Normalized Differential Scattering Probability

[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?

 

[eljose79]

1. The reason why $\langle f | f \rangle$ and $\langle i | i \rangle$ must be in the denominator is because they are the inner products of the final and initial states with themselves, respectively. In quantum mechanics, the probability of finding a particle in a particular state is given by the square of the inner product of that state with itself. In this case, we are looking at the probability of scattering, so the denominator includes the inner products of the initial and final states with themselves.

2. The reason why $\int \frac{dp}{2\pi}=\frac{1}{L}$ is due to dimensional analysis. In quantum mechanics, the momentum operator is given by $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$, where $\hbar$ is the reduced Planck's constant. The dimensions of momentum are $[p]=\frac{[L]}{[T]}$, where $[L]$ is the dimension of length and $[T]$ is the dimension of time. In order for the integral to have the correct units, we need to have a factor of $\frac{1}{L}$ in front of the integration variable. The $2\pi$ comes from the fact that we are using a convention where the momentum eigenstates are normalized as $\langle p | p \rangle = (2\pi)^3(2\omega_p)\delta^3(0)$, which leads to the factor of $2\pi$ in the integration.

3. The continuum limit of having discrete points and wavenumbers can be understood as follows. In quantum mechanics, we can represent a state as a superposition of discrete momentum eigenstates. In the continuum limit, we have an infinite number of these eigenstates, which can be represented as a continuous function of momentum. The integral over momentum then becomes a sum over all possible discrete points in the continuum limit, which leads to the normalization condition of $\int d\Pi = 1$. This is similar to taking the limit of a Riemann sum to calculate an integral in classical mechanics.