Scattering amplitude and LSZ formula

Question

I'm arriving at a contradiction.

To calculate the scattering amplitude, one usually follows the prescription given by the Feynman rules that you only consider fully connected diagrams with the required number of incoming and outgoing external legs (See Peskin & Schroeder pg 111 where they say: Only fully connected diagrams contribute to the $T$ matrix).

By fully connected, one means that you consider only graphs from which you can get from one line to any other line (See page 3 of this document).

On the other hand, we have the LSZ formula, which says that the scattering amplitude is given by the residue (as the momenta go on-shell) of the corresponding correlation function. For example, in $\phi^4$ theory, \begin{align} &\mathcal{M}(p_a,p_b \to k_1, k_2) \delta^{(4)}(p_a + p_b-k_1 -k_2) \sim \nonumber\\ &\lim_{p_a^2,p_b^2,k_1^2,k_2^2 \to m^2} (p_a^2 - m^2)(p_b^2 - m^2)(k_1^2 - m^2)(k_2^2 - m^2)G(p_a,p_b,-k_1,-k_2). \end{align}

But these two prescriptions seems to give a contradiction. Consider in $\phi^4$ theory, the $\mathcal{M}(4 \to 4)$ scattering. We have this diagram (ok if someone could draw the diagram that'll be great),

\begin{align} \text{X} \text{X} \end{align}

which consists of two separate $2 \to 2$ scattering processes.

This diagram is not fully connected, so we should ignore it by the first prescription, yet, it does not evaluate to $0$ under the LSZ formula, so we should include it.

Physically it makes sense that the leading order contribution to a $4 \to 4$ process is given by two separate $2 \to 2$ ones, but the fully connected prescription misses that out.

So, is there a caveat to the fully connected rule of drawing Feynman diagrams, since I believe the LSZ formula is mathematically true and physically reasonable?

Answer 1

T-matrix elements are not the same as fully connected diagrams. Remember T-matrix is defined as $$T=S-I,$$ where $S$ is the S-matrix and $I$ is the identity matrix. S-matrix part includes all possible diagrams except vacuum bubbles and unamputated diagrams. Now we just need to ask, does subtracting an identity matrix remove all the disconnected diagram? Obviously no, because $$\langle p_1,\ldots,p_m|I|k_1,\ldots,k_n\rangle=\delta_{mn}\sum_{\sigma}\prod_i\delta^4(p_{\sigma(i)}-k_i),$$ where $\sigma$ is a permutation of indices, and I have assumed bosons to avoid sign changes upon permutations. So identity matrix corresponds to diagrams with straight lines not only disconnected, but also containing no vertices, however the S-matrix contains some disconnected diagrams containing vertices, so only a subtraction of $I$ won't take them away.

$2\to2$ scattering is special, because if you actually try to draw diagrams with 4 external lines, it's either disconnected and containing no vertices(since we have excluded vacuum bubbles and unamputated diagrams), or just fully connected, so in this case the substraction of $I$ will take away all disconnected diagrams. This is why for $2\to2$ scattering, we just need to calculate fully connected diagrams. Peskin & Schroeder is potentially confusing because they never go into situations with more than 4 external lines.

Textbooks do not talk much about disconnected diagrams because they can be trivially calculated from its connected components, this silence might be another source of confusions(again, I'll advocate Weinberg here).

In conclusion, there is simply no such thing as "fully connected diagram prescription"(unless you want a fancy name for the $2\to2$ scattering case), and after these clarifications there should be no "contradiction" as described by OP.