# Eta prime meson as SU(3) singlet

1. Jan 4, 2014

### gildomar

I understand that when the quark theory was being developed that SU(3) was used to explain the mesons that were ultimately found to be composed of the up, down, and strange quarks. I also get that the SU(3) is grouped as an octet and a singlet, with the eta prime meson being the singlet. But I'm a little fuzzy on the why of some of them, since the book I'm reading doesn't exactly make it clear. That is, why is the SU(3) grouping that of an octet and a singlet? What differentiates the octet from the singlet? Why is the eta prime meson the singlet and not one of the other mesons?

2. Jan 4, 2014

Staff Emeritus
The idea is that there is a pure singlet eta-prime, and the octet possesses mixed symmetrry. However, since the eta and the eta-prime have the same quantum numbers, they mix, and the physical eta and eta-prime are mixtures of singlet and octet. Because of this, you will often hear this group of mesons called a nonet.

3. Jan 4, 2014

### The_Duck

Here is why the light pseudoscalar form an octet and a singlet. The quarks $u, d, s$ live in the $3$ representation of SU(3). The antiquarks $\bar u, \bar d, \bar s$ live in the $\bar 3$ representation. The mesons made of one of these quarks and one of these antiquarks will fill out a direct product representation $3 \otimes \bar 3$. This direct product representation, it turns out, can be decomposed as the direct sum of two irreducible representations, the $8$ and the $1$.

Their transformation properties under SU(3). For example, $u \bar u - d \bar d$ is not invariant under all SU(3) transformations, while $u \bar u + d \bar d + s \bar s$ is. The first is part of the octet, the second is the singlet.

What exactly are you asking here? How we know that the $\eta'$ is the singlet, and that the $\pi^0$ isn't, say?

4. Jan 4, 2014

### gildomar

@The_Duck: Yes, that's what I mean. And the other stuff clears things up a little.

5. Jan 5, 2014

### andrien

Under a rotation in flavour space ,you interchange u,d or s .The singlet state is uu-+dd-+ss- which does not change under an interchange while if you will take any of the octet one and interchange u by d and so forth,you will find that octet transform into each other.This will never be the case for singlet,it will never mix with the other octet one.The way to write it is $3 \otimes \bar 3$=8+1,you find particles in the irreducible represntation of this flavour SU(3),OCTET AND SINGLET.Same is there for baryons for which you have
$3 \otimes 3 \otimes 3$=1+10+8+8,you will find a decuplet for baryons too along with singlet and octet.For mesons you can decompose tensor Tij into a traceless part and trace itself which are octet and singlet respectively,you can not reduce it further.In the baryon case,you can just make a young tableau to find irreducible representation.

6. Jan 5, 2014

### The_Duck

OK, here's my take on how we identify which meson is which in the $3 \otimes \bar 3 = 8 \oplus 1$.

First, note that we expect there to be a very light triplet of mesons with the following quark content: $u \bar d, u \bar u - d \bar d, d \bar u$. These should have almost identical mass, because the SU(2) symmetry relating u and d quarks is extremely good, much better than the SU(3) symmetry that includes the s quark. Looking at the quark contents, the electric charges of these mesons should be $+1, 0, -1$. Looking at the pseudoscalar meson spectrum, we indeed find a triplet of three very light mesons with almost identical masses and charges $+1, 0, -1$. They are the pions, $\pi^+, \pi^0, \pi^-$. So we have figured out the quark content of the pions.

Next, consider $u \bar s$ and $d \bar s$. These should be two mesons that are somewhat heavier than the pions, because the s quark is somewhat heaver than the u or d. But the $s \bar u$ and $s \bar d$ should be quite close to each other in mass, because the $u$ and $d$ both have tiny masses. The electric charges of these mesons should be $+1$ and $0$. We can indeed find a pair of mesons in the spectrum satisfying these properties: the kaons $K^+$ and the $K^0$. Their antiparticles $s \bar u$ and $s \bar d$ are the $K^-$ and $\bar K^0$.

By identifying the $\pi^+, \pi^0, \pi^-, K^+, K^-, K^0,$ and $\bar K^0$, we've knocked out 7 of the 9 linearly independent quark contents in the $3 \otimes \bar 3$. The remaining two are

$\eta_8 = u \bar u + d \bar d - 2 s \bar s$

and

$\eta_1 = u \bar u + d \bar d + s \bar s$.

We write them in this kind of weird way so that each of them has definite transformation properties under SU(3). The first one is part of the octet, and can be transformed into any of the previous 7 by an SU(3) transformation. The last one is a singlet, and is completely invariant under any SU(3) transformation. That's why I called them $\eta_8$ and $\eta_1$. We're left to figure out how to identify these two quark contents with the actual particles we observe experimentally, which are called the $\eta$ and the $\eta'$. Both of these are electrically neutral, as they should be.

Here things are a little messier, as Vanadium 50 said above. The problem is that SU(3) symmetry is not exact. If it were exact, here is how things would turn out. The $\eta_8$ and $\eta_1$ would be the physical particles. Originally, it was thought that both the $\eta_8$ and the $\eta_1$ should be light, because they should both be "pseudo-Goldstone bosons" of spontaneously broken symmetries (as the 7 other light pseudoscalar mesons are), and pseudo-Goldstone bosons are very light. But eventually it was realized that for technical reasons the $\eta_1$ does not actually correspond to a spontaneously broken symmetry--there is an "anomaly"--and so the singlet $\eta_1$ should be significantly heaver than the octet $\eta_8$.

So you say: "aha, looking at the particle masses I see that the $\eta'$ is much heaver than the $\eta$, so $\eta' = \eta_1$ and $\eta = \eta_8$." Well, to some approximation this is true, but actually the SU(3) symmetry on which the above considerations were based isn't exact. The consequence is that in reality the $\eta'$ is some linear combination (superposition) of the $\eta_1$ and $\eta_8$, and so is the $\eta$. I think it is true to say that the $\eta'$ is *mostly* $\eta_1$ and the $\eta$ is *mostly* $\eta_8$, which conforms with the idea that the $\eta_1$ should be rather heavier than the $\eta_8$.

Last edited: Jan 5, 2014
7. Jan 5, 2014

### gildomar

For the SU(3) transformations, how accurate would be think of them as rotations of the of a vector in "flavor space"? That is, any one of the mesons in the octet could be changed into another one in the octet by a rotation of the vector that defines the meson within "flavor space"? And then the singlet wouldn't be able to be gotten by any kind of rotation of the octet vectors, kinda similar to spin states with a triplet and singlet.

8. Jan 5, 2014

### The_Duck

This is accurate, as long as you keep in mind that the "flavor vectors" are *complex* vectors (i.e., vectors with three complex components), which is why the group is SU(3) instead of SO(3).

Yes, right! And if SU(3) were an exact symmetry, physics would have to be unchanged under these rotations. So for instance if SU(3) were an exact symmetry, all the mesons in the octet would have exactly the same mass. The extent to which they actually have different masses is a measure of how badly the SU(3) symmetry is broken by the differences between the u, d, and s quark masses.

Exactly. You're right to make an analogy to spin representations. When we add the angular momenta of two spin-1/2 particles, we do the SU(2) group representation decomposition $2 \otimes 2 = 3 \oplus 1$, decomposing into a triplet and a singlet. This is the SU(2) analog of the SU(3) formula $3 \otimes \bar 3 = 8 \oplus 1$.

9. Jan 5, 2014

### gildomar

Thanks! That clears things up, since I was having a hard time visualizing what was going on of how the octet mesons were related in comparison to the singlet. Even though it was basically the same thing that andrien said, I think that I had to explicitly think of it in vector terms first in order to make sense of it, since it's much easier to understand as just rotations of a (complex) vector in flavor space (broken symmetry notwithstanding).

I suddenly find it amusing trying to explain to nonphysicists that the concept of a "vector in flavor space" is a real thing (more or less) in quantum mechanics :tongue: