# A Normalized Differential Scattering Probability

1. May 5, 2016

### spaghetti3451

The normalized differential quantum-field-theoretic probability $dP$ of scattering is given by

$dP=\frac{|\langle f |S|i\rangle|^{2}}{\langle f|f\rangle\langle i|i\rangle}d\Pi,$

where $|i\rangle$ is the initial state, $|f\rangle$ is the final state, $\langle f|S|i\rangle$ are the elements (in the basis of the $|i\rangle$ and $|f\rangle$ states) of the $S$-matrix and $d\Pi$ is the region of final state momenta at which we are looking.

The goal of this post is to investigate the normalization of $dP$ and the form of the region $d\Pi$ of final state momenta.

The $\langle f | f \rangle$ and $\langle i | i \rangle$ in the denominator come from the fact that the one-particle states, defined at fixed time, may not be normalized to $\langle f | f \rangle = \langle i | i \rangle = 1$. In fact, such a convention would not be Lorentz invariant.

Also, there are two constraints on the form of $d\Pi$:

1. $d\Pi$ must be proportional to the product of the differential momentum, $d^{3}p_{j}$, of each final state.

2. $d\Pi$ must integrate to $1$.

Therefore, $d\Pi$ must be of the form

d\Pi=\prod\limits_{j}\frac{V}{(2\pi)^{3}}d^{3}p_{j}.

My Question:

1. I understand that $\langle p | p \rangle = (2\pi)^{3}(2\omega_{p})\delta^{3}(0)$ so that $\langle p | p \rangle \neq 1$. But why does the fact that $|p\rangle$ is not unit-normalized mean that $\langle f | f \rangle$ and $\langle i | i \rangle$ must be in the denominator of the formula for $dP$?

1. In order for $\int d\Pi = 1$, I notice that $\int \frac{dp}{2\pi}=\frac{1}{L}$ via dimensional analysis and a $2\pi$ convention. Can you please explain how dimensional analysis and a $2\pi$ convention lead us to $\int \frac{dp}{2\pi}=\frac{1}{L}$? Is this a heuristic derivation?

2. The normalization $\int d\Pi = 1$ is also the natural continuum limit of having discrete points $x_{i}=\frac{i}{N}L$ and wavenumbers $p_{i}=\frac{2\pi}{L}\frac{i}{N}$ with $i=1, \dots , N$. Can you please explain in detail?

2. May 10, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?