# 5 rep and 10 rep etc

1. Oct 1, 2005

### robousy

Lets say I am looking at SU(5) and I draw a (young taleux) box and I say thats the 5 rep of SU(5).

Then I draw 2 boxes stacked on top and I say thats the 10 rep of SU(5)...

What does this mean!!!?

I know how to calculate the dimension of a young tableux but I don't know what it means in terms of physics and particles.

Can anyone shed some light on this for me??

2. Oct 3, 2005

3. Oct 3, 2005

### samalkhaiat

In order to have a well-defined behaviour with respect to a symmetry group, an OBJECT must belong to a representation (multiplet). If the representation is reducible, the OBJECT is composite, in the sense that it can be broken into more elementary objects belonging to irreducible representation. So, elementary "PARTICLES" must belong to irreducible representations. i.e. they must be grouped (classified) in multiplets of the symmetry group. Each multiplet is charactrized by the eigen values of CASMIR OPERATORS of the symmetry group.
IN THE NEXT POST I WILL CONSIDER FLAVOUR-SU(3) AS AN EXAMPLE.
NOW I NEED TO DRINK SOME COFFEE.

4. Oct 3, 2005

### samalkhaiat

OK, I DRANK MY COFFEE.
THE DEFINING (FUNDAMENTAL) REPRESENTATION OF FLAVOUR-SU(3) IS CARRIED BY 3-COMPONENT OBJECT(spinor),
q = (u,d,s) [a box in young tableaux Dim=3]
note that flavour su(3) is an approximate symmetry because the masses of the u,d and s quarks are not equal.if we ignor this, then q transforms as,
q ----> qU,
where U is 3x3 unitary matrix with unit determinant i.e. belongs to su(3).
In contrast to su(2), su(3) has a cojugate REP [3*,or 2 boxes on top of each other] which is a different irreducible REP of su(3). It contains the antiquarks:u*,d*& s*.
q* is a 3-component column, it transform as
q*---> U*q*.
we can get higher-dimensional IRR. REP. by taking the tensor products of the 3 and 3*, and projecting out the invariant subspces. In physics, this corresponds to qq* picture of mesons (psedoscalar & vector). YOU CAN DO IT. JUST USE YOUNG TABLEAUX. SO:
3 X 3* = 8 + 1
Which means
qq* = mesons octet + mesons singlet
of course the octet is traceless 3x3 matrix, and the singlet is (uu*+dd*+ss*)
NOW YOU DO THE SAME WORK FOR BARYONS=qqq.
3 x 3 x 3 = 8 + 8 + 10 + 1 .

regards

sam

5. Oct 4, 2005

### robousy

Thanks so much for the response.

I am starting to 'get' things - or rather, they are becoming slowly clearer.

You mentioned that q is an fundamnetal rep of SU(3) - is this the same as irreducible?

Also - I understand that the {3} rep is just a box in the youngs tab. (dim = 3)

and that the 3* is 2 boxes on top of each other, and also has dim = 3.

...which represents the antiquarks.

however, you are free to create any number of types/shapes of boxes.
Do ALL of these combinations correspond to something?

eg, do 7 vertical boxes (dim = 1) correspond to a physical interaction?

Thanks again for taking the time out to answer me.

Last edited: Oct 4, 2005
6. Oct 4, 2005

### Norman

You cannot have more rows than the value of N in S(N) for a given theory. Most books on Lie Algebras will explain Young Tableau. Just remember that you always have to connect back to the physical world when talking about this stuff. When you are talking about antiquarks in SU(3) are you talking color SU(3) or flavor SU(3)? These are questions you need to consider when trying to relate the math (representation theory) to physics.
Cheers,
Ryan

7. Oct 4, 2005

### robousy

SU(3) flavor...well for this example - but thats just to get a handle on the physics.

So the questions should NOT be what particle does a given arrangement give me, but should be what is the arrangement of the particle set that I know exists in my model (eg standard model).

8. Oct 4, 2005

### Haelfix

Yea and thats model dependant, for instance its not uncommon to invoke a 'horizontal' symmetry group relating generations of particles (flavors but mixed between leptons and fermions).

In some string theories this can be quite complicated, for instance SO(10) horizontal might show up, but then you have suppressed (sometimes twisted) sectors and you don't span the full representation.

But yes usually when we are drawing Young Tableux's in physics we are either talking about gauge groups or the poincare group (or rather its double cover).

And then its all about how you identify particles therein. For instance in SO(10) its nice because fermions and leptons both sit in one and the same irreducible rep (+1 exotic, the sterile neutrino). It can get a little confusing sometimes, b/c its often the case that thats *not true*. For instance the Higgs sector is often bumped up into much higher representations (instead of the 16 of SO(10) its more like 127 of SO(10).

So you do have to keep in mind the model dependancy in what you are talking about, otherwise it really is quite nonsensical. The literature often jumps between models almost at will (in the same paragraph) and it can be confusing keeping things straight.

9. Oct 4, 2005

### CarlB

10. Oct 5, 2005

### robousy

Thanks so much Carl, Haelfix, Norman and samalkhaiat. This is really clearing things up.

While we are on the subject I have a couple more questions.

I was also wandering the significance of RANK.
I know that in any gauge group the rank corresponds to the number of commuting generators, which make up the Cartan Subalgebra - but these are just words to me.

What is the significance of this algebra and why is RANK an interesting quantity.

I think it might be something to do with constructing charge from the generators but Im not sure.

Thanks in advance again.

...Also, carl in your last post you quoted an interesting example where you had reducible reps on one side and irreducible reps on the other side of the equation. Sorry for being slow here but what was noteworthy about this...was it the fact that reducible can always be broken down into irreducible?

11. Oct 5, 2005

### CarlB

The Rank is the same as the number of quantum numbers you need to describe the states in any particular irrep. For example, in SU(2), the rank is 1 and you need only one quantum number to describe, for example, spin in the z direction. That is, if the irrep you're talking about is spin-2, then there is only one quantum number needed to distinguish between the elements, and that number can be 2, 1, 0, -1 or -2.

In SU(3), the rank is 2, so instead of a linear set of states, each irrep will require two quantum numbers and the set of states is planar.

I'm not sure what the mathematical requirements are for which reps are reducible and which are not, but for the usual goups we deal with, I think that reducible groups can always be put into irreducible form. But mostly I was making an unnecessary editorial comment.

What I'm trying to say about reps being more important than irreps, is that the irreps are just a way of solving a mathematics problem. Nature is not fundamentally described by irreps. She's described by representations, as with the SU(2) orbital angular momentum example.

We use irrep group theory because we don't know what the rep is. As soon as we find the rep, the irrep theory will melt away the same way that the old matrix mechanics melted away when Schroedinger's wave equation appeared.

SU(3) is a sign of our ignorance of the physics of quarks, not a success that we can improve by guessing yet another irrep. The irreps are what happen when you break a physics situation down to its mathematical nuggets, they are not the physics situation itself. To advance physics past the standard model, I believe we have to look for representations of the physics not irreps of it.

Carl

12. Oct 6, 2005

### samalkhaiat

Yes, mathematically the fundamental REP is irreducible.
Look, as I mentioned in post#4, the FUN REP of su(n) is carried by n-component object i.e n-vector in the abstract space of su(n). Apart from the trivial REP (= Idintity = 0-vector = scalar = etc), there is no ([n-m])-vector in this space. So the space spaned by the FUN REP is smallest irreducible representation space.
Remember; Young assigns a box for the FUN REP, and young tableaux has no half box.
That was math. In physics, flavour-su(3) for example, we say that the (u,d,s) quarks are the FUNDAMENTAL block of matter, i.e truly irreducible elemantary particle.
THE DISTINCTION IS THIS;
FUN REP is IRR REP, But IRR REP need not be FUNDAMENTAL ONE.

13. Oct 6, 2005

### samalkhaiat

CARL,
Are you suggesting that we should study the block of matter not the building block of matter?
Sir, the matter we see in nature is reducible to its irreducible & fundamental parts, fermions and bosons.
You and me are made of protons and neutrons which are members of the IRREDUCIBLE octet,{8}, in;
q x q x q = {10} + {8} + {8} + {1}
And all particles of the IRREDUCIBLE REPRESENTATIONS {10},{8},{8} and {1}discovered 40 years ago.
The fact that we,now,have QCD is down to one particle in the{10} IRR REP.(DELTA++ which is (uuu) quark-state, unless we colour the quarks, the delta++ violates Paule's principle, this lead to QCD).
As for "the old matrix mechanics" , Do you know, in QFT, we use ,
dA/dt = [ iH , A] (the Heisenberg equation of the matrix mechanics),
more than Schrodinger's equation.So matrix mechanics did not "melt away". But, What is the connection of THIS to the theory of representations (IRR or not) of some symmetry group, which is the subject of this thread?

14. Oct 7, 2005

### CarlB

This is exactly what I was referencing. The old matrix mechanics disappeared from those places that the Schroedinger equation could explain. I am saying that the modern QFT is likely similar to the old matrix mechanics, that is, doomed to melt away when a wave version comes along that ties the various irreps back together into a single wave theory.

The comment is with respect to the question of where do irreducible representations come from. Why are they so useful in physics? If one looks at the Schroedinger equation and compares it to the old matrix mechanics, one sees that the irreducible representations correspond to solutions of systems of differential equations that possess certain symmetries.

For example, the solutions to the differential equation:

$$(-\frac{\hbar^2}{2m}\grad^2 + V(r))\psi = i\hbar\partial_t \psi(r,t)$$

where $V(r)$ is a central force can be classified according to irreducible representations of SU(2), however, the heart of the physics is the above, not SU(2) in the following sense: If you know the above equation, you know everything about all the solutions. If all you know is that the solutions form irreducible representations of SU(2) your understanding is rather barren.

Carl

15. Oct 7, 2005

### Haelfix

I agree with Carl in the following sense. Its always better to know explicit wavefunctions for a theory. For instance, know one knows the precise wavefunction of say the charm quark, only a functional approximation.

But then again, we probably never will, calculating it involves a many body process and its doubtful we will ever be able to completely untangle that mess.

We do know a few things about the wavefunction though, and thats due to symmetry and quantum field theory (the best many body formalism ever invented) and it has allowed us to make predictions and calculate in most conceivable spots.

Now, as to why the schroedinger picture is better than the Heisenberg picture I will remain silent. At a physicists level of rigor (and under strong mathematical axioms) they are entirely equivalent and the choice is merely one of calculational ease.

16. Oct 8, 2005

### samalkhaiat

Look Carl, you should know that in 1926 Hilbert proved that schroginger equation is, mathematically & physically, equivalent to Heisenberg equation.
In QM we CHOOSE to work with Schrodinger equation simply because differential equations are easier than matrix equations. Ironically, the opposite is true in QFT.
I want to know, where did you hear or see that some one used solutions of Schrodinger equation to show SU(2) symmetry?
If this was true, then we did not need to bother ourselves and invent the great theory of Quantum fields. Let me explain why. Suppose you are right.
So you should be able to explain why the proton & the neutron have small mass difference, After all the (P,N) form an SU(2)-doublet. Can you? No. Can any great physicist, we know, explain that? No, no body can.
Indeed, NEITHER Schrodinger EQ NOR Heisenberg EQ can explain this fact. Only QFT with its symmetry principles can explain N-P mass difference.
(since this thread is about the connection between particles and representations, I will, may be tomorrow, explain how does QFT explain the above fact).
DO you know that the non-relativistic Schrodinger EQ you mentioned in your last post (i.e QM) does not explain ordiary spin, let alone isospin, indeed spin introduced into QM in an AD HOC manner. Only Dirac equation contain the spinors naturally.
I hope you are aware of the difference between spinors & isospinors.
Spinors belong to a Lie groups acting on the spacetime manifold, like Lorentz group SO(1,3)(this is why spin doesn't show up in Schrodinger equation).
Isospinors belong to the isospin group SU(2) which act in the internal space of elementary particles.
So,CARL DONN'T EXPECT TOO MUCH FROM A VERY LIMITED CONCEPT.
WE CAN NOT IMPOSE ANY LIMIT ON NATURE, SHE DOES THAT HERSELF.
I THINK WE ARE GOING IN THE RIGHT DIRECTION BY PROGRESSING FROM QUANTUM MECHANICS TO QUANTUM FIELD THEORY.

REGARDS

Sam

Last edited: Oct 8, 2005
17. Oct 9, 2005

### CarlB

Of course I'm aware of this. It's the most basic part of the history of the field. Nevertheless, the fact is that Schroedinger's equation is going strong while the old matrix mechanics is barely taught.

This is partly true and partly false. There are parts of QFT that can be written in differential equations. In particular, QFT was developed by condensed matter theorists who had difficulty solving Schroedinger's equation. However, the QFT for elementary particles has no differential equation underneath it, except to the extent that one can copy the Dirac equation over for each particle. To unify the particles in a differential equation requires that we have a single equation where the various particles are excitations and this has not yet been found.

I didn't quite say that anyone "used solutions of Schroedinger's equation to show SU(2) symmetry," though it's certainly possible. What I wrote was that the solutions to Schroedinger's equation "can be classified according to irreducible representations of SU(2)." This is a well known fact. For the readers who didn't take an undergraduate QM class:

Spherical Harmonics ...
The functions in this table are placed in the form appropriate for the solution of the Schrodinger equation for the spherical potential well, but occur in other physical problems as well.

http://hyperphysics.phy-astr.gsu.edu/hbase/math/sphhar.html

The symmetry properties of the two-sphere are given by the Lie groups SO(3) and its double-cover SU(2). The spherical harmonic transform under the integer-spin representations of these groups; they are a part of the representation theory of these groups.
http://www.algebra.com/algebra/about/history/Spherical-harmonics.wikipedia

It's interesting that elementary particle theorists didn't invent the theory of quantum fields. It was invented by condensed matter theorists who wanted to solve problems that were written using Schroedinger's equation but where the number of particles was too large to be easily solved. It was only then that the method was copied over into elementary particles.

From the point of view of the condensed matter theorists, the Schroedinger equation is more fundamental than QFT, which is only an effective theory. If the discovery of QFT had preceded that of QM, perhaps today's elementary particle theorists would be more inclined to look for a version of QM that would make the current standard model QFT, into an effective theory.

The N and P are known to be composite particles. It's well known that the masses of the elementary leptons enter into the standard model as arbitrary parameters. A better example would be the SU(2) rep of the electron and its neutrino. The standard model does not provide an explanation for their masses. Arxiv.org is littered with papers attempting to reduce the number of arbitrary mass parameters in the standard model.

Of course I know these things. I took a 990 in the GRE advanced test in Physics. I'm abd PhD in Physics with a concentration in elemetary particles and fields at U. Cal., Irvine.

Carl

18. Oct 10, 2005

### samalkhaiat

19. Oct 10, 2005

### CarlB

A unification would be a single differential equation. The condensed matter theorist's version of QFT is an effective field theory based on a single Schroedinger's equation. What I would like is a single differential equation underlying the standard model.

Ooooops. What I should have written was that the concepts of renormalization and spontaneous symmetry breaking, which are so critical to the standard model's QFT, were first discovered by condensed matter theorists working with a field theory that was based on Schroedinger's equation. They weren't even doing relativistic work.

I agree that algebra is the key, but my guess is that it will be a continuous algebra. I'm going to the PANIC meeting in Santa Fe and will give a poster on how one can break the symmetry of the Geometric Algebra and get a result with the correct symmetry for the leptons. It's basically Clifford algebra stuff. Hope to see you.

I don't think we're at all in disagreement here. To me, the Dirac equation is a differential equation. It is written in terms of an algebra, the Dirac algebra, which is equivalent to a Clifford algebra $\mathcal{CL}(3,1)$. In other words, the algebra defines a differential equation.

Any rep of SO(3) is also a rep of SU(2) and therefore possesses SU(2) symmetry as well as SO(3) symmetry. The concept you're thinking of is "faithful", and I agree that SO(3) reps are not faithful reps of SU(2). But they are, in fact, reps. But I didn't mean to distinguish between Schroedinger's, Pauli's and Dirac's equations.

There is an excellent introduction to the relationship between these equations by Hestenes. In it, he shows that Schroedinger's equation is not the equation for a spinless particle, but instead is the equation for a spin-1/2 particle with the spin stuck in one position. It's a fascinating read if you're interested in algebra:

Consistency in the formulation of the Dirac, Pauli, and Schroedinger theories
Consistency with the Dirac theory is shown to imply that the Schroedinger equation describes not a spinless particle as universally assumed, but a particle in a spin eigenstate. The bearing of spin on the interpretation of the Schroedinger theory discussed.
http://modelingnts.la.asu.edu/pdf/Consistency.pdf [Broken]

Despite being weaned on spinors, I'm beginning to suspect that the density operator formalism is more fundamental. With density operators, the arbitrary phases in a spinor are cancelled out so it seems like you end up collapsing an SU(2) rep back down to an SO(3) rep. However, it turns out that you don't actually lose complex phases as can be seen if you consider products of Stern-Gerlach type projection operators. For example:

$$(1+\sigma_z)(1+\sigma_x)(1+\sigma_y)(1+\sigma_z)$$
$$= \frac{1+i}{4}(1+\sigma_z)$$

Thus the effect of running a spin+1/2 beam of fermions through the above sequence of Stern-Gerlach filters would be to introduce a complex phase of 45 degrees (and a reduction in intensity).

I've only been thinking about this off and on for the last few weeks so please comment.

Some readers may be unfamiliar with density operators so:
This indistinguishability makes the density operator a perhaps more intuitive representation of a state than the state vector. For example, a state vector has an arbitrary overall phase factor, so two vectors differing by a phase factor represent the same state. Still, when the vector is put together with its dual vector to form a pure state density operator, the phase factor vanishes because of the complex conjugation. Hence, all state vectors representing the same state are represented by the same density operator.
http://arxiv.org/PS_cache/quant-ph/pdf/0408/0408094.pdf [Broken]

I'm not sure what you're asking for here. In geometric algebraic language, the massless Dirac equation is simply

$$\nabla \Psi = 0$$

which is about as simple a differential operator as one can hope for. The above differs from the usual Dirac equation in that it includes a set of four (independent) spinors in one equation. If you want to pick out just one of these, which is identical to the Dirac equation, then you multiply on the right by your favorite idempotent:

[tex] (\nabla \Psi) \iota = \nabla (\Psi \iota) = \nabla \psi = \gamma^\mu\partial_\mu \psi = 0[/itex]

You can add mass back into these equations if you like. See equation 2.41 in this introductory article for details:
http://arxiv.org/PS_cache/quant-ph/pdf/0509/0509178.pdf [Broken]

Oh, and "abd" means "all but dissertation". It means I left after I passed the PhD qualifying exams but without a dissertation. It seems to be a habit, as I'm also abd PhD in Math. at U. Washington.

Carl

Last edited by a moderator: May 2, 2017
20. Oct 11, 2005

### Haelfix

Carl, concerning geometric algebra. I reviewed it briefly at some point, and asked various colleagues about it. It seems a rather priveledged methodology that really prefers dimension 3 and 2. I don't quite see what the construction buys you in dim 4+

What does it buy you over the usual presentation of clifford algebra and spin geometry (eg spinor bundles, Pin groups, Dirac operators, Heat kernels, index theory, etc)

A lot of it just seemed like it was trying to retain notions of cross products which as we all know is a priviledged concept.

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