Why the Gell-Mann matrices in the SU(3)-model need to be trace orthogonal?
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1$\begingroup$ Could you please flesh out your question? What are the Gell-Mann matrices? are they uniquely defined objects, or merely a collection that satisfies certain conditions? $\endgroup$– Yemon ChoiCommented Feb 23, 2012 at 21:27
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$\begingroup$ 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/Gell-Mann_matrices $\endgroup$– HAJVCommented Feb 23, 2012 at 21:48
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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 JagyCommented Feb 23, 2012 at 22:42
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$\begingroup$ 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 ChoiCommented Feb 23, 2012 at 23:04
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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. Gell-Mann 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 StoicaCommented Feb 23, 2012 at 23:22
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1 Answer
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Let us imagine we are building the Gell-Mann 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 $i = 1,\ldots,n^2-1$ 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 Gell-Mann 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.
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$\begingroup$ I see this question has been asked again elsewhere. mathoverflow.net/questions/89431/… $\endgroup$ Commented Mar 11, 2012 at 22:29
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$\begingroup$ 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. $\endgroup$– HAJVCommented Mar 13, 2012 at 21:58