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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.

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    $\begingroup$ 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. $\endgroup$ – dmckee Sep 5 at 22:08
  • $\begingroup$ @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. $\endgroup$ – adiselann Sep 5 at 22:34
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    $\begingroup$ Related: physics.stackexchange.com/q/341422 $\endgroup$ – dmckee Sep 5 at 22:59
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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.

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

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