Show the following reaction cannot occur (from introductory high energy course)

Click For Summary
SUMMARY

The reaction π- + d → n + n + π0 cannot occur due to violations of conservation laws, specifically concerning intrinsic parity and spin configurations. In this scenario, both pions have an intrinsic parity of -1, while deuterium has a parity of +1 and neutrons also possess a parity of +1. The spins of the particles involved are 0 for pions, 1 for deuterium, and 1/2 for neutrons, leading to a contradiction when considering the final state of two identical fermions and the additional π0. The analysis suggests that the final state parity does not match the initial state, confirming the impossibility of the reaction.

PREREQUISITES
  • Understanding of intrinsic parity in particle physics
  • Familiarity with conservation laws in high energy physics
  • Knowledge of spin configurations of subatomic particles
  • Basic concepts of quantum mechanics, particularly regarding identical fermions
NEXT STEPS
  • Review intrinsic parity and its implications in particle interactions
  • Study conservation laws in high energy physics, focusing on parity conservation
  • Learn about spin statistics and their role in particle reactions
  • Examine examples of similar reactions in "Introduction to High Energy Physics" by Perkins, particularly section 3.3.1
USEFUL FOR

Students and educators in high energy physics, particularly those studying particle interactions and conservation laws. This discussion is beneficial for anyone tackling complex problems involving intrinsic parity and spin in particle physics.

enjoi668
Messages
9
Reaction score
0

Homework Statement


This is from Introduction to High Energy Physics by Perkins, question 3.1
Show that the reaction π- + d → n + n + π0 cannot occur for pions at rest.

the π's are pions, the d is a deuterium, and the n's are obviously neutrons.

Homework Equations



Conservation laws

I started from the knowledge that the intrinsic parity of the pions are both = -1, and that the internal parity of d and n is =1.

I also know the spins to be 0 for the pions, 1 for deuterium, and 1/2 for the neutrons

The Attempt at a Solution


Our professor suggested the pions are both in s-wave configurations if they are at rest (?)

I tried following the arguments that ultimately decided the intrinsict parity of pi-, i.e.
something like this

http://quantummechanics.ucsd.edu/ph130a/130_notes/node323.html

to try to find some kind of contradiction if an extra π0 is included, however I didn't seam to get anywhere as that argument is built upon the fact that the final state consists of two identical fermions...

I've been working on this for about an hour, so I'm really just looking for any ideas on how to start this problem off or if I'm going about it the right way and I'm just missing something
 
Last edited:
Physics news on Phys.org
I have the third edition of Perkins. I hated that book. :wink:

Your approach is probably the right one. I haven't worked it out, but I think you're on the right track. In my copy of Perkins, there's a section on pion parity, section 3.3.1, which has a similar analysis to the one you linked to. You can look at the final state as the neutron pair, taken as a whole, plus the pion. When you do that, how does that affect the parity of the final state?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
1
Views
3K
Replies
1
Views
11K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 2 ·
Replies
2
Views
6K