-x')## and the Fourier transform of the delta function, you will find the correct normalisation of the completeness relation.

• vanhees71 and bhobba
So, do you mean that if ##|p\rangle## were the state normalised such that ##\langle x|p\rangle = \frac{1}{(2\pi)^{3/2}} e^{ip\cdot x}##, then the correct normalisation of the completeness relation is ##1 = \int d^{3}p |p\rangle \langle p|##?

Orodruin
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Yes. If you construct the state like that, the completeness relation would not have the factors of 2pi. The states would then be normalised such that ##\langle p|p'\rangle = \delta(p-p')##.

Last edited:
The orthonormality condition ##\langle p|p'\rangle = \delta(p-p')## looks more natural than ##\langle p|p'\rangle = 2 \pi \delta(p-p')##.

Don't the factors of ##2 \pi## violate the normalisation condition [the probability adding up to one] when ##p=p'##?

vanhees71
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That's convention regarding the fourier transformation. In the high-energy physics community one usually uses the convention
$$\tilde{f}(\omega,\vec{p})=\int_{\mathbb{R}^4} \mathrm{d}^4 x \exp(\mathrm{i} \omega t-\mathrm{i} \vec{p}\cdot \vec{x}),$$
$$f(t,\vec{x})=\int_{\mathbb{R}^4} \mathrm{d}^4 p \frac{1}{(2 \pi)^4} \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{p} \cdot \vec{x}),$$
i.e., one lumps all ##2 \pi## factors to the momentum measure.

Orodruin
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The orthonormality condition ##\langle p|p'\rangle = \delta(p-p')## looks more natural than ##\langle p|p'\rangle = 2 \pi \delta(p-p')##.

Don't the factors of ##2 \pi## violate the normalisation condition [the probability adding up to one] when ##p=p'##?

Neither the states ##|p\rangle## nor ##|x\rangle## are normalisable.

Thanks!