# R-parity and conservation of angular momentum

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1. Aug 24, 2015

### Ken41

Assuming R-pairity and thus the creation/destruction of supersymmetric particles happens in pairs,
how is angular momentum conserved when a particle and its supersymmetric partner have different spin by 1/2?

2. Aug 24, 2015

### fzero

It depends on the process, but angular momentum will be conserved in ways that are already familiar from SM processes. For example, in the SM, we can have processes like $gg\rightarrow q \bar{q}$. At tree-level, there is a diagram

$g$ ^^^^^^^^^^^^^^^^---------------------- $q$
$\hspace{3.65cm}$ |
$\hspace{3.65cm}$ |
$\hspace{3.65cm}$ |
$g$ ^^^^^^^^^^^^^^^^---------------------- $\bar{q}$

Angular momentum is conserved because the initial state can have $J = 0,1,2,\ldots$ and so can the final state.

In the MSSM, the same diagram can describe gluino production $gg\rightarrow \tilde{g} \bar{\tilde{g}}$. It is also easy to add additional vertices to produce a 1-loop diagram with squarks or a squark + $q\bar{q}$ in the final state

3. Aug 24, 2015

### Ken41

Thanks for responding! Is it obvious why in your example, gg→g̃ barg̃ , ( where NN->SS) , the reverse reaction (SS->NN) would not be allowed? The R- parity would not change, so CPT stays the same...

4. Aug 24, 2015

### fzero

The reverse reaction would be allowed. The R-parity of $\tilde{g}$ is $-1$, but the R-partity is multiplicative, so for a system of two of them it is $1$. In general, R-parity is conserved by particle-antiparticle annihilation for all SUSY particles.

5. Aug 25, 2015

### ohwilleke

More generally, any Feynman diagram's truth or falsity is never changed by rotating the diagram.

6. Aug 29, 2015

### Blackforest

Do you know this document: "mT2: the truth behind the glamour", arXiv:hep-ph/0304226v1, 23 April 2003? Perhaps you might be interested by it.