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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?
##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?