1. Jan 29, 2009

### ismaili

Dear All,

In chap 6, he introduced "Roots and Weights". What I didn't understand is the discussion of section 6.2 about the adjoint representation. He said: "The adjoint representation, is particularly important. Because the rows and columns of the matrices defined by $$[T_a]_{bc} = -if_{abc}$$ are labeled by the same index that labels the generators, the states of the adjoint representation correspond to the generators themselves."
The sentence with underline is the point that I didn't understand. Why states of the adjoint representation correspond to the generators? And then he denotes the state correspond to an arbitrary generator $$X_a$$ as $$|X_a\rangle$$, moreover,
$$\alpha|X_a\rangle + \beta|X_b\rangle = |\alpha X_a + \beta X_b\rangle$$
Could anybody show me why any state in the adjoint representation would correspond to a generator? Thanks a lot!

2. Jan 30, 2009

### nrqed

Consider a column vector "V" which would be in the adjoint representation. Then we would apply the matrices T_a on v to produce a new vector "v prime". Writing this explicitly, we would have

$$v'_b = (T_a)_{bc} v_c = -i f_{abc} v_c$$

Now consider the following. Instead of incorporating the coefficients V_c of the initial vector in a column vector, construct the matrix $V_c T_c \equiv \vec{v} \cdot \vec{T}[/tex]. This is now a matrix that represents the state with components V_c (instead of a column vector). Now, we want to apply a transformation (using T_a) to this state. Instead of just applying the matrix T_a to our "state", we will say that to do a transformation, we must take the commutator of the state with the matrix producing the transformation. So we say the the transformed "state" is given by (warning: I am doing this by memory, I am pretty sure I will get some minus signs wrong) $$\vec{v}' \cdot \vec{T} = [T_a, v_c T_c] = i v_c f_{acd} T_d$$ where I have avoided using an index "b" on the right side to make things more clear. Ok, now, let's say that we want the coefficient [itex] v_b'$ (to compare with the formula obtained above using column vectors). This is the coefficient of the matrix $T_b$ on the left side. So we must set d=b on the right side too. We get

$$v'_b = i v_c f_{acb}$$

Using the antisymmetry of the structure constants, we finally get

$$v'_b = - i f_{abc} v_c$$

which is the same result that we obtained using states being column vectors.

So in this representation, we can use the generators themselves to build the states!
But again, this also implies that we use commutators to "apply operators to the states".

3. Jan 30, 2009

### ismaili

Thank you very much!

So, given a state under some basis in the adjoint representation(then this state is a column vector $$v_a$$), equivalently we can get the adjoint representation by forming the linear combination of generators, $$v_aT_a$$, to be the states, and by adopting the transformation rule that defined by commutator as you described.
So, in a sense there is a map from states to generators in adjoint representation

$$v_a \rightarrow v_aT_a$$

And probably this map is one-to-one so that Georgi used the generator to label the state, $$|X_a\rangle$$, where this state corresponds to generator $$X_a$$.

I think basically I got the idea. (Is this so trivial? Georgi didn't state it at all...)

-------------------------------

However, when I read section 6.1 just now, I found something I don't really understand.
He said "Cartan generators can be simultaneously diagonalized. After diagonalization of the Cartan generators, the states of the representation $$D$$ can be written as $$|\mu,x,D\rangle$$, where
$$H_i|\mu,x,D\rangle = \mu_i|\mu,x,D\rangle\quad---(*)$$
($$H_i$$ is the hermitian Cartan generator) and $$x$$ is any other label that is necessary to specify the state
."

My question is, the $$|\mu,x,D\rangle$$ should be the basis for the states of the representation, right? Since not all states in the representation can have eq(*), for example, linear combination of $$|\mu,x,D\rangle$$ can not have eq(*). So, I don't understand why every state in the representation can be written as $$|\mu,x,D\rangle$$?

Thanks for any illumination!

4. Jan 30, 2009

### nrqed

You are welcome. Well, the first time I "got it", I did not find it a trivial concept.

I have to say that I never really found Georgi's book to be very clear. I tried to learn from him a few times and ended up using other references.

Yes, he means the basis states.

5. Feb 10, 2009

### Atakor

Hi,
I just got this book (Georgi's) .. I dont find it very clear too.
Could you tell me what are these ''other references'' you found useful ?

I have also "Relativity, Groups, Particles - Saxl/Urbantke".. looks interesting.

Thanks.

6. Feb 13, 2009

### nrqed

As a starting point I think that nothing beats Greiner's book ''Symmetries'' (I think that the full title is ''Quantum Mechanics: Symmetries''). There is an appendix that discusses roots, weights, the Cartan classification and so on and it is very clear. If you find good references, let us know!

7. Feb 13, 2009

### JosephButler

Other useful references are:

* Cahn's semisimple lie algebras book. It's now available free from the author at http://phyweb.lbl.gov/~rncahn/www/liealgebras/book.html , or alternately via a very inexpensive Dover edition.

* Jan Gutowski's "Part III: Symmetries and Particle Physics" notes, available on his webpage: http://www.mth.kcl.ac.uk/~jbg34/Site/Dr._Jan_Bernard_Gutowski.html [Broken]

Last edited by a moderator: May 4, 2017
8. Feb 16, 2009

### Atakor

Hi,
Thanks for the ref's.

I asked also a theoretical physicist who's basically "speaking groups"..
according to him and his friends, there's no good self-contained references for group theory (for physicists)..
but he thinks that Georgi's is the best one.. when used with Group Theory and Its Application to Physical Problems- Morton Hamermesh.
I just ordered the latter one..