# Relativistic normalisation of delta functions

So in general with a Fourier transform you have something like:

$$\int d^3x\ e^{i\vec{q}\cdot\vec{x}}\ e^{-i\vec{p}\cdot\vec{x}} = \int d^3x\ e^{-i(\vec{p}-\vec{q})\cdot\vec{x}} = (2\pi)^3\ \delta^3(\vec{p}-\vec{q}). \tag{1}$$

So then you add a $\frac{1}{(2\pi)^3}$ when you do the inverse so you get:

$$\int \frac{d^3\vec{p}}{(2\pi)^3}\ (2\pi)^3\ \delta^3(\vec{p}-\vec{q})\ e^{i\vec{p}\cdot\vec{x}} = e^{i\vec{q}\cdot\vec{x}}\tag{2}$$

as desired.

However, in texts dealing with relativistic physics (particularly QFT) the Lorentz invariant phase space element is given by:

$$\frac{d^3\vec{p}}{(2\pi)^3} \frac{1}{2E_p}.\tag{3}$$

This is fine, and I understand how to derive this. They then say that the correct normalisation for the delta function is:

$$2E_p\delta^3(\vec{p}-\vec{q})\tag{4}$$

which makes sense because it is Lorentz invariant etc.

My problem is what should I do when presented with an integral like:

$$\int d^3x\ e^{-i(\vec{p}-\vec{q})\cdot\vec{x}}\tag{5}$$

in a relativistic context (i.e., $\vec{p}$ and $\vec{q}$ are the 3D part of 4-momenta bound by the relativistic dispersion relation which is what gives rise to the extra term in the phase space element). Does it evaluate as normal:

$$\int d^3x\ e^{-i(\vec{p}-\vec{q})\cdot\vec{x}} \overset{?}{=} (2\pi)^3\ \delta^3(\vec{p}-\vec{q})\tag{6}$$

or does it somehow invent an extra $2E_p$:

$$\int d^3x\ e^{-i(\vec{p}-\vec{q})\cdot\vec{x}} \overset{?}{=} (2\pi)^3\ 2E_p\ \delta^3(\vec{p}-\vec{q})\tag{7}$$

The former feels correct mathematically (why would the mathematics care about the wider context), but then what's the point of the whole hoo-ha about normalising delta functions with a $2E_p$? Is that literally just for defining the CCRs of annihilation and creation operators so they cancel nicely with the phase space normalisation?

N.B. I'm not in any way concerned about the normalisation for the phase space volume that makes total sense.

The "corrected" delta function is Lorentz invariant. This means that if $p$ and $q$ are related by a Lorentz transformation then

$$\delta(\vec p) \neq \delta(\vec q)$$

but

$$E_p\, \delta(\vec p) = E_q\, \delta(\vec q)$$

as can be seen by considering the integral

$$\int \frac{d^3p}{E_p} E_p\, \delta^3(\vec p) f(p)$$

with $f$ any scalar function, and using the fact that the measure is invariant.

This does not mean that the mathematical identity is somehow changed. Math doesn't care about your physics: the integral of just an exponential is just a delta function. We put the $E_p$ in the delta function because we also want to put it in our integrals to make the 3D measure invariant, but if you end up with an integral without the $E_p$, it's not going to magically appear.