# Why do Gell-Mann matrices have this normalization?

This might be a stupid question, but why is the normalization of the Gell-Mann matrices (basis of the $$\mathrm{su}(3)$$ Lie algebra) chosen to be $$\mathrm{trace}(\lambda_i\lambda_j)=2\delta_{ij}$$ instead of just $$\delta_{ij}$$ without the factor $$2$$? In most of linear-algebra, basis vectors are normalized to $$1$$ (or not normalized at all). Why not in the context of Lie Algebras? Is there a way of looking at this which makes the factor $$2$$ seem natural?

On a related note, some physics texts change the normalization by defining "the generators of the $$\mathrm{SU}(3)$$ group" as $$T_i=\frac{1}{2}\lambda_i$$. But these just fulfil $$\mathrm{trace}(T_iT_j)=\frac{1}{2}\delta_{ij}$$ which seems just as unnatural to me. (And the difference between these two normalization conventions just cost me an hour of chasing a missing factor $$4$$ in a long calculation. Which is why I'm asking this question xD).

History. The Gell-Mann matrices are an extension/generalization of the Pauli spin matrices for su(2), and $$\lambda_{1,2,3}$$ identify with these, so obey the same trace relation.
• well, Pauli matrices have the same "problem" of $\mathrm{tr}\sigma_i\sigma_j=2\delta_{ij}$ (also $[\sigma_i\sigma_j]=2i\epsilon_{ijk}\sigma_k$). But changing their normalization would destroy the rather nice property $\sigma_i^2=1$ (which implies Eigenvalues$=\pm 1$ and Determinant$=-1$). These properties even kinda carry over to Gell-Mann, whose eigenvalues are $1,-1,0$ (except for $\lambda_8$). Thanks, Cosmas. – Simon Aug 18 '20 at 14:39
• @Cosmas Zachos: Actually, it would be nice if you could expand your answer a bit, because I wonder why only in gauge theory the normalisation Lie-algebra generators is done, but not in representation theory. Why is this so? Why does it matter in gauge theory? Secondly, the trace of 2 generators looks like a killing form $B(x,y) = tr(ad_X, ad_Y)$. Curiously for a compact group like $SU(2)$ I would expect a negative normalisation $tr(t_i, t_j)=-\frac{1}{2}\delta_{ij}$, and in some books it is indeed like this, but why not here ? – Frederic Thomas Aug 18 '20 at 21:47