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What is is the starting point to measure the amplitude in an $s$-diagram like the one below where two particles collide and create a propagator followed by the final product?

I know that in $t$-diagram I should start from the opposite side to the direction of propagation but in this case, the particles are propagating to the same point, meaning:

enter image description here

What I mean is:

Should I start by adding the maths expressions from the top right to bottom left, as in $Z_L(p_3)$, followed by the vertex and then by the mathematical expression for $Z_L(p_4)$ followed by the propagator?

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  • $\begingroup$ Yeah we are going to need a lot more context to give a straight answer here, even if a particle name might seem obvious, you should define all things written down for the sake of completeness of the post. I am going to make the most likely assumption you are indeed asking about Z bosons here? $\endgroup$ – Triatticus Mar 17 at 18:40
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As the diagram is rather general, i.e. it is unknown if the particles $Z_L(p_i)$ are distinguishable or undistinguishable, particles or anti-particles, bosons or fermions, one can only make very general statement. Anyway, it is possible to start translating the Feynman-diagram wherever you want, as composing the mathematical expression is essentially multiplication and multiplication is commutative. The expression for the scattering amplitude would be in the most general form ($g_i$ with $i=1,2$ are the coupling constants which are not necessarily the same). I assume a scalar coupling:

$$i{\cal M} = J(1,2)\frac{-i g_1 g_2}{s -m^2_{H_0}}J(3,4)$$ where $J(1,2)$ is the "current" (caveat: possibly this "current" is not conserved, here this does not matter) of the particles $Z_L(p_1)$ and $Z_L(p_2)$, and $J(3,4)$ is the "current" of $Z_L(p_3)$ and $Z_L(p_4)$. If the particles are not distinguishable, another diagram has to be added where the outgoing particles are swapped with respect to the ingoing particles.

BONUS:

As this diagram is supposed to be s-channel process, $Z_L(p_1)$ and $Z_L(p_2)$ would annihilate, and $Z_L(p_3)$ and $Z_L(p_4)$ would be created. So in case of fermions, the "currents" would be something like:

$$J(1,2) = \overline{v}(-p_2)u(p_1) \quad \text{and} \quad J(3,4) = \overline{u}(p_4)v(-p_3)$$

but I don't guarantee that this expression is 100% correct, it is just for giving you an idea.

EDIT:

Actually, the products of the bispinors, for instance $\overline{u}\cdot v$, are not commutative, so one could define a rule in which order they should be written up (Nevertheless $J(1,2)$ and $J(3,4)$ can be still commutated). However, it would be only valid for a $s$-channel process, for the other channels other rules would have to be applied.

ANOTHER EDIT:

The signs of the momenta depend on the assumptions that $p_1$ corresponds an in-going particle and $p_2$ corresponds to an outgoing particle. Finally, $p_3$ is considered as an in-going particle and $p_4$ corresponds to an out-going particle. If the direction of a particle is swapped, the sign has to be swapped correspondingly.

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  • $\begingroup$ @gfgc: I edited my post in order to give more details on the order to apply on bispinors, I also fixed a bug in $J(1,2)$. $\endgroup$ – Frederic Thomas Mar 18 at 11:59
  • $\begingroup$ Thank you for your help again, in that last point, related to my other question, just to check if I understood you edit, what you mean is: We can write $J(1,2) ....J(3,4)$ and we could also write $J(3,4) ....J(1,2)$ in an s-channel process as long as 3,4 and 1,2 are kept in this order, and not changed to 4,3 ,2,1 or 2,3,4,1. $\endgroup$ – user256673 Mar 18 at 13:16
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    $\begingroup$ @gfgc: I precise : If $J(1,2)=\overline{v}(-p_2)u(p_1)$, the adjoint bispinor always has to be on the left side of a bispinor product. But actually, I assumed that the $Z_L(p_1)$ is a fermion, but it could also be an anti-fermion (and the $Z_L(p_2)$ a fermion), then $J(1,2) = \overline{v}(-p_1)u(p_2)$, the same is true for $J(3,4)$. In the diagram posted arrows are missing, they would show the right choice. Anyway, the chosen process is quite rare, I guess, only $\overline{t}t$ are aproppiate fermion candidates for it. For bosons best candidates would be $W^{\pm}$ I guess. $\endgroup$ – Frederic Thomas Mar 18 at 14:07
  • $\begingroup$ Thank you so much for your help $\endgroup$ – user256673 Mar 18 at 14:14

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