Optical theorem applied to forward scattering of a single particle

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.

• 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 $i$.
– CAF
Commented Aug 13, 2018 at 16:53
• @CAF Thanks, but I cannot see from first principles why the sum over $n$ must be restricted to $n \neq i$. Could you please explain?
– jj_p
Commented Aug 13, 2018 at 17:27

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, $$\text{Im}\, M_{i, i} = m_i \sum_n \Gamma(i \rightarrow n)$$

• Thanks for your answer, but it's not clear to me whether the restriction $n \neq i$ follows from first principles (after all in my derivation above we just inserted a complete set of states in the form of 1), or are you saying $T_{ii}=0$? but then $M_{ii} \neq 0$? sorry for the confusion..
– jj_p
Commented Aug 14, 2018 at 13:52
• Remember $n$ here is a kinematically permissible intermediate state between in and out. For a single particle initial state $i$ and you want to describe a forward ‘scattering’ $f=i$, these $n$ states are just corrections to the propagator. The contributions to the imaginary part of an amplitude are regions in phase space when intermediate states go on shell
– CAF
Commented Aug 14, 2018 at 14:15
• I agree that is the correct conclusion, but where does one see it in the derivation I provided above? in other words, are you saying that $T_{i,i}$ is zero?
– jj_p
Commented Aug 14, 2018 at 16:56
• The case $i=f$ with no intermediate states belongs in $1$. The case $i=f$ and with intermediate states (ie you want to describe a forward scattering, or in one particle case like yours, corrections to propagator) belongs in $T$. It is this $T$ used in deriving the optical theorem and by construction of the $S= 1 + iT$ split, the $n$ states must be non trivial intermediates.
– CAF
Commented Aug 14, 2018 at 17:09
• Is it perhaps clearer now? The decomposition of S is just into trivial and non trivial events between i and f.
– CAF
Commented Aug 14, 2018 at 17:10