SU(3) Cartan Generators in Adjoint Representation

nigelscott
Messages
133
Reaction score
4
I am trying to work out the weights of the adjoint representation of SU(3) by calculating the 2 Cartan
generators as follows:

I obtain the structure constants from λa and λ8 using:

ab] = ifabcλc

I get:

f312 = 1
f321 = -1
f345 = 1/2
f354 = -1/2
f367 = -1/2
f376 = 1/2

f845 = √3/2
f854 = -√3/2
f876 = -√3/2
f867 = √3/2


I construct the matrices from:

(T1)ab = -if3ab and (T2)ab = -if8ab

I then diagonalize the resulting matrices and get the eigenvalues::

T1 = diag(-1/2,-1/2,1,0,0,-1,1/2,1/2)

T2 = diag(-√3/2,-√3/2,0,0,0,0,√3/2,√3/2)


Which is close but not correct. I think the problem may be with diagonalizing the matrices individually
rather than simultaneously as the eigenvectors don't match. Before, I begin that journey, however, I wanted
to samity check what I am doing. I am following Zee's lecture which can be found at:

https://www.youtube.com/watch?v=u-g9hzDByJ8 minute ~ 5

Any help would be appreciated.
 
Physics news on Phys.org
May I ask you why you want to do this? There's a reason why textbooks use ##\mathfrak{sl}(2)##, resp. ##\mathfrak{su}(2)##, as example. ##\mathfrak{su}(3)## is a case where it really has to be considered to write a computer program instead. I assume the CSA is spanned by the diagonal matrices ##i \lambda_3, i\lambda_8## and would as a first step try to find a basis for the Borel subalgebra (maximal solvable subalgebra). Then you know at least the ascending and descending elements. You can find an educated guess here. To work out the entire example with structure constants is quite troublesome, so I would remain as general as long as possible, i.e. write down the equations by using basis vectors, rather than structure constants. Besides all this, I guess the representations could be found on the internet.
 
Yes, your point is well taken. I realize that this is a tedious approach and the eigenvalues can be found more easily using [Hi,Eα] = αiEα where Eα are the I, U and V spin operators. However, my approach should work correct? In the defining rep the 3 x 3 Cartan generators share the same eigenbasis and so I would expect that the 8 x 8's would too. However, I am not seeing that. Either my structure constants are wrong or my approach is not valid. I used λ not iλ in my calculations. Could that be a problem?
 
nigelscott said:
Yes, your point is well taken. I realize that this is a tedious approach and the eigenvalues can be found more easily using [Hi,Eα] = αiEα where Eα are the I, U and V spin operators. However, my approach should work correct? In the defining rep the 3 x 3 Cartan generators share the same eigenbasis and so I would expect that the 8 x 8's would too. However, I am not seeing that. Either my structure constants are wrong or my approach is not valid. I used λ not iλ in my calculations. Could that be a problem?
It shouldn't be a problem. I just find it confusing to work with Lie algebras and their representations and the theorems available, but without having one. The Gell-Mann matrices are simply not skew Hermitian, which should affect the eigenvalues, so caution is required.

Wikipedia has the structure constants, so you can check them. I probably would just calculate the ##[\lambda_\alpha,\lambda_\beta]##. Should be easy and the imaginary factor can be easily applied as well, so you can have both in parallel, or look them up.

I still find it more convenient to work with ##[H^i,E_\alpha].v = E_\alpha.v## to get to ##H^i(E_\alpha.v)=(\alpha^i+\mu_i)E_\alpha.v## rather than the whole thing in coordinates. I only renamed the eigenvalues ##\mu_i## to avoid confusion with the Gell-Mann matrices. And if the representation isn't the natural one, you'll need this more general approach anyway. However, you said to use only the adjoint representation. In this case: why don't you just do matrix multiplications? ##\operatorname{ad} i\lambda_\alpha . i\lambda_\beta = - [ \lambda_\alpha , \lambda_\beta ] = \sum f_{\alpha \beta}^k (i\lambda_k)## gives you eight nice ##8 \times 8## matrices. I only would avoid to calculate ##\operatorname{Ad}## by exponentiation.

##\operatorname{ad}i \lambda_3## and ##\operatorname{ad} i \lambda_8## two are commuting semisimple linear transformations, so it should be no problem to verify the basis vectors of the eigenspaces.
 
I am still struggling with interpreting the results I get. For ad( H1) and ad( H2) I get:

adjoint.jpg

Diagonalization using WolframAlpha gives:

ad(H1): diag(-1,-1/2,-1/2,,0,0,1/2,1/2,1)

ad(H2): diag(0,0,0,0,-√3/2,-√;-3/2,√;3/2,√;3/2)

1. All the weights are there but not in the correct order.
2. 6/8 eigenstates of ad(H1) and ad(H2) match. 2 are different.

I still can't figure out what is going on.

WolframAlpha diagonalization links:

ad(H1): http://www.wolframalpha.com/input/?...,0},+{0,0,0,0,0,-i/2,0,0},+{0,0,0,0,0,0,0,0}}

ad(H2): http://www.wolframalpha.com/input/?...0,0,0,0,0,i√3/2,0,0},{0,0,0,0,0,0,0,0}}
 

Attachments

  • adjoint.jpg
    adjoint.jpg
    16.3 KB · Views: 1,071
Here is what I have.

I used the basis conventions at the end of https://www.physicsforums.com/insights/representations-precision-important/ , i.e. the numeration of the ##\lambda -##matrices, as well as
$$\left\{ \,H_1:=T_3=\frac{1}{2}\lambda_3\; , \;H_2:=Y\; , E_{\pm\alpha} :=iT_{\pm}\; , \;E_{\pm\beta} :=iU_{\pm}\; , \; E_{\pm(\alpha +\beta)}:=iV_{\pm}\;\,\right\}$$

Because it's far easier to calculate commutators, I set ##e_{ij}## to be the matrix with a single ##1## at position ##(i,j)## and ##0## elsewhere. This gave me
$$H_1=\frac{1}{2}e_{11}-\frac{1}{2}e_{22}\; , \;H_2=\frac{1}{3}e_{11}+\frac{1}{3}e_{22}-\frac{2}{3}e_{33} $$
$$
E_\alpha =ie_{12}\; , \;E_{-\alpha}=ie_{21}\; , \;E_{\beta}=ie_{23}\; , \;E_{-\beta}=ie_{32}\; , \;E_{\alpha + \beta}=ie_{13}\; , \;E_{-\alpha - \beta}=ie_{31}
$$
in the usual notation of the Cartan subalgebra and the corresponding root vectors.

If I made no sign errors and no typos, I had the following multiplications:
##\operatorname{ad}H_1.(E_{\alpha},E_{-\alpha},E_{\beta},E_{-\beta},E_{\alpha + \beta},E_{-\alpha - \beta}) = (E_{\alpha},-E_{-\alpha},-\frac{1}{2}E_{\beta},\frac{1}{2}E_{-\beta},\frac{1}{2}E_{\alpha + \beta},-\frac{1}{2}E_{-\alpha - \beta})##
##\operatorname{ad}H_2.(E_{\alpha},E_{-\alpha},E_{\beta},E_{-\beta},E_{\alpha + \beta},E_{-\alpha - \beta}) = (0,0,E_{\beta},-E_{-\beta},\sqrt{3}E_{\alpha + \beta},-\sqrt{3}E_{-\alpha - \beta})##
which looks good so far. I also have ##[E_{\pm \alpha\, , \,E_{\pm \beta}}] = \pm i \cdot E_{\pm (\alpha + \beta)}## as a kind of parity check, and I didn't calculate the rest.

If the ##E_\alpha## should be made into ##SU(3)## elements, so probably ##E_\alpha \longrightarrow i \cdot \lambda = E_\alpha + E_{-\alpha}\; , \; E_{-\alpha} \longrightarrow E_\alpha - E_{-\alpha}## etc. should be the right choices. Otherwise with the choices I've made, we only get an isomorphic version of it, i.e. still the same Lie algebra, but not skew-Hermitian anymore. But as a Lie algebra, my non-Hermitian basis is eventually easier to handle.
 
Last edited:
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top