Why the Gell-Mann matrices in the SU(3)-model need to be trace orthogonal?
[Math] Why the Gell-Mann matrices in the SU(3)-model need to be trace orthogonal
lie-algebraslie-groupsmp.mathematical-physicsrt.representation-theory
Related Solutions
I'm no expert, and I haven't asked Professors Gell-Mann or Ne'eman, but with their choice of matrices, L_3 measures the familiar Heisenberg iso-spin quantum number, while L_8 measures the then-novel hypercharge. Mixing the operators would mix the quantum numbers.
Pastur and Vasilchuk have extended the result of Diaconis and Evans for $a_{2k}$ from $2k\leq n/2$ to $2k\leq n-1$:
$$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2)\;\;\text{for}\;\;2k\leq n-1\quad\quad[*]$$
As suggested by Liviu Nicolaescu, for small $n$ you might directly integrate over the probability distribution of the eigenvalues in $O_n$, which you can find here. (This is the socalled "circular real ensemble" of random matrices.)
$\bullet$ For $n=2$ one reproduces the result $$a_{2k}=\frac{1}{\pi}\int_0^{\pi/2}dx\,(2\cos x)^{2k}=\tfrac{1}{2}(2k)!(k!)^{-2}$$ you quoted above, which agrees with $[*]$ for $k=1$ and $k=2$. (For $k=3$ it gives $10$ instead of $15$.)
$\bullet$ For $n=3$ one obtains $$a_{2k}=\frac{1}{2\pi}\int_0^{2\pi}dx\,(1-\cos x)(1+2\cos x)^{2k}$$ which evaluates to $$a_2=1,\;\; a_4=3,\;\; a_6=15,\;\; a_8=91,\;\;a_{10}=603,\;\;a_{12}=4213$$ in agreement with the series of Qiaochu Yuan's answer. The formula $[*]$ for $a_{2k}$ holds for $k\leq 3$ (for $k=4$ it gives $105$ instead of $91$).
$\bullet$ For $n=4$ one finds $$a_{2k}=\frac{1}{8\pi^2}\int_0^{2\pi}dx\int_0^{2\pi}dy\,(\cos x-\cos y)^2(2\cos x+2\cos y)^{2k}$$ $$\quad\quad+\frac{1}{2\pi}\int_0^{2\pi}dz\,(1-\cos^2 z)(2\cos z)^{2k}$$ which evaluates to $$a_2=1,\;\;a_4=3,\;\;a_6=15,\;\;a_8=105,\;\;a_{10}=903,\;\;a_{12}=8778.$$ Now the formula $[*]$ for $a_{2k}$ holds for $k\leq 4$ (for $k=5$ it gives $945$ instead of $903$).
$\bullet$ Continuing with $n=5$, we have $$a_{2k}=\frac{1}{2\pi^2}\int_0^{2\pi}dx\int_0^{2\pi}dy\,(1-\cos x)(1-\cos y)(\cos x-\cos y)^2(1+2\cos x+2\cos y)^{2k}$$ which evaluates to $$a_2=1,\;\;a_4=3,\;\;a_6=15,\;\;a_8=105,\;\;a_{10}=945,\;\;a_{12}=10263.$$ The formula $[*]$ for $a_{2k}$ holds for $k\leq 5$ (for $k=6$ it gives $10395$ instead of $10263$).
I am now tempted to $$\text{conjecture}\quad\quad a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2)=(2k-1)!!\;\;\text{for}\;\;k\leq n\quad\quad[**]$$ thus doubling the range of validity of $[\ast]$. Is it true?
Best Answer
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.