# Are amplitudes for inverse processes related to each other?

The (generalized) optical theorem is presented in the book of Peskin and Schroeder (An introduction to Quantum Field Theory - chapter 7-Radiative Corrections:Some Formal Developments) as follows

$$-i (\mathcal{M}(a \to b) - \mathcal{M}^*(b\to a))= \sum_{f}\int d\Pi_f \mathcal{M}^*(b \to f)\mathcal{M}(a \to f)$$,

where the sum runs over all possible sets $$f$$ of final-state particles.

My main question is: are $$\mathcal{M}(a \to b)$$ and $$\mathcal{M}^*(b\to a)$$ somehow related?

I was naively thinking about using

$$T -T^\dagger = 2i\, Im(T) \to \langle b|T -T^\dagger|a\rangle = 2 i\, Im(\mathcal{M}(a \to b)) =\mathcal{M}(a \to b) - \mathcal{M}^*(b\to a)$$

but, if this is true, why would Peskin and Schroeder avoid using this compact form?

I will preface by saying that I am not an expert in this, and if there are things I am missing or misstating I would very much appreciate comment. There are often a lot of $$(2\pi)$$'s and delta functions floating around that I have chosen not to include here, so this argument is more heuristic manipulation of the relevant matrices and matrix elements. Note that there is NO implied summation on repeated indices, all sums are explicitly labeled.

You write,

$$T-T^\dagger = 2i \text{Im}(T)$$

but this is not correct (it is close to correct though). $$T$$ is a matrix not a number, so $$T-T^\dagger$$ corresponds to the anti-hermitian part of the $$T$$ matrix. If we consider matrix elements,

$$T_{ab}-(T^\dagger)_{ab}=T_{ab}-T^*_{ba}$$

now for a diagonal element, corresponding to a state $$a$$ scattering into itself (AKA not scattering) then we get,

$$T_{aa}-T^*_{aa}=2i\text{Im}(T_{aa})$$

since this is no longer a full operator, but just a component (a complex number).

I argue that your first equation is the most general statement relating amplitudes for processes and inverse processes, and it follows only from assuming the S-matrix is unitary. Below I will outline the general idea it is saying, and maybe how to think of it in a more intuitive way.

I think it is interesting to then add in the idea of time reversal invariance and see what that implies about these processes.

Unitarity The most robust thing that relates scattering amplitudes is unitarity. You have already used unitarity and the first equation you quote is derived from requiring the $$\mathcal{S}$$ matrix to be unitary. I would say in words that the equation is saying "The difference between scattering amplitude from $$a\rightarrow b$$ to $$b\rightarrow a$$ is related to how many intermediate states $$f$$ the system can transition to".

This statement is much more straightforward if we consider forward scattering (AKA how much of our beam does NOT scatter), then we consider $$a\rightarrow a$$. The probability for forward scattering is,

$$|\mathcal{S}_{aa}|^2 = |1+iT_{aa}|^2=(1-iT_{aa}^\dagger)(1+iT_{aa})=1+i(T_{aa}-T^\dagger_{aa})+|T_{aa}|^2=1-2\text{Im}(T_{aa})+|T_{aa}|^2$$.

The first and third terms are strictly positive, so for any decrease in the forward scattering there has to be an imaginary part of $$T_{aa}$$. Now the equations above give "the optical theorem":

$$\text{Im}(T_{aa})=\sum_f\int d\Pi_f T_{af}^* T_{fa}=\sum_f\int d\Pi_f|T_{fa}|^2\propto \sigma_a$$,

where the last equality is taking "the probability for $$a$$ to scatter to $$f$$ summed over all final states $$f$$ is propotional to the cross section". This whole equation simply states "the amount of beam that doesn't go straight is equal to the amount that scatters", which follows straightforwardly from unitarity (unitarity meaning every state gets mapped into another state).

We see that this is just a special case of the first equation that you write, which in general relates difference $$T_{ab}-T^*_{ba}$$ to the absorptive part of the scattering process.

T-Reversal (and CPT) There is one more physical property that is useful for this discussion and it is time reversal. Following the discussion here: https://www.sciencedirect.com/science/article/pii/0550321371904603 (also relevant https://www.sciencedirect.com/science/article/pii/0550321387906857)

If we denote the "time reversal" of state $$a$$ as $$\tilde{a}$$, meaning something like "flip the direction of the momenta, and direction of the spins", then we can relate the process of state $$a\rightarrow b$$ to $$\tilde{b}\rightarrow\tilde{a}$$ (remember time reversal also swaps initial and final state). Time reversal invariance means $$T_{ab} = T_{\tilde{b}\tilde{a}}$$. This sort of resembles what you are asking about, but it is not exactly what you ask about.

We can use your first equation and apply T-invariance to extract the relevant information: \begin{align} T_{ab}-T^\dagger_{ab} &= T_{ab}-T^*_{ba}\\ &= T_{ab}-T^*_{\tilde{a}\tilde{b}} \end{align}

Now using your first equation (which follows from unitarity)

$$T_{ab}-T^*_{\tilde{a}\tilde{b}}=i\sum_f\int d\Pi_f T_{bf}^* T_{fa}$$

If we define the following $$\alpha_{ab}=\sum_f\int d\Pi_f T_{bf}^* T_{fa}$$, then we can get the following (by pushing around the terms and squaring both sides):

\begin{align} |T^*_{\tilde{a}\tilde{b}}|^2 &= |T_{ab}-i\alpha_{ab}|^2\\ |T_{\tilde{a}\tilde{b}}|^2 &= |T_{ab}|^2 +i( T_{ab}\alpha^*_{ab}-T_{ab}^*\alpha_{ab})+|\alpha_{ab}|^2\\ |T_{ab}|^2 - |T_{\tilde{a}\tilde{b}}|^2&=2\text{Im}(T_{ab}\alpha_{ab})-|\alpha_{ab}|^2 \end{align}

Meaning that a difference in the rate of $$a\rightarrow b$$ compared to $$\tilde{b}\rightarrow \tilde{a}$$ is only generated by absorptive parts of the scattering process. This is why absorptive effects can mimic "$$T$$-violating" signals, even when there is no $$T$$ violation in the Hamiltonian. This discussion immediately generalizes to $$CPT$$ if we replace $$\tilde{a}$$ as the "$$T$$-reversal image of $$a$$", and instead consider $$\tilde{a}$$ as the $$CPT$$ image of $$a$$ (flip spins, and replace particle with anti-particle).

as a footnote: I'm not entirely sure why it's ok to assume that $$T$$-reversal doesn't induce a phase between the transition elements. Although I think if you allow a phase difference it is inconsequential to the final results.

as a second footnote: I actually get the last line should say $$2\text{Im}(T_{ab}\alpha^*_{ab})$$ which does not replicate what the authors wrote. It is probably my mistake, and I think it is inconsequential to the main idea.

(Sorry for the inexistent formatting, I am not used to writing in LaTex outside my usual environment.) Consider the matrix representation of an operator $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ Take its Hermitian conjugate, $$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$ then the difference is $$\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \neq 2 i \Im \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ Wouldn't this suffice as a counter-example?

• I added the correct coding to your answer and I tried to correct what I thought was an error, but you should check to make sure it's what you intended. Commented Mar 2, 2023 at 15:52
• I think you are interpreting $T$ as "any operator", but it is very specifically related to the $S$ matrix by $S=1+iT$. Note that the $S$ must be unitary, and 1 + i*(the matrix you write) does not make a unitary matrix
– TEH
Commented Mar 2, 2023 at 22:32