Why Group Theory Works: SU(6) & Quarks

[RedX]

Why is it that the group SU(6) gives you the correct wave function for hadrons made up of the u, d, and s quarks? The 6 elements in the fundamental representation would be u up, u down; d up, d down; s up, s down. What's the meaning behind SU(6)?

Also, for 6 quarks, would it be SU(12)?

 

[tom.stoer]

For N_f quarks flavours you have to use SU(N_f). That means for the three quark flavours u,d,s, you have to use N_f=3.

 

[martinbn]

tom.stoer, isn't there spin involved? Two spin states per flavour and that's why SU(6).

 

[RedX]

For N_f quarks flavours you have to use SU(N_f). That means for the three quark flavours u,d,s, you have to use N_f=3.

If you just consider u,d,s, then SU(6) with the 6 components u +-, d+-, s+-; where + and - refer to spin up or spin down; gives the correct wavefunctions including spin.

If you just consider SU(3) with u,d,s, then this doesn't take into account spin, and you get weird things such as separate octets for spin 0 mesons and spin 1 mesons, or for baryons the fact that one of the octets is missing from 3x3x3=1+8+8+10.

But for some reason when you use SU(6) with the spins for u,d, s, then all that is solved (I think).

But why SU(N) or SU(2N)? I understand that you want to say that all particles are the same, just viewed from a different set of axis, and group theory provides such a transformation of the axis. But why not U(N) or U(2N)?

I guess the answer is that other groups don't reproduce the multiplet structure. But is it a coincidence that SU(N) works for N=3, and then when we found 4 quarks, SU(4) worked, and then 5 quarks, SU(5). Then we decided to add spin, and it so happened SU(6) worked to give the right multiplet structure for u,d,s? Does nature seem to like SU(N) structures?

There is one catch though. If u,d,s is represented by SU(6), then adding color doesn't change it to SU(18). It seems nature prefers SU(6)xSU(3), so that although you can no longer write the state in SU(3) as :

$$|flavor>|spin>$$

and instead must consider SU(6) as:

$$|flavor+spin>$$

when including color you can still write as a direct product:

$$|flavor+spin>|color>$$

as opposed to SU(18):

$$|flavor+spin+color>$$

 

[tom.stoer]

Sorry, but "6" has nothing to do with "2*3" due to spin and flavor!

Spin and isospin do not go together into one single group. Spin is due to the Spin(4) ~ SU(2)*SU(2) ~ SO(3,1) symmetry of spacetime. So taking spin into account you get SU(N_f) * Spin(4).

In QCD you have to introduce another quantumnumber, called color. Again it comes in a three-dim. fundamental rep. of an SU(3) group, but now this is SU(3_c). So in total you get SU(3_c) * SU(N_f) * Spin(4).

Considering SU(6) is not due to SU(6) = SU(2*3) where 2 is related to spin and 3 is related to flavor. You get SU(6) = SU(N_f) by taking into account all N_f=6 flavors u,d,c,s,t,b.

btw.: you must never do something like SU(M) * SU(N) ~ SU(M*N); this is wrong as can be seen by looking at the dimension the groups; dim SU(N) = N²-1; therefore for SU(M) * SU(N) you get dim [SU(M) * SU(N)] = M²-1 + N²-1 whereas for SU(M*N) you get dim SU(M*N) = (MN)²-1

 

[RedX]

Spin and isospin do not go together into one single group. Spin is due to the Spin(4) ~ SU(2)*SU(2) ~ SO(3,1) symmetry of spacetime. So taking spin into account you get SU(N_f) * Spin(4).

In QCD you have to introduce another quantumnumber, called color. Again it comes in a three-dim. fundamental rep. of an SU(3) group, but now this is SU(3_c). So in total you get SU(3_c) * SU(N_f) * Spin(4).

Considering SU(6) is not due to SU(6) = SU(2*3) where 2 is related to spin and 3 is related to flavor. You get SU(6) = SU(N_f) by taking into account all N_f=6 flavors u,d,c,s,t,b.

btw.: you must never do something like SU(M) * SU(N) ~ SU(M*N); this is wrong as can be seen by looking at the dimension the groups; dim SU(N) = N²-1; therefore for SU(M) * SU(N) you get dim [SU(M) * SU(N)] = M²-1 + N²-1 whereas for SU(M*N) you get dim SU(M*N) = (MN)²-1

Well just to show you that I'm not completely crazy, here's Wikipedia:

http://en.wikipedia.org/wiki/Quark_model#Baryons

"approximate symmetry is called spin-flavour symmetry".

The 56 comes from the symmetric state in SU(6): 6*7*8/3!

For some reason they're taking the direct product of spin and flavor (what you're calling isospin) and combining them into one larger group SU(6), and then they are saying that: SU(6)$$\supset$$SU(3)xSU(2). They are assuming just u,d, and s quarks here.

 

[tom.stoer]

I see that for the very first time.

What would happen for u,d quarks with SU(2) isospin and SU(2) spin? Would you get SU(4) spin-isospin? and how do proton, neutron etc. fit into SU(4)? What happens to higher flavor groups, i.e. full for all SU(6).

What about the dimension?

dim SO(3,1) = dim SU(2) + dim SU(2) = 2*3 = 6
dim SU(3) = 3²-1 = 8

=> dim [SO(3,1) * SU(3)] = 6 + 8 = 14

but

dim SU(6) = 6²-1 = 35

 

[RedX]

I have no idea. Just to give you an idea of things you can do in this SU(6) model, consider this wavefunction, where arrows refer to spin.

$$(ud-du)(\uparrow \downarrow-\downarrow \uparrow)$$

In order to get a spin up proton, you need to add a u-quark with an uparrow.

Now normally if you just consider the direct product of flavor and spin, then your choices are limited on how you can add the u quark. In fact it would be:

$$(\mbox{mixed symmetry permutation with u,u, d})(\mbox{spin part})$$

But with the SU(6) model you have the freedom to put the u-up quark anywhere! So the first term in the expression:

$$(ud-du)(\uparrow \downarrow-\downarrow \uparrow)$$

is $$|u\uparrow d\downarrow>$$, coming from multiplying the ud in the first factor with the $$\uparrow \downarrow$$ in the second factor. What you do now is insert a u-up in every spot in this term, to get:

$$| u\uparrow u\uparrow d\downarrow>+| u\uparrow u\uparrow d\downarrow>+| u\uparrow d\downarrow u\uparrow>$$

Now do this for all the terms in

$$(ud-du)(\uparrow \downarrow-\downarrow \uparrow)$$

and you get the proton wavefunction with spin, and I don't think it can be written as a direct product of flavor and spin (note that the total wavefunction done this way is symmetric, but color makes the baryon antisymmetric)! What's amazing is that it predicts that the d-quark is twice as likely to be spin down than spin up! So a spin up proton most likely has its d-quark spin down. With this knowledge, you can calculate the magnetic moment of a spin up proton as the sum of the magnetic moment of the quarks, but with the configuration where the d quark is spin down twice as likely:

I showed the calculation here:

https://www.physicsforums.com/showthread.php?t=479196

Anyways, what also makes this confusing is the fact that if you swap a muon and an electron, then you don't get a negative sign. Why is it that if you swap two different quarks you get a negative sign?

 

[RedX]

I tried the calculation again, this time with:

$$(ud+du)(\uparrow \downarrow+\downarrow \uparrow)$$

or the direct product of two symmetric states instead of two antisymmetric states, then I symmetrized it all with a u-up quark. It can be easily seen that you get the d quark with spin down by multiplying the first term with the third term, and the 2nd term with the 4th term in the above product, and then symmetrize evertying with a u-up quark. So changing the + signs above into minus signs shouldn't change the probability the d-quark spends as spin up or spin down. Anyways, the result is:

$$2(u\uparrow u \uparrow d \downarrow+ u \uparrow d \downarrow u\uparrow + d \downarrow u\uparrow u \uparrow ) \pm (u\uparrow u \downarrow d \uparrow+u\downarrow u \uparrow d \uparrow+u\downarrow d \uparrow u \uparrow + u\uparrow d \uparrow u \downarrow+ d\uparrow u\uparrow u \downarrow+ d\uparrow u \downarrow u\uparrow)$$

where the + occurs when choosing:
$$(ud+du)(\uparrow \downarrow+\downarrow \uparrow)$$ and the - occurs when choosing:
$$(ud-du)(\uparrow \downarrow-\downarrow \uparrow)$$

I guess it makes sense that there are two ways to do it and hence the +-. In a manner of speaking, you really have 4 particles: u-up, u-down, d-up,d-down. So when you symmetrize everything you have to choose whether you want the single d-quark to be up or down, so this forms two groups.

This is still mesmerizing, but the fact is that it reproduces the correct magnetic moments of all hadrons. The probability that the down quark is spin down can be read from above and is proportional to 2^2*3=12, and spin up is 1^2*6, so that the probability that the d quark is spin down is twice as likely as spin up.

 

[tom.stoer]

I guess it has something to do with the fact that the dimensions of the fundamental representatons match exactly:

dim rep. SU(2) * dim rep. SU(3) = 2*3 = 6
dim rep. SU(6) = 6

Besides that I guess it's a nice accident that cannot be generalized to higher flavor groups.