# How can the Cartan-Weyl basis of su(2) be a basis if it does not consist of antihermitian operators?

Consider a Lie algebra. The ladder operators (i.e. root vectors, or eigenvectors of the Cartan subalgebra with respect to the adjoint representation) form a handy basis of the algebra called a Cartan-Weyl basis (not unique).

But in the following example of su(2) they are not antihermitian, while su(2) requires that.

Consider su(2). As a vector space it can be the real vector space of traceless antihermitian matrices. The Cartan subalgebra is spanned by

$H = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$.

The ladder operators are

$\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ -2 & 0 \end{pmatrix}$.

But they are not antihermitian. How is it that they can be used as a basis for the su(2) algebra?

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I am not a mathematician, so please be patient when explaining/correcting any misconceptions that I might have. –  Krastanov Feb 21 '13 at 14:27
I have always been content to work with Lie algebras over algebraically closed fields, so I may not know a formally correct way of stating it, but IIRC the ladder operators only exist in the complexification of the real Lie algebra $su(2)$. And surely $su(2)$ has real dimension 3 only? –  Jyrki Lahtonen Feb 21 '13 at 14:33
@JyrkiLahtonen, could you elaborate what you meant by the ladder ops existing only in the complexification of su(2). Concerning the number of dimensions, you are completely right, I was thinking of su(3), I will correct it. –  Krastanov Feb 21 '13 at 14:37
The ladder operators exist in the real Lie algebra $sl(2)$ of traceless real 2x2-matrices. What (not positive about this) happens is that when we "extend the scalars" from reals to complex numbers, these two Lie algebras become the same. Formally $$su(2)_\mathbb{R}\otimes\mathbb{C}\cong sl(2)_\mathbb{R}\otimes\mathbb{C}.$$ More informally, just start looking at complex linear combinations of the matrices in a basis over the reals. When we look at a complex representation of $su(2)$, we can extend the Lie-algebra action to this complexified version just by "extending the action linearly". –  Jyrki Lahtonen Feb 21 '13 at 14:51
(cont'd) Therefore the ladder operators will be there whenever the space of the representation is complex. For example, in quantum mechanical presentations of spin/angular momentum the ladder operators $L^{\pm}=L_x\pm iL_y$ are not real linear combinations of the elements $L_x, L_y$ representing the usual (real) basis elements of $su(2)$ but rather represent the corresponding elements in the complexification $su(2)_\mathbb{R}\otimes \mathbb{C}.$ That's my educated guess as to why the problem you observed is not a concern. I'm afraid, I don't have the confidence to write this as an answer :-) –  Jyrki Lahtonen Feb 21 '13 at 14:56

## 1 Answer

As Jyrki mentioned in the comments, the root vectors Lie in the complexified algebra. This always happens for semi-simple Lie algebras. Suppose $\mathfrak g$ is a compact semi-simple Lie algebra (meaning the Lie groups it corresponds to are compact) and $G$ the simply connected Lie group corresponding to $\mathfrak g$. In your case $\mathfrak g = su(2)$ and $G = SU(2)$.

Any representation of $G$ preserves a hermitian metric: take any metric $\langle \cdot,\cdot\rangle_0$ on the representation space and average over $G$ by integrating-- $$\langle v,w\rangle := \int_G \langle gv,gw\rangle_0.$$ Since $G$ preserves $\langle \cdot,\cdot\rangle$, the Lie algebra acts anti-self-adjointly, i.e. as anti-hermitian operators. Any anti-hermitian operator has purely imaginary eigenvalues.

Now if you apply this to the adjoint action of $\mathfrak g$ on itself, you see that $H$ has purely imaginary eigenvalues, so that it's eigenvectors must lie in the complexification of $\mathfrak g$.

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Thanks, Eric and Jyrki. I guess the most important thing to remember is that "... the root vectors Lie in the complexified algebra. This always happens for semi-simple Lie algebras.". –  Krastanov Feb 21 '13 at 16:13