# I $SU(2)$ generators in $1$, $2$ and $3$ dimensions

1. Mar 16, 2017

### spaghetti3451

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?

2. Mar 17, 2017

### Orodruin

Staff Emeritus
1. The two-dimensional representation is the fundamental representation. You can find out what the matrices J in its Lie algebra are by writing $e^{-itJ}$ and requiring that this matrix is part of SU(2). This will give you a set of matrices of which you can pick a basis.

For the three-dimensional irrep, it is the symmetric part of the tensor product representation. You can construct the generators explicitly by constructing a basis for the tensor product space and checking how the representations act on this basis.

2. I will give you a hint: All 1x1 matrices commute so [A,B]=0 for any A and B.

3. Mar 24, 2017

### spaghetti3451

So, are the matrices all equal to (0) for the one-dimensional representation?

4. Mar 24, 2017

### Orodruin

Staff Emeritus
Yes. It is the trivial representation.

5. Mar 24, 2017

### spaghetti3451

It is also called non-faithful. Is this because every element of $SU(2)$ is being mapped to the same element, that is, $1$?

6. Mar 24, 2017

### Orodruin

Staff Emeritus
The three-dimensional representation is not faithful either - but it is not trivial. It maps $A$ and $-A$ to the same matrix.

The trivial representation by definition maps all group elements to the identity.

Also, you have a mistake it the third three-dimensional generator above - it should not be the identity.

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