Adjoint representation correspondence?

[ismaili]

Dear All,

I'm reading Georgi's text about Lie algebra, 2nd edition.
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!

 

[nrqed]

Dear All,

I'm reading Georgi's text about Lie algebra, 2nd edition.
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 [itex]V_c T_c \equiv \vec{v} \cdot \vec{T}[/tex]. This is now a matrix that represents the <i> state </i> with components V_c (instead of a column vector). <br /> <br /> 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 <i> commutator </i> of the state with the matrix producing the transformation. So we say the the transformed "state" is given by <br /> <br /> (warning: I am doing this by memory, I am pretty sure I will get some minus signs wrong)<br /> <br /> $$\vec{v}' \cdot \vec{T} = [T_a, v_c T_c] = i v_c f_{acd} T_d$$<br /> <br /> where I have avoided using an index "b" on the right side to make things more clear.<br /> <br /> Ok, now, let's say that we want the coefficient $v_b'$ (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<br /> <br /> $$v'_b = i v_c f_{acb}$$<br /> <br /> Using the antisymmetry of the structure constants, we finally get<br /> <br /> $$v'_b = - i f_{abc} v_c$$<br /> <br /> which is the same result that we obtained using states being column vectors. <br /> <br /> So in this representation, we can use the generators themselves to build the states! <br /> But again, this also implies that we use commutators to "apply operators to the states".[/itex]

 

[ismaili]

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 [itex]V_c T_c \equiv \vec{v} \cdot \vec{T}[/tex]. This is now a matrix that represents the <i> state </i> with components V_c (instead of a column vector). <br /> <br /> 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 <i> commutator </i> of the state with the matrix producing the transformation. So we say the the transformed "state" is given by <br /> <br /> (warning: I am doing this by memory, I am pretty sure I will get some minus signs wrong)<br /> <br /> $$\vec{v}' \cdot \vec{T} = [T_a, v_c T_c] = i v_c f_{acd} T_d$$<br /> <br /> where I have avoided using an index "b" on the right side to make things more clear.<br /> <br /> Ok, now, let's say that we want the coefficient $v_b'$ (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<br /> <br /> $$v'_b = i v_c f_{acb}$$<br /> <br /> Using the antisymmetry of the structure constants, we finally get<br /> <br /> $$v'_b = - i f_{abc} v_c$$<br /> <br /> which is the same result that we obtained using states being column vectors. <br /> <br /> So in this representation, we can use the generators themselves to build the states! <br /> But again, this also implies that we use commutators to "apply operators to the states".[/itex]

[itex] <br /> Thank you very much!<br /> <br /> 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. <br /> So, in a sense there is a map from states to generators in adjoint representation <br /> <br /> $$v_a \rightarrow v_aT_a$$<br /> <br /> 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 $$X_a$$.<br /> <br /> I think basically I got the idea. (Is this so trivial? Georgi didn't state it at all...)<br /> <br /> -------------------------------<br /> <br /> However, when I read section 6.1 just now, I found something I don't really understand. <br /> He said "<i>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 <br /> $$H_i|\mu,x,D\rangle = \mu_i|\mu,x,D\rangle\quad---(*)$$<br /> ($$H_i$$ is the hermitian Cartan generator) and $$x$$ is any other label that is necessary to specify the state</i>."<br /> <br /> My question is, the $$|\mu,x,D\rangle$$ should be the <b><u>basis</u></b> 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 $$|\mu,x,D\rangle$$? <br /> <br /> Thanks for any illumination![/itex]

 

[nrqed]

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.

 

[Atakor]

Hi,

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.

 

[nrqed]

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!

 

[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

 

[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..