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?

  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/51993, short answer is, "Only fully connected diagrams contribute to the T matrix" is simply not true, it is only true for "$2\to2$" scattering, if you have more particles in the initial or final states, fully connected diagrams only contribute to the so called "connected part of S-matrix" which is not identical to matrix elements of T-matrix. Weinberg Vol I on scattering theory has a good introduction. $\endgroup$
    – Jia Yiyang
    Commented Oct 27, 2013 at 15:01
  • 2
    $\begingroup$ @JiaYiyang I just thought about this further, in my example two $2 \to 2$ scattering, there is still a delta function $\delta^{(4)}(p_1 + p_2 - k_1 - k_2)$ aside from the overall delta function $\delta^{(4)}(p_1 + p_2 + p_3 + p_4 - k_1 - k_2 - k_3 - k_4)$ so in this restricted phase space of 8 particles the first delta function has support on measure $0$ anyway (since we need the restrictive condition that 2 incoming particles' momenta equal 2 particular outgoing ones') so its contribution vanishes, so maybe that's why the fully connected prescription is still right? $\endgroup$
    – nervxxx
    Commented Oct 28, 2013 at 2:33

1 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.

  • $\begingroup$ Thank you for your answer, but for (II), it is obvious physically and mathematically that your LHS must contain a momentum conserving delta function! Recall the mathematical identity $\delta(x)\delta(y) = \delta(x+y)\delta(y)$. So let's apply it to $2 \to 2$ scattering in free scalar theory, which has only 2 (disconnected propagators). So if $p_1$ goes to $k_1$ and $p_2$ goes to $k_2$, we have $\sim \delta^{(4)}(p_1-k_1)\delta^{(4)}(p_2-k_2) = \delta^{(4)}(p_1+p_2-k_1-k_2)\delta^{(4)}(p_1-k_1)$ which is your overall momentum conserving delta function. $\endgroup$
    – nervxxx
    Commented Oct 29, 2013 at 14:09
  • $\begingroup$ Also, just to clarify, I don't mean there should be a delta function Outside of $\langle \cdots |S|\cdots\rangle$, but rather there is a delta function within the matrix elements of $S$, so you can factor $\langle \cdots |S|\cdots\rangle$ into delta function $\times$ something else (which turns out to be the so called invariant matrix element $\mathcal{M}$) $\endgroup$
    – nervxxx
    Commented Oct 29, 2013 at 14:17
  • $\begingroup$ @nervxxx: you are right, I'll fix this. $\endgroup$
    – Jia Yiyang
    Commented Oct 29, 2013 at 14:21
  • $\begingroup$ @nervxxx: Actually after your clarification I don't quite get your confusion anymore, indeed as you showed $\delta^{(4)}(p_1 + p_2 + p_3 + p_4 - k_1 - k_2 - k_3 - k_4)\delta^{(4)}(p_1 + p_2 - k_1 - k_2)$ restrict us on a phase space surface of lower dimension than just an over all delta function, but this only has to do with cross sections, while here we are only talking about S-matrix elements. So it seems part I of my answer suffices to answer you question, there is simply no contradiction, unless you insist there is a "fully connected diagram" prescription, which is wrong. $\endgroup$
    – Jia Yiyang
    Commented Oct 29, 2013 at 15:02

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