Pair production in boson-boson collision

Click For Summary
SUMMARY

The discussion centers on calculating the amplitude for the boson-boson collision process \(\psi + \psi \rightarrow e^- + e^+\) using the interaction Lagrangian \(L_I (x) = -g \overline{\psi}(x) \psi (x) \phi (x)\). The key issue addressed is the determination of relative signs in the contributions from two Feynman diagrams, where the final fermions are swapped. It is established that there is no relative minus sign between the contributions due to the nature of the outgoing fermions being non-identical, despite the swapping of incoming particles. The propagators for both diagrams remain unchanged.

PREREQUISITES
  • Understanding of quantum field theory and interaction Lagrangians
  • Familiarity with Feynman diagrams and their interpretation
  • Knowledge of fermionic and bosonic particle statistics
  • Basic proficiency in calculating matrix elements in particle physics
NEXT STEPS
  • Study the derivation of Feynman rules for scalar and fermionic fields
  • Learn about the implications of particle statistics on scattering amplitudes
  • Explore the role of propagators in quantum field theory calculations
  • Investigate examples of second-order processes in perturbation theory
USEFUL FOR

Particle physicists, graduate students in quantum field theory, and researchers analyzing boson-boson interactions and scattering processes.

member 251684

Homework Statement



I'm given this interaction Lagrangian:

L_I (x) = -g \overline{\psi}(x) \psi (x) \phi (x)

where g is the coupling constant, \psi is an electron and \phi is a neutral scalar field. I have to calculate the amplitude for process
\psi + \psi \rightarrow e^- + e^+ that is to say that I have to calculate the matrix element

\langle e^+ (\vec{p}_3) e^- (\vec{p}_4) |S| \phi (\vec{p}_1 ) \phi (\vec{p}_2)\rangle


Homework Equations



I'm stuck determining the relative sign of the contributions. I know that there are two Feynman diagrams contributing to the matrix element and that the final fermions are swapped in the two diagrams, so there's a - (minus) sign in the relative contributions.

On the other side, I find that in the first graph the propagator is given by

\langle 0 | T [\overline{\psi}(x) \psi(y)] |0 \rangle

while in the second graph the propagator is given by

\langle 0 | T [\psi (x) \overline{\psi}(y)] |0 \rangle

I just need to know if this change in the propagator adds a new - (minus) sign, so that the relative contributions end up having the same sign or if they ends up having opposite sign.

The Attempt at a Solution



I know thath this is a second order process in g and I know also that the final state is produced by a fermionic propagator between the neutral bosons. I've calculated many similar processes but this question on relative signs of the two contributions is really blocking me.

Sorry for my English, I'm not a native speaker. Thank you in advance for you help.
 
Physics news on Phys.org
There is no relative minus sign. Physically, this is because the two outgoing fermions are not identical. Or, you can see the two diagrams as related by swapping the two incoming particles, which are identical; but these are bosons, so there is no relative sign.
 
Oh well. I've to think about it. I just tought that "swapping" two fermions always lead to a minus sign...
But, about the propagator, it stays the same in both cases?

Thank you for your help
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
1
Views
2K