Why the GellMann matrices in the SU(3)model need to be trace orthogonal?

1$\begingroup$ Could you please flesh out your question? What are the GellMann matrices? are they uniquely defined objects, or merely a collection that satisfies certain conditions? $\endgroup$ – Yemon Choi Feb 23 '12 at 21:27

$\begingroup$ Those are a representation of the generators of the special unitary group SU(3). I could explain, but not better than you can find at en.wikipedia.org/wiki/GellMann_matrices $\endgroup$ – HAJV Feb 23 '12 at 21:48

1$\begingroup$ We do answer questions at grad student level, even some at undergraduate level. However, we do not really do tutorials. Also, it is desirable that you put in more effort than us. This is a site for professional researchers. The additional reasons for you to put in more background are to show that you have done more work than absolutely nothing, and that you already know more than absolutely nothing. So far you are batting zero. $\endgroup$ – Will Jagy Feb 23 '12 at 22:42

$\begingroup$ mathoverflow.net/howtoask Since not everyone has the same background as you, or uses the same terminology, it is good manners to explain such things. For instance, now that you have provided a link, it seems that the key words are "Lie algebra". $\endgroup$ – Yemon Choi Feb 23 '12 at 23:04

2$\begingroup$ My first guess is that the trace orthogonality property for rep 3 is related to the Killing form (which is defined by the adjoint rep 8). Maybe this makes them ideal for making the root diagrams represent particles with their properties. GellMann started with diagrams representing hadrons, and he choose his matrices so that their action transforms one hadron in another, by increasing or decreasing certain numbers associated to the particles. So, in fact, the representation was provided by nature. $\endgroup$ – Cristi Stoica Feb 23 '12 at 23:22
Let us imagine we are building the GellMann matrices. For us, SU($n$) is the group of $n \times n$ unitary matrices with determinant 1. An element $U$ can be written $U = e^{i \alpha_i T_i}$, where $i = 1,\ldots,n^21$ and the $\alpha_i$ are real. The $T_i$ are a basis for the algebra. They must be traceless and hermitian. Consider $\mathrm{tr}\ T_i T_j$. This matrix is real and symmetric. Diagonalize it. Call your new basis $\lambda_i$. Normalize so $\mathrm{tr}\ \lambda_i \lambda_j = 2\delta_{i j}$. From here the GellMann matrices are further specified by the condition that the Pauli matrices sit nicely inside, etc.
Why diagonalize $\mathrm{tr}\ T_i T_j$? Of course, it is because we want a nice orthogonal basis. Would you rather have as a basis for two space {(1,2),(3,4)} or {(1,0),(0,1)}? The orthogonal representation is also simply related to the raising and lowering operators used to build reps and decompose product reps. Notice that we could choose another basis $X_i = M_{i j}\lambda_j$ where $M$ is invertible and real. This does not change the algebra but, for example, the structure constants will be different, the relation of the basis to the raising and lowering operators will be complicated, and the $X_i$ will not be orthogonal in general.

$\begingroup$ I see this question has been asked again elsewhere. mathoverflow.net/questions/89431/… $\endgroup$ – user26872 Mar 11 '12 at 22:29

$\begingroup$ Thanks Oenamen, Christi Stoica and Art Brown. Indeed, key point is the orthogonality of tr(TiTj) of course. Oenamen's last line and indeed Art's suggestion gave the answer I was looking for. As I understand it now, that orthogonality is linked to the definition of independent quantum numbers. $\endgroup$ – HAJV Mar 13 '12 at 21:58