# Particle notation: What does the asterisk stand for?

For instance, what's the difference between $$\Sigma^0$$ particles and $$\Sigma^{0*}$$ particles? I know they have the same quark configuration $$sdu$$, so how exactly are they different?

Some context: I'm reading the 2nd chapter of Halzen's Quarks and Leptons. I found the sigma baryons (with and without the asterisk) on Figure 2.8. More precisely, I found the $$\Sigma$$ particles on the spin 1/2 graph, and the $$\Sigma^*$$ particles on the spin 3/2 graph.

• Star superscripts can mean different things in different context. I suspect that "excited state" and "off-shell" are the most common meanings, but without some more data people will be shooting in the dark. – dmckee Sep 5 at 22:08
• @dmckee thanks for the quick comment. I'm just starting out with the study of the quark model. For example, in the electromagnetic decay $\Sigma^*(1385)^- \rightarrow \Sigma^- \gamma$, I wonder what (1385) means (I guess it's the mass) and what the star/asterisk means. Perhaps there's still more context needed, but thank you anyway. – adiselann Sep 5 at 22:34
• – dmckee Sep 5 at 22:59

In case of the $$\Sigma$$ particles the $$^*$$ means "excited" (i.e. having a higher energy, thus a higher rest mass). For other particles the $$^*$$ may have other meanings (see the posted comments below).

According to Wikipedia "Sigma baryon":

The $$\Sigma^{0*}$$ baryon has total angular momentum $$J=\frac{3}{2}$$ and rest mass $$1383\text{ MeV}/c^2$$.
The $$\Sigma^0$$ baryon has total angular momentum $$J=\frac{1}{2}$$ and rest mass $$1192\text{ MeV}/c^2$$.

Each quark ($$u$$, $$d$$ and $$s$$) has a spin angular momentum of $$\frac{1}{2}$$.
Grossly simplified you can imagine the particles like this:

• In the $$\Sigma^{0*}$$ baryon all 3 quarks are spinning in the same sense, giving a total angular momentum of $$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}$$.

• In the $$\Sigma^0$$ baryon 2 quarks are spinning in one sense, and 1 quark is spinning in the opposite sense, giving a total angular momentum of $$\frac{1}{2}+\frac{1}{2}-\frac{1}{2}=\frac{1}{2}$$.

Explaining why $$\Sigma^{0*}$$ has a higher energy than $$\Sigma^0$$ would be a more difficult story.

• For mesons, such as $K^*, D^*$,... the asterisk denotes vector as opposed to spinless mesons lacking the asterisk... – Cosmas Zachos Sep 6 at 0:44
• 1. You can have excited states without the star. 2. Scalar flavoured mesons (as opposed to pseudoscalars) have stars, e.g. the $K^*_0(1430)$. – dukwon Sep 6 at 5:36

In the PDG notation, a star is not synonymous with an excited state. Yes, any state with a star is excited, but not all excited states have a star. It depends on the spin and parity.

From the PDG naming scheme for hadrons in the section on flavoured mesons:

1. If the spin-parity is in the "normal" series, $$J^P = 0^+, 1^-, 2^+,...$$, a superscript "$$^*$$" is added.

In other words, the star tells you the parity is positive for even spin and negative for odd.

They don't seem to define rules for starred baryons, and the only one I can find is the $$\Sigma_b^*$$.