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.

  • 4
    $\begingroup$ 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 --- ex-moderator kitten Sep 5 '19 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 '19 at 22:34
  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/341422 $\endgroup$ – dmckee --- ex-moderator kitten Sep 5 '19 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.

| cite | improve this answer | |
  • $\begingroup$ For mesons, such as $K^*, D^*$,... the asterisk denotes vector as opposed to spinless mesons lacking the asterisk... $\endgroup$ – Cosmas Zachos Sep 6 '19 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 '19 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 $\Sigma_b^*$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.