Optical theorem applied to forward scattering of a single particle

Question

I'm slightly confused. Write the S-matrix as $$ S = 1 + i T $$ Unitarity implies $$ T - T^\dagger = i T^\dagger T $$ In scattering from $|i\rangle$ to $|f\rangle$, $$ T_{f,i} - T^\dagger_{f,i} = i \sum_n T^\dagger_{f,n} T_{n,i} $$ where the sum stands for both a summation over number of particles (say we only consider one type of scalar particle) and integration over their phase space, and (at least for $|a \rangle \neq |b \rangle$) $$ T_{a,b} = \langle a | T|b \rangle = (2\pi)^4 \delta^{(4)} (p_a -p_b) M_{a,b} $$ Therefore $$ M_{f,i} - M^*_{i,f} = i \sum_n (2\pi)^4 \delta^{(4)} (p_n -p_i) M^*_{n,f} M_{n,i} $$ In particular for forward scattering $$ 2 \operatorname{Im} M_{i,i} = \sum_n (2\pi)^4 \delta^{(4)} (p_n -p_i) |M_{n,i}|^2 $$ Consider the case of a single particle. What happens in the last sum over $n$ to the term with $n=1$? I'm confused as to how we do account for such $\delta(0) |M_{i,i}|^2$ term (if present) when we say that $\operatorname{Im} M_{i,i} =0$ if the particle is stable.

Answer 1

When you talk about a single particle initial state $| i \rangle $, the sum over $n$ on the rhs then translates into the physically reliasable decay modes of $i$. If the particle is stable, $M_{n,i} = 0$ for all $n$ and if it’s unstable, the sum delimiter explores all phase space decay products and in particular $n$ is never $i$.

Recall the decay rate is given by $$\Gamma(i \rightarrow n) \sim \frac{1}{2 m_i} \int \text{d}\Pi_n (2 \pi)^4 \delta^4(p_i-p_n) |M_{n,i}|^2,$$ where $1/(2m_i)$ is the relativistic normalisation of states in the rest frame of the decaying particle with the branching fraction of a given decay species given by $B(i \rightarrow n) = \Gamma(i \rightarrow n)/\sum_n \Gamma(i \rightarrow n)$. In this sum we never include the case where $n=i$.

It may help intuitively to realise the evaluation $n=i$ within the $1$ of $S=1+iT$, i.e we have $|f \rangle = |i \rangle$ therein by definition with nothing happening in between. All contributions with $n≠i$ are encoded in the transition amplitude $T$, transition meaning something non trivial actually happens.

The forward 'scattering' optical theorem with a single particle initial state $| i \rangle$ can then be written as, using the above decay rate, $$\text{Im}\, M_{i, i} = m_i \sum_n \Gamma(i \rightarrow n)$$