# Completeness relation

1. Oct 17, 2015

### 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.

2. Oct 17, 2015

### Orodruin

Staff Emeritus
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.

3. Oct 17, 2015

### 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 $1 = \int d^{3}p |p\rangle \langle p|$?

4. Oct 18, 2015

### Orodruin

Staff Emeritus
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: Oct 18, 2015
5. Oct 18, 2015

### 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 $p=p'$?

6. Oct 18, 2015

### 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.

7. Oct 18, 2015

### Orodruin

Staff Emeritus
Neither the states $|p\rangle$ nor $|x\rangle$ are normalisable.

8. Oct 20, 2015

Thanks!