# $t_1$, $t_2$, $t_3$ Hermitian generators of $SU(2)$

What is the exact $SU(2)$ representation to which these Hermitian generators belong? $$t_a=\{t_1,t_2,t_3\}=\left\{\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},\,\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix},\,\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \right\}$$

I'm a bit puzzled with this, these are not the generators of the triplet representation of $SU(2)$ (the triplet in $SU(2)$ is the adjoint rep which is real and constructed with the structure constants), nevertheless they are used as if they were in much literature. What are really these generators? They seem like the extension to 3 dimensions of the Pauli matrices. The 3 dimensional fundamental rep (this does not make sense to me)? some kind of non-irreducible representation?

• They are the generators in the adjoint representation, just a different basis. You can show that the adjoint representation matrices are related to these via a similarity transformations. Oct 2 '13 at 13:44
• Thanks! I've seen the change of basis before, but did not know that the apparently "complex" behaviour in one basis does not apply if I can find with a unitary transformation a basis where the generators are real. Oct 2 '13 at 16:14

As noted in the comments, these matrices are a basis for the adjoint representation of $SU(2)$ (the image of which representation is, as you likely know, $SO(3)$) but relabeled by a similarity transformation $S$ where:

$$S = \left( \begin{array}{ccc} -\frac{1}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\ \end{array} \right)$$

which works as follows. I rescale your three t's so that they are skew-Hermitian (so they are a basis for the Lie algebra $\mathfrak{su}(2) = \mathfrak{so}(3)$) and I also re-order them so that, $(t_1, t_2, t_3)$ is a right-handed basis (in the order you have them, they are not):

$$t_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & i & 0 \\ i & 0 & i \\ 0 & i & 0 \end{pmatrix},\, t_2 = \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -i \end{pmatrix},\, t_3=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$$

Now we have the standard basis vectors in the Lie algebra acted on by the adjoint representation:

$$s_1 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{array} \right),\, s_2 = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),\, s_3=\left( \begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$$

then the similarity transformation above does the trick: $t_j = S s_j S^\dagger$.

These are not really analogous to the Pauli matrices (as you speculate), even though all of the above form bases for a faithful representation of the Lie algebra $\mathfrak{su}(2) = \mathfrak{so}(3)$. Recall that the Pauli matrix (multiplied by $i$) Lie algebra exponentiates to $SU(2)$, whereas all the bases discussed here exponentiate to $SO(3)$, the former being the double cover of the latter.

A last way to think of this problem (this is a good one to keep up your sleeve for when you come across occasionally unwonted, weird and whacky representations), even if you can't straight away see that there is a similarity transformation, is through the Universal Enveloping Algebra (see Wikipedia page with this name). Once you've checked (which you can nowadays do in a few minutes with a software like Mathematica) that the commutation relationships fulfilled by the given matrices are indeed those of the $\mathfrak{su}(2) = \mathfrak{so}(3)$ Lie algebra, you know that the matrix representation you have been give must be isomorphic to some subset of the universal enveloping algebra, namely, the freest algebra possible tethered by the commutation relationships defining the Lie bracket. In the $\mathfrak{su}(2) = \mathfrak{so}(3)$ case it is not hard to show (although I can't put my hand on the proof right at the moment) that every member $H$ of such a freest possible algebra in this case fulfills the relationship

$$H\,(H + r^2 I) = 0\quad\quad\quad(1)$$

where $I$ is the unital element of the ring and $r^2$ some positive real, but that only in some matrix representations of the relevant Lie algebra do we have the stronger condition:

$$H + r^2 I = 0\quad\quad\quad(2)$$

So you can put your matrices into Mathematica and quickly test that indeed (1) is fulfilled but (2) is not. So this is indeed the universal enveloping algebra, i.e. the $3\times3$ real matrix algebra $\mathfrak{so}(3)$ rather than the more constrained one of $2\times2$ matrix algebra $\mathfrak{su}(2)$. There must therefore be some linear transformation between the given algebra and the standard $3\times3$ real matrix algebra that preservers the Lie brackets. It is readily shown that this must be a similarity transformation.