# Sum of spin for chiral particles?

1. Sep 10, 2014

### arivero

How does the spin of a pair of particles work if both particles are known to be chiral? generically if I sum the spins of two different (EDIT: spin 1/2, indeed ;-) particles I expect to get a triplet with S=1

$\uparrow\uparrow$,
$\uparrow\downarrow+\downarrow\uparrow$,
$\downarrow\downarrow$

and a S=0 singlet $\uparrow\downarrow-\downarrow\uparrow$,

and in the case of identical particles only the singlet survives, isn't it?

Now, that happens if both particles are non-identical but of the same chirality? Do we still have the four states? Only the $\uparrow\uparrow$ combination survives? or we have a "massless S=1" entity, with both $\uparrow\uparrow$, and $\downarrow\downarrow$?

And for different chiralities?

Last edited: Sep 10, 2014
2. Sep 10, 2014

### DrDu

I can only guess: Even non-relativistically for massive particles, the spin of the compound particle is not the singlet or triplet you are constructing but the sum of your spin part and the orbital angular momentum in the rest frame. For relativistic and hence especially for massless particles, this is the same. As long as the particles are not flying exactly in the same direction, you can find a frame where the total momentum vanishes. In this frame you can decompose the wavefunction of each particle into vector spherical harmonics and sum the total angular momenta of both particles.

3. Sep 10, 2014

### muppet

I'm not 100% sure on this... but your post clearly assumes you're talking about spin-1/2 particles, which don't have a spin-zero component anyway, so think the addition of angular momentum should proceed in exactly the same way.

However, if I think about a collision occurring in the centre-of-mass frame, demanding that both particles have the same chirality implies that the projections of their spins onto the collision axis are anti-parallel, as they're travelling in different directions, so I suspect this would cut you down to at most two possible final states.

As an aside, I'd be cautious about interpreting the final state as being "massless" even if its constituents are!

4. Sep 10, 2014

### arivero

by the way, have you noticed how the signs of the spin triplet (orbital states) are different of the isospin triplet (pions)? I guess that it some difference between the adjoint of SU(2) and the fundamental of SO(3).

5. Sep 11, 2014

### arivero

6. Sep 11, 2014

### Avodyne

In general you have all four states, whether or not the particles are chiral.

7. Sep 11, 2014

### DrDu

I probably don't quite understand what you are talking about. There are no massless and hence also chiral spin 1/2 particles in nature, so the question is somewhat hypothetical.
Even if they existed, as they would be massless, you have to treat them relativistically and you have to take into account the orbital momentum of the component particles in the construction of the total spin of the compound particle.

8. Sep 11, 2014

### arivero

Hmm yes, really is not SO(3) the player here, but the Lorentz group.

Let me think with Dirac spinors first. What I am asking thus is about the decomposition of

$$[(\frac 12,0) \oplus (0,\frac 12)] \otimes [(\frac 12,0) \oplus (0,\frac 12)]$$

in spin 0 and spin 1 states, that in the non relativistic limit should approach some states in the representations of SO(3)

and then the same question when we only have chiral particle, this is, for

$$(\frac 12,0) \otimes (0,\frac 12)$$

or

$$(0,\frac 12) \otimes (0,\frac 12)$$

Yep, thinking relativistically the question seems easier to answer. The problem comes when you want to separate particle and antiparticle.

Last edited: Sep 11, 2014
9. Sep 11, 2014

### DrDu

That's not where I wanted to go. Consider a positronium with l=1 and s=1in it's rest frame. The total spin I of the positronium can take the values I=|l-s| .. l+s =0,1,2. If the two electrons anihilate into a pair of (hypothetical massless) neutrinos, total spin I of the new particle will be the same. but in general you won't be able to write I as a vectorial sum of l and s.

10. Sep 11, 2014

### arivero

by the way, googling around I see that 1S0 can not decay to (massless) neutrino+antineutrino, but the orthopositronium, 3S1, can. Not sure which are the rules for the cases you suggest, that should be for 3P0 3P1 3P2, isn't it?

11. Sep 12, 2014

### DrDu

I am not too familiar with the selectrion rules for the decay of positonium. What I wanted to say is that given a term symbol like 3p0 or SLJ in general, S and L are only approximate quantum numbers and often don't make sense at all for relativistic particles. The only quantum number which is exact and conserved is J. J is the spin of the compound particle. So in calculating the spin of a compound particle made up from massless particles, you always have to take into account the orbital angular momentum of the particles and not only their helicity.