- #1

- 1,344

- 33

##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?