# Construction of SU(3) multipletts, implication of Gell-Mann-Nishijima

1. Jul 21, 2008

### blue2script

Hi all!

I am currently preparing for an oral exam in quantum field theory and particle physics and I have some problems with the SU(3)-Hadron Multipletts and the relation to the Gell-Mann-Nishijima equation: First, for SU(2) Multipletts you take your Casimir-Operator J, some commuting operator $$J_3$$ and index your states by $$\left|j,m\right\rangle$$. Then, in principle, you get all states of a multiplett by starting with some highest $$J_3$$-state and take the $$J_-$$ operator to generate all states. You know that all states with $$j < m$$ must be zero.

Now, the treatment of the SU(3) multipletts is a lot different and I have only a vague idea why that is the case. For the full SU(3) flavour symmetry you index the states by means of the two diagonal generators $$T_3 = F_3, Y = 2/\sqrt 3 F_8$$ where $$T_3$$ is the third component of the isospin and Y is the hypercharge. Than you introduce the ladder operators $$T_\pm, U_\pm, V_\pm$$. What I don't understand is: You label the states by $$t_3, y$$. Why? Sure, both generators commute, but where are the eigenvalues of the two Casimir-Operators of the SU(3)? In text-books it is argued that the ladder operators above change $$t_3, y$$ and so you construct your nice diagrams in this plane. But why for example do I get a triangle decuplett? How do I know which states give zero under the action of one of the ladder operators?

Why don't I just take the Casimir-Operators of the SU(3) (which I can't find in any book...), take two generators, calculate the eigenvalues and dependences of the eigenvalues of the four commuting generators and construct my decuplett? Especially: What is the implication of the Gell-Mann-Nishijima equation? What does it mean that the eigenvalues of $$T_3, Y$$ give the charge of the particle?

Thank you very much for answering my questions!! Hope I clarified them good enough!

Blue2script

2. Jul 21, 2008

### nrqed

For SU(2), the eigenvalue of teh Casimir operator (which is $$\tau^2 \equiv \tau_1^2 + \tau_2^2 + \tau_3^2$$ ) is $$j_{max}(j_{max}+1)$$ and the raising/lowering operators simply change the value of $$j_3$$ by one unit. We identify a state by providing the values of $$j_{max}$$ and $$j_3$$. A multiplet is identified by providing the value of $$j_{max}$$ only.

I know you know all this, I am just repeating in order to show the corresponding situation in SU(3).

In SU(3), one of the Casimir operators is simply the sum of the squares of the Gell-Mann matrices. I don't remember off the top of my head what the other one is.

Now, a multiplet is identified by providing two numbers, obviously, which are nonnegative integers (the normalization is chosen so that it comes out this way). Let's call them p and q.

The relations that is the equivalent of $$j = j_{max}(j_{max}+1)$$ for SU(2) are

$$p = (T_3)_{max} + \frac{3}{2} Y_{max}$$
and
$$q = (T_3)_{max} - \frac{3}{2} Y_{max}$$

You can of course invert to get the Ymax and (T3)max.

For example, the multiplet (p,q) = (1,0) is the fundamental quark representation (looks like a triangle pointing down) with states

$$(T_3,Y) = (1/2, 1/3), (-1/2,1/3),(0,-2/3)$$

(here the highest weight state is (1/2,1/3))

The other fundamental multiplet is the antiquark multiplet (p,q) = (0,1) with states

$$(T_3,Y) = (0,2/3),(1/2,-1/3),(-1/2,-1/3)$$

Other representations may be built out of tensor products of those two.

For example, consider the tensor product $$(1,0) \otimes (0,1)$$
Graphically, this is done by placing at each vertex of the (1,0) graph the second grpah (i.e. its origin). If you do that you end up with 6 states at the vertices of an hexagon plus three states at the origin (so, of course, 9 states). But this is not an irreducible representation.

So what you do is exactly the same as when you do tensor products with SU(2). For example, when you do the tensor product $$1/2 \otimes 1/2$$ you get four states but they break up into two ireeducible representations corresponding to spin 0 and spin 1. How can we tell that? what we do is that we take the highest weight state (with $$(S_z)_{total} = 1$$ ) and we keep applying the lowering operator. We generate this way the three states corresponding to spin one. Then we build the spin 0 state by making it orthogonal to the S_z=0 component of spin one.

For SU(3), you do the same. You take the highest weight state of $$(1,0) \otimes (0,1)$$. This state is the one with $$(T_3)_{max},Y_{max} = 1,0$$ so it corresponds to a multiplet with (p,q) = (1,1).
Now you keep applying the raising and lowering operators to generate all the states. You end up with eight of them so the (p,q) = (1,1) is an octet.

There remains one linear combination of the states at the origin (T_3=Y=0) which can be built which is orthogonal to the T_3=Y=0 states of the octet. This combination is a singlet which obviously corresponds to (p,q) = (0,0). So the octet is the analogue of the triplet state when you combine two spin 1/2 angular momenta and the singlet is the analogue of the spin 0 state which is also a singlet of course.

So one has

$$(1,0) \otimes (0,1) = (1,1) \oplus (0,0)$$

If you do instead $$(1,0) \otimes (1,0)$$ you get this time the sum of a sextet plus a triplet (well, an antitriplet to be exact):

$$(1,0) \otimes (1,0) = (2,0) \otimes (0,1)$$

Hope this helps

Patrick

3. Jul 21, 2008

### nrqed

I gave details in my previous post but let me emphasize one point. You are right that it is NOT enough to simply give Y and T_3 to specifiy a state completely! This would be like saying a particle has $$s_z = -3/2$$, it could be that the particle is spin 3/2 but it could be spin 2, 5/2, etc.

In addition to giving the values of Y and T_3 one must also specify what multiplet one is working with (the equivalent of providing the spin of the particle for SU(2)). Often people don't do that in SU(3) because the context is clear (if one is working with the octet multiplet or the fundamental 3 or $$\overline{3}$$ and so on). But to be precise, one should also provide two additional numbers (two since SU(3) has rank two and therefore two Casimir operators). Those are the parameters p and q that I explain in my other post. So an SU(3) state is completely specified by providing the values of p,q,T_3 and Y.

Hope this helps

Patrick

4. Jul 22, 2008

### blue2script

Hi Patrick,

thank you very much for your long reply! It clarified a lot to me! I followed along your text and constructed the quark-triplett and the "meson"-oktet. However, there are still some open questions that I stumbled over:

1) You wrote that p,q take over the role of j from SU(2) and that p,q can be written as

$$p = (T_3)_{max} + \frac{3}{2} Y_{max}$$
$$q = (T_3)_{max} - \frac{3}{2} Y_{max}$$

However, what I find for the anti-quark-triplett (p,q) = (0,1) is the following:

$$(T_3,Y) = (0,2/3) \Rightarrow p' = 1, q' = -1$$
$$(T_3,Y) = (1/2,-1/3) \Rightarrow p' = 0, q' = 1$$
$$(T_3,Y) = (-1/2,-1/3) \Rightarrow p' = -1, q' = 0$$

So, what I mean is that the the p,q-values range both between -1 and 1. If e.g. p was something like j, than all p values should be zero.

Another thing is the formula above: the maximum values in the triplett are $$(T_3)_{max} = 1, Y_{max} = 2/3$$. This would give $$p = 2, q = 0$$. Thus the triplett should correspond to (2,0)? Surely not but I guess I miss an important point. It would be great if you could help me out here!

2) Do you know where I can find the second Casimir operator as well as the eigenvalues of the ladder operators of the SU(3)?

3) I know that T3 is the third component of the Isospin and belongs to SU(2). Moreover I found that the strangeness belongs to some U(1) symmetry. Ok, it may be the Nothercharge of U(1). But then one says (e.g. Cheng and Li, p. 115) that both symmetries are not exact but one finds Q = T3 + Y/2 is conserved and belongs to SU(2)T\otimes U(1)Y symmetrie or some similar, higher symmetry. This turns out the be the flavour SU(3). However, I wonder from what group-theoretical point we get the hypercharge Y? In a first approximation it is Y = B + S where B is the baryon number. I guess this hypercharge comes from some inexact U(1) symmetry, right? But then, what group-theoretical role plays the charge here? It seems to play the role of the Nother-charge of the flavour SU(3)? Is that right? If yes, how can I see it is the charge and not anything else? But then, behold, we later expand the hypercharge to top, charm, bottom and so on - however the SU(3) only belongs to u,d,s. A SU(6)-flavour wouldn't make sense, would it?

You see, I am somehow confused - although your post helped me a lot! I am really surprised not to find any reference or similar questions in the net. Am I the only one having trouble with that?

Anyway, any answer to any one of the three questions above would help me a lot!! I hope I wrote down my problems clear enough. A big thanks in advance!

blue2script

PS: Sorry for the bad math-expressions in 3); the tex-system didn't work properly.

5. Jul 22, 2008

### nrqed

I am in a rush right now so I can't get to all your questions. I am not sure how you you got a T_3 max of one!!

You should not calculate p and q using all the values of the states, youmust use only the highest weight state to find p and q. Like in SU(2), you use only j_max to calculate j_max(j_max+1)!. Here the highest weight state is (1/2,-1/3) which gives p = 0, q=1.

I will have to write later

6. Jul 22, 2008

### blue2script

Ok, this is right, it is not Tmax = 1 but +1/2. Then you say that the highest weight is given by the state with the highest T3-value, right? Ok, but why not the maximum of Y? (the maximum would not be unique, ok, but anyway...). And then, why is a state like (1/2,5/3) not allowed? I could get there with some ladder operator. What restricts me, to make question 1) more concrete?

blue2script

7. Jul 22, 2008

### nrqed

If I recall correctly, the rule is tha the highest weight is defined the following way: you take the state with the highest T_3. If there are several such states then you pick the one with the highest Y.

I don't see how you could get from the highest state (1/2,-1/3) to any thing "higher"". No raising operator will get you to (1/2, 5/3). Try it.

8. Jul 22, 2008

### blue2script

If we start with (1/2,-1/3), use U+ and V+ and you will get to (1/2,5/3), don't you? The U+ lowers t3 by 1/2 units and raises y by one unit whereas V+ raises t3 by 1/2 units and again raises y by one unit. So the question is: What restricts me not to do this?

Thank you! If you have some time, maybe you could also take a look at my questions above. Hope I don't get too much on your nerves...

Thank you!

blue2script

9. Jul 22, 2008

### nrqed

Well, let me turn the question around: when we compose two spin 1/2, the highest weight state has S_z =1, right? What prevents me from applysing S_+ on that and getting states with z components equal to 3/2, 2, 5/2 and on and on??

If you answer this question for SU(2) it will tell you the answer for SU(3)

Of course you are not getting on my nerves! I am glad to help. But I haven't used any of that stuff in many years so I am very rusty and I am not sure I will be able to answer all your questions. I will try to get back to them when I have a few minutes free.

Patrick

10. Jul 22, 2008

### nrqed

Just to make something clear: when we say (T_3)_max and Y_max, we mean the values corresponding to the highest weight state. You can't take a value of T_3 from one state and a value of Y from a different state (I just realized that this is waht you were doing in an earlier post). No, you must identify the highest weight state and then take the values of T_3 and Y of that state (even if that might mean that the Y you are using is not the maximum value of all the Ys appearing in the multiplet!!).

For example, if you look at the (0,1) multiplet, the highest weight state is the one with T_3,Y = (1/2, -1/3) so the Y_max you use to calculate p and q is -1/3. But note that there is another state that has a Y of 2/3!! But we still use Y_max = -1/3 in the calculation. SO you should understand the notation Y_max as meaning "the Y value of the highest weight state" and not as meaning "The largest value of Y you can find in all the states". You see what I mean?

So the question is how to identify the highest weight state. And this is what I mentioned in an earlier post: the rule is to look for the state with highest T_3. If it happens that several states have the same highest value of T_3, then you pick of those the one that has the highest Y.

11. Jul 22, 2008

### blue2script

Hmm... in the SU(2) case the Casimiroperator J prevents me from applying S_+ on the highest weight state: We know that the eigenvalues of J_3 can only range between -1 ... 1 (for your example) and so S_3 has to give zero for the S_z = 1 state. But what is the range for the t_3 and y eigenvalues? That is sort of the essence of my question.

Yeah, I guessed that you mean that. But then: It seems as a rule of thumb to me. I can't see where this comes from. Ok, maybe that is to deep analysis so at this level I would also be satisfied with some sort of rules if they neetly fit. So, taking this as a rule of thumb, I am ok on this part. Do you have some better explanation?

What else could I say here than thanks again for taking your free time to help some stranger in the internet ;)! No, really, the discussion with you helps me a lot to clarify these details! It has helped me more than any textbook so far!

Thank you! With best greetings from germany,
Blue2script

12. Jul 22, 2008

### nrqed

I am assuming you meant S_+ instead of S_3 in that sentence?

But there is nothing preventing someone from applying S+ on the Sz=+1 state. The question is : why don't we get a state with Sz=+3/2 then? since we know that S+ raises the value of the z component?
In some sense, you are avoiding the question by stating the final answer that Sz ranges from -1 to +1. But if a student (or a prof!) asks you: what happens if I apply S+ on the highest weight and you say "it gives zero", they could ask you "why does it give zero? Why does it give zero when applied to the highest weight but not zero when applied to, say, the Sz=0 state?

Applying S+ to the highest weigth does not give zero by definition. One can actually perform the calculation and pretend that one has forgotten that it is the highest weight, and the result will come out to be zero. My question is: what makes this happen?

That's a good question. I admit that I don't know the general formula. Unfortunately, I am in the process of moving so most fo my book sare packed in boxes, including the books that could tell me the answer. Most physics books do not give general formula because in physics one usually work only with a few specific multiplets with the quantum numbers of all the states known explicitly.

well, it comes from the full group-theoretical analysis. I guess that one way to convince oneself is to check that all the raising operators will annihilate this state (and this goes back to the question I asked a bit above). I guess that this is the best argument I can think of to motivate the rule.

But you could also see it as a rule of thumb. It's a bit like when we couple two angular momentum j1 and j2 and we say that jtotal ranges from j1+j2 to |j1-j2|. This is not trivial to prove but we soon start using it as a rule of thumb.

This very nice of you to say. It's always a pleasure to help someone who has good questions and who is quick at learning.

Talking about Germany, one of the best book on QCD is the one by Greiner (well, ALL his physics books are truly excellent). He shows a lot of details in the calculations (showing steps that most books just skip completely) and explains things really really well.

Patrick

13. Jul 23, 2008

### blue2script

Yes, sure!

Ok, that is true. The right answer would be that the eigenvalue of the ladder opertor is zero for this state, right? But then: I don't know what the eigenvalues of the ladder operator of the SU(3) are.

Do you have in mind in which books I could look into?

In my search for literature I stumbled over the book of H. Georgi. I think I will go through this book after my exams. He uses the mathematical language of Lie algebras including roots, weights, etc. From the abstract

www.itap.physik.uni-stuttgart.de/lehre/vorlesungen/gtsem/HSGruppen_Berger.pdf

I got some general idea of how the construction works. However, this topic will never be part of my exams (it is far away from elementary particle physics I guess) and it would cost me a day to go through the chapter in Georgi in detail. Thus, I will have to delay this discussion until after my exams. It would also be interesting to see if there is a way to represent the construction in the physical language, so without roots and weights and stuff. But anyway, I think from the pdf above I got the general idea and for the moment I will stick to that.

Haha... in what part of Canada are you? In the english or the french part? For the Greiner: People love or hate him. I for myself have not read to much of him but I definitely will since he seems to have written the only comprehensive and modern introduction into QCD and I also have some oral exam in this topic in mid-october.

But now, before we close the discussion of the SU(3): Do you have an idea about the connection of the Gell-Mann-Nishijima relation with the eigenvalues of the t_3 and hypercharge eigenvalues? What is the deeper reason that both can be combined to give the charge of the state?

This is the last question that bothers me in this context. Good luck for your move! I moved myself just a few days ago ;).

With best wishes to canada et salut!

Blue2script

14. Jan 20, 2009

### cimgo1903

do you have computing of all casimir operators of su(3) in mathematica? i need mathematica codes.

15. Jan 20, 2009

### strangerep

blue2script, cimgo1903,

The two SU(3) Casimirs are given in Greiner & Muller's textbook
"Quantum Mechanics - Symmetries" (p138 in my edition).

As nrqed said, the first Casimir $C_1$is the sum of the squares of
the generators:

$$C_1 ~:=~ \sum_{i=1}^8 F_i^2$$

and the 2nd Casimir is

$$C_2 ~:=~ \sum_{ijk} d_{ijk} F_i F_j F_k$$

where the $d_{ijk}$ are the symmetric
"structure coefficients" from the anti-commutation
relations among the SU(3) generators. G&M give details.

For any physicist trying to understand the "deep group theoretical
reasons" why multiplets are the way they are, G&M's book is very
good. The "reasons" become surprisingly transparent when you see
a pedestrian derivation of the kind that Greiner & Muller provide.
E.g., the derivation of SO(3)/SU(2) quantum numbers j,m in section
2.1 is quite easy to follow. They also give an extensive but
understandable tour of how all the bits of SU(3) fit together in ch7.

16. Jan 21, 2009

### cimgo1903

thanks for everything, strangerep.