What is the significance of the factor (2π)^3 in the completeness relation?

[spaghetti3451]

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.

 

[Orodruin]

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.

 

[spaghetti3451]

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]

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 $\langle p|p'\rangle = \delta(p-p')$.

 

[spaghetti3451]

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]

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]

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.

 

[spaghetti3451]

Thanks!