The squares of the Gell-Mann matrices, $\lambda_1^2$, $\lambda_2^2$, and $\lambda_3^2$, all have the bizarre value $2/3 * I + \lambda_8 / \sqrt{3}$.

Likewise, $\lambda_8^2=2/3 * I - \lambda_8 / \sqrt{3}$. (Equivalently, $\lambda_3^2 + \lambda_8^2 =4/3 * I$.)

Is there a simple way to deduce these results from the fact that $\lambda_8$ is orthogonal to all other $\lambda_n$, and has trace $0$ (or some other of its properties) WITHOUT using the specific Gell-Mann representation matrices?

If the numerical values are due to historical choices of normalisations, is there at least a simple way to prove that both $\lambda_3^2$ and $\lambda_8^2$ must be linear combinations of the identity and of $\lambda_8$ ?

  • $\begingroup$ Is $*$ intentional or do you mean $\times$ (given by $\times$)? $\endgroup$
    – Shaun
    Apr 6, 2023 at 11:17
  • $\begingroup$ The " * I " simply means "times the identity". $\endgroup$
    – KlausK
    Apr 6, 2023 at 12:59
  • 3
    $\begingroup$ Note that "the" multiplication table for a Lie algebra depends on, hence has more to do with, the chosen basis ("generators") than with the Lie algebra as such. A Cartan Weyl basis would have a much nicer table. So in my mind the question asks why physicists like those Gell-Mann matrices so much even though they have that weird multiplication table. $\endgroup$ Apr 6, 2023 at 14:28
  • $\begingroup$ @Torsten Oh, I did not know this. Does it mean that there is no answer to my question? $\endgroup$
    – KlausK
    Apr 6, 2023 at 14:39

1 Answer 1


Following Torsten's suggestion:

A Cartan-Weyl is just a basis of a Cartan subalgebra of $\mathfrak{g}$ plus root vectors so the multiplications are relatively easy to work out if you know the root system (which is $A_2$). However, it will really be a basis of the complexification $\mathfrak{sl}(3,\mathbb{C})$ since $\mathfrak{su}(3)$ is compact, so you would need to adjust it slightly. In fact what you can do in the compact case is take (appropriately chosen) root vectors $X_\alpha, X_{-\alpha}$ and make a basis instead with $X_\alpha - X_{-\alpha}$ and $ i(X_\alpha + X_{-\alpha}) $.

So lets apply this to our example. Firstly, I will be using the maths convention that $\mathfrak{su}(3)$ consists of anti-Hermitian matrices, rather than Hermitian ones as physicists often use. To swap to the physics convention just multiply everything by $i$ and insert an $i$ into the definition of the commutator. Lets start with the complexification $\mathfrak{sl}(3,\mathbb{C})$ of tracefree matrices. The natural Cartan subalgebra to use here is the set of diagonal matrices, so a natural basis is:

$$H_{\alpha_1} = \begin{pmatrix} 1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0\\ \end{pmatrix},\quad H_{\alpha_2} = \begin{pmatrix} 0 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\\ \end{pmatrix}.$$

Then root vectors are simply matrices with only one non-zero entry (off-diagonal) e.g.:

$$X_{\alpha_1} = \begin{pmatrix} 0 & 1 & 0\\0 & 0 & 0\\0 & 0 & 0\\ \end{pmatrix},\quad X_{\alpha_2} = \begin{pmatrix} 0 & 0 & 0\\0 & 0 & 1\\0 & 0 & 0\\ \end{pmatrix}, X_{\alpha_1 + \alpha_2} = \begin{pmatrix} 0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\\ \end{pmatrix}.$$

Note the basic property of such a basis is that for every diagonal matrix $H$, $[H,X_\alpha] = \alpha(H)X$ (here $\alpha_1(H_1) =2, \alpha_1(H_2) = -1$ and so on), while $[X_\alpha, X_\beta]$ is in the span of $X_{\alpha+\beta}$ (you have to work the coefficient out in general as this depends on our choice of scale but easy to do in our example). Note also that $X_{-\alpha}$ can be chosen here to be the transpose of $X_\alpha$ (more generally I would use a Cartan involution $\tau$ and take $X_\alpha + \tau X_\alpha, i(X_\alpha - \tau X_\alpha)$)

Now as I said this is not a basis of $\mathfrak{su}(3)$. Indeed not one of our elements is anti-Hermitian. But as I suggested above if we take

$$ Y_{\alpha_1} = X_{\alpha_1} - X_{-\alpha_1} = \begin{pmatrix} 0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 0\\ \end{pmatrix}, Z_{\alpha_1} = i(X_{\alpha_1} + X_{-\alpha_1}) = \begin{pmatrix} 0 & i & 0\\i & 0 & 0\\0 & 0 & 0\\ \end{pmatrix}$$

and so on we now have a very natural basis (we must also replace $H_1, H_2$ by $iH_1,iH_2$) and commutators can be worked out by hand or from the root data in the more general case. Hey presto this is almost exactly the Gell-Mann matrices in fact (after multiplying by $i$ to make them Hermitian because physicists) but replacing $H_{\alpha_2}$ with another diagonal matrix which is orthogonal to $H_{\alpha_1}$ with respect to the Killing form (or equivalently trace form) and normalised so that the trace of its square is $2$. Wikipedia claims that the reason for this choice is so that the exponentials will generate the whole group $SU(3)$ (I am confused by this as that should be true for any basis). It also says this is so the 3 natural $\mathfrak{su}(2)$ subalgebras are normalised the same as the Pauli matrices. But then that is true in my basis and is even neater, so I'm still not sure what the point is.

I see several questions here and here asking about this without much more answer than it's convenient to have an orthogonal basis.

One final thing to note is that squaring these matrices or doing any multiplication that is not a commutator takes us out of the realms of Lie algebras. Thus it is a little hard to talk about their squares or similar in Lie theoretic terms except perhaps as regards the Casimir operators.

But more simply looking at them as matrices it is perhaps obvious that $Y_{\alpha}^2 = - X_\alpha X_{-\alpha} - X_{-\alpha}X_\alpha = Z_{\alpha}^2$ ($X_\alpha^2 =0 $ for all $\alpha$) and are diagonal so naturally in the span of our diagonal basis plus the identity.


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