• ismaili
In summary, the conversation discussed the adjoint representation in Lie algebra and its relation to generators. It was explained that in this representation, states can be constructed using the generators and transformations can be applied using commutators. It was also mentioned that there is a one-to-one map between states and generators in the adjoint representation.
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!

ismaili said:
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!

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".

nrqed said:
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".

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!

ismaili said:
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...)

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.

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

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!

Yes, he means the basis states.

Hi,
nrqed said:
I tried to learn from him a few times and ended up using other references.
I just got this book (Georgi's) .. I don't 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.

Atakor said:
Hi,

I just got this book (Georgi's) .. I don't 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.

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!

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

Last edited by a moderator:
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..

## 1. What is the adjoint representation correspondence?

The adjoint representation correspondence is a mathematical concept that describes the relationship between a Lie group and its Lie algebra. It states that for every element in the Lie group, there is a corresponding element in the Lie algebra, and vice versa. This correspondence is important for studying the properties and structure of Lie groups and Lie algebras.

## 2. How is the adjoint representation correspondence defined?

The adjoint representation correspondence is defined by a map between the Lie group and the Lie algebra, known as the adjoint map. This map takes an element from the Lie group and maps it to an element in the Lie algebra, and vice versa. The adjoint map is a homomorphism, meaning it preserves the group structure, and it is also a linear map.

## 3. What is the significance of the adjoint representation correspondence?

The adjoint representation correspondence is significant because it allows us to study the properties of Lie groups and Lie algebras in a unified framework. It also allows us to translate problems and concepts between the two structures, making it easier to understand their relationship and the properties of both.

## 4. How is the adjoint representation correspondence used in physics?

In physics, the adjoint representation correspondence is used to describe the symmetries and transformations of physical systems. Lie groups and Lie algebras are used to represent these symmetries, and the adjoint representation correspondence helps translate between the two. This is particularly useful in quantum mechanics and particle physics.

## 5. Are there any limitations to the adjoint representation correspondence?

One limitation of the adjoint representation correspondence is that it only applies to Lie groups and Lie algebras. It cannot be extended to other mathematical structures. Additionally, the adjoint map may not be bijective, meaning there may not be a one-to-one correspondence between elements in the Lie group and the Lie algebra. This can make it challenging to fully understand the relationship between the two structures in some cases.

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