# What is the difference between the lambda particle and the neutral sigma particle?

1. May 15, 2010

### Jack_O

Hi this isn't actually a homework question more my own curiosity/confusion. They have the same quark constituents and http://en.wikipedia.org/wiki/File:Baryon_octet.png" [Broken] for both but mass and decay rate are different, please explain. (if the lambda is an excited state of the sigma why isn't it on the baryon decuplet rather than octet?)

Last edited by a moderator: May 4, 2017
2. May 15, 2010

### diazona

Re: What is the difference between the lambda particle and the neutral sigma particle

It's isospin. Take a look at Wikipedia's list of baryons; the sigma has isospin 1 and the lambda has isospin zero. So (and I do hope I'm remembering this correctly), the lambda particle has a wavefunction that is flavor-antisymmetric between the u and d quarks, whereas the sigma has a wavefunction that is flavor-symmetric between those two quarks.

3. May 15, 2010

### Jack_O

Re: What is the difference between the lambda particle and the neutral sigma particle

Thanks I didn't notice that, so it's a singlet. How come their isn't an excited state (lambda star) on the http://en.wikipedia.org/wiki/File:Baryon-decuplet-small.svg" [Broken] accompanying the sigma star?

When you say lambda 'flavour-antisymmetric' do you mean the u and d quarks in lambda have the same colour while in sigma they are different?

Last edited by a moderator: May 4, 2017
4. May 15, 2010

### diazona

Re: What is the difference between the lambda particle and the neutral sigma particle

Flavor, not color. It's always the case that the three quarks in a baryon have different colors. And the symmetric-vs.-antisymmetric thing has to do with different signs of different parts of the wavefunction, although after writing the rest of this post, I'm not so sure what I said before was right.

Anyway, let's see if I can work out a better way to explain this. We're talking about various ways to combine the u, d, and s quarks. Since they're quantum entities, combining them is not just a simple matter of putting u, d, and s in a box. In the technical lingo, the u, d, and s quarks are a basis of the fundamental representation of the SU(3) flavor symmetry group, so in order to figure out the combinations, you have to take the product of three of these representations and decompose it into the sum of irreducible representations which gets you the decuplet and octets, or something like that... I'm still working on figuring out the whole group theory thing

Let's start over. Think about spin in quantum mechanics. You add a spin-1/2 particle to another spin-1/2 particle and you get a symmetric, spin-1 triplet and an antisymmetric, spin-0 singlet.
$$\begin{pmatrix}\uparrow \\ \downarrow\end{pmatrix}\otimes\begin{pmatrix}\uparrow \\ \downarrow\end{pmatrix} = \begin{pmatrix}\uparrow\uparrow \\ \uparrow\downarrow + \downarrow\uparrow \\ \downarrow\downarrow\end{pmatrix} \oplus (\uparrow\downarrow - \downarrow\uparrow)$$
The "up" and "down" states form a doublet, a basis for a 2D space. When you multiply them like this, you get a triplet (three states, a basis for a 3D space) and a singlet (basis for a 1D space).

Quark flavor (by this I mean up vs. down vs. strange etc.) works much the same way. The difference is, instead of just two "basic" states, spin up and spin down, you have three: up, down, and strange. (Actually there are six, for the 6 flavors of quarks, but if you pretend there are only three for simplicity the procedure still works.) So multiplying them out is going to be a bit more complicated. Here's the product of two representations:
$$\begin{pmatrix}u \\ d & s\end{pmatrix}\otimes\begin{pmatrix}u \\ d & s\end{pmatrix} = \begin{pmatrix}uu \\ \frac{ud + du}{\sqrt{2}} & \frac{us + su}{\sqrt{2}} \\ dd & \frac{ds + sd}{\sqrt{2}} & ss\end{pmatrix}\oplus\begin{pmatrix}\frac{ud - du}{\sqrt{2}} & \frac{us - su}{\sqrt{2}} \\ \frac{ds - sd}{\sqrt{2}}\end{pmatrix}$$
(the best way to visualize it is with equilateral triangles, but that should get the gist across) Then you need to multiply it by another representation, because there are three quarks. This is a mess. Here's just part of the first term:
$$\begin{pmatrix}uu \\ \frac{ud + du}{\sqrt{2}} & \frac{us + su}{\sqrt{2}} \\ dd & \frac{ds + sd}{\sqrt{2}} & ss\end{pmatrix}\otimes\begin{pmatrix}u \\ d & s\end{pmatrix} = \begin{pmatrix}uuu \\ \frac{udu + duu + 2uud}{\sqrt{6}} & \frac{usu + suu + 2uus}{\sqrt{6}} \\ \frac{udd + dud + 2ddu}{\sqrt{6}} & \frac{dsu + sdu + usd + sud + uds + dus}{\sqrt{6}} & \frac{2ssu + uss + sus}{\sqrt{6}} \\ ddd & \frac{2dds + dsd + sdd}{\sqrt{6}} & \frac{dss + sds + 2ssd}{\sqrt{6}} & sss\end{pmatrix}\oplus\cdots$$
(actually that's not even complete, because we need to symmetrize the product over quark orderings, but enough math for now) You may recognize the baryon decuplet in there, and if I'd had the patience to go on, one baryon octet would appear. Then from multiplying the other term from two equations above (the one with the three antisymmetric flavor combinations) by another set of uds quarks, I'd get the other baryon octet and a singlet.

But we have yet to incorporate spin! Because remember, each of these quarks has a spin in addition to a flavor. (I'll spare you the details ) Now, the spin part of the wavefunction, like the flavor part, can be either symmetric or antisymmetric. Baryons always have an antisymmetric wavefunction, by the spin-statistics theorem (they're fermions), but the color part of the wavefunction, which is always a singlet and always antisymmetric, completely accounts for that. So for flavor and spin, we pick the combinations that are symmetric overall: either symmetric flavor and symmetric spin, or antisymmetric flavor and antisymmetric spin. Somewhere in the mathematical details it emerges that the baryons in the decuplet can only be spin-3/2; it has to do with the fact that the only spin wavefunction with total spin 3/2 is symmetric. Similarly you find that the octets are spin 1/2. (The two different octets wind up representing the same physical particles) And the flavor singlet, the lambda particle, is also spin 1/2.