Why can an electron-positron collider only produce particles with spin 1?

[lark]

I read a statement in an article that an electron-positron collider can only make particles with spin 1, not spin 0, if just one particle is generated in the collision.
Why would that be? (Maybe there are more provisos to that. )
Laura

 

[lark]

Apparently if the electron and positron have spin in opposite directions, the magnetic moments would repel each other. I don't know if this is the explanation.
Laura

 

[Physics Monkey]

Can you tell us what the article was?

Naively, the Yukawa type coupling $$\bar{\psi} \psi \phi$$ would permit the annihilation of the fermion and anti-fermion described by $$\psi$$ into the scalar described by $$\phi$$ as long as $$\phi$$ is massive.

Perhaps the article was considering a special situation of some type?

 

[humanino]

Naively, the Yukawa type coupling $$\bar{\psi} \psi \phi$$ would permit the annihilation of the fermion and anti-fermion described by $$\psi$$ into the scalar described by $$\phi$$ as long as $$\phi$$ is massive.

Precisely how such a collider would produce the Higgs...

 

[lark]

Can you tell us what the article was?

Naively, the Yukawa type coupling $$\bar{\psi} \psi \phi$$ would permit the annihilation of the fermion and anti-fermion described by $$\psi$$ into the scalar described by $$\phi$$ as long as $$\phi$$ is massive.

Perhaps the article was considering a special situation of some type?

It was an article in Physics Today about finding the http://blogs.physicstoday.org/update/2008/07/bottomonium-ground-state-in-th.html" , the meson which is composed of the bottom quark and bottom antiquark.
They can make excited spin-1 bottomonium but the spin-0 state only appears as a decay from the spin-1 state. By emitting a photon. The way the article put it, it sounds like a general principle is involved.
Laura

 

[Vanadium 50]

The process for producing quarkonium exclusively (plus nothing else) is $e^+ + e^- \rightarrow \gamma^* \rightarrow (q\overline{q})$. The $(q\overline{q})$ pair needs to have the same quantum numbers as the photon, 1--.

If you allow inclusive production, you can have reactions like $e^+ + e^- \rightarrow (q\overline{q}) + \gamma$, which opens up many other quantum numbers for the $(q\overline{q})$ pair.

 

[lark]

The process for producing quarkonium exclusively (plus nothing else) is $e^+ + e^- \rightarrow \gamma^* \rightarrow (q\overline{q})$. The $(q\overline{q})$ pair needs to have the same quantum numbers as the photon, 1--.

I see. So the electron-positron annihilation makes two photons, and one of them might turn into a bottomonium meson?

What does $\gamma^*$ mean? Virtual photon?

Laura

 

[lark]

So in general when an electron collides with a positron, if it's only producing a single particle, that particle would result from a photon, so it would have to have spin 1?
Laura