${\bf su}(2)$ generators in 1, 2 and 3 dimensional matrix representations

Question

The ${\bf su}(2)$ Lie algebra in a representation $\bf R$ is defined by

$$[T^{a}_{\bf R},T^{b}_{\bf R}]=i\epsilon^{abc}T^{c}_{\bf R},$$

where $T^{a}_{\bf R}$ are the $3$ generators of the algebra.

In 2 dimensions, these generators are the Pauli matrices

$$T^{1}_{\bf 1} = \frac{1}{2}\begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}, \qquad T^{2}_{\bf 1} = \frac{1}{2}\begin{pmatrix}0 & -i\\ i & 0 \end{pmatrix}, \qquad T^{3}_{\bf 1} = \frac{1}{2}\begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}.$$

In 3 dimensions, these generators are

$$T^{1}_{\bf 2} = \frac{1}{\sqrt{2}}\begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \qquad T^{2}_{\bf 2} = \frac{1}{\sqrt{2}}\begin{pmatrix}0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{pmatrix}, \qquad T^{3}_{\bf 2} = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$

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1. How can you derive the generators in $2$ and $3$ dimensions?

2. What are the generators in $1$ dimension?

Answer 1

1. Using the ladder operators, as described e.g. in the wikipedia page.

2. Hint: $1\times 1$ matrices commute.

Answer 2

Another approach:

1. Construct the $2\times 2$ representation from first principles directly from the commutation relationships, as I do here. Clean up the $i$ factor to make them Hermitian rather than skew Hermitian;

2. Since the Lie algebra $\mathfrak{su}(2)$ is centerless, the adjoint representation is an isomorphism ($\ker(\mathrm{ad}) = \mathscr{Z}$ is a fact you should remember). Moreover, the image of $\mathfrak{su}(2)$ under $\mathrm{ad}$ gives you a $3\times 3$ representation because $\mathfrak{su}(2)$ is three dimensional. Thus compute the images $\mathrm{ad}(i\,\sigma_j)$ of the matrices $\sigma_j$ in (1) under the adjoint representation and you're done for $3\times 3$ after adding the $i$ factor to make them Hermitian;

3. $1\times 1$ matrices can only represent the unique, one dimensional Lie algebra $(\mathbb{R},\,+)$, so, as noted elsewhere, "it can't be faithful".

The technique in (1) gives you representatives of equivalence classes of representations modulo similarity transformations, and an analysis of this technique shows that it is exhaustive, i.e. the equivalence class contains all possible solutions. Thus, the application of (3) will give you different matrices from the ones you cite, but which can be transformed into yours with the appropriate similarity transformation.