spinor representation

Question

I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then How can we represent a spinor using matrix ?

Answer 1

$SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,

$$ | z_1 |^2 + |z_2|^2 = 1,$$

invariant.

The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,

$$ \left(\begin{array} \ z_1 \\ z_2 \end{array}\right),$$

but there is nothing to prevent us from representing these complex numbers in a $2\times 2$ matrix.

In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as

$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \qquad |\alpha|^2 + |\beta|^2 = 1,$$

notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.

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To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.

$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \left( \begin{array} \ z_1 \\ z_2\end{array} \right) = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 \\ \beta z_1 + \bar{\alpha} z_2\end{array}\right)$$

Now consider the other transformation:

$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \left( \begin{array} \ z_1 & - \bar{z_2} \\ z_2 & \bar{z_1} \end{array}\right)$$

$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 & -\alpha \bar{z_2} - \bar{\beta} \bar{z_1} \\ \beta z_1 + \bar{\alpha} z_2 & -\beta \bar{z_2} + \bar{\alpha} \bar{z_1} \end{array}\right) $$

$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 & \overline{ -\bar{\alpha} z_2 - \beta z_1} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{-\bar{\beta} z_2 + \alpha z_1 } \end{array}\right) $$

$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 & -\overline{\left( \bar{\alpha} z_2 + \beta z_1\right)} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{ \alpha z_1 -\bar{\beta} z_2 } \end{array}\right) $$

$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 & -\overline{\left( \beta z_1 + \bar{\alpha} z_2 \right)} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{ \alpha z_1 -\bar{\beta} z_2 } \end{array}\right) $$

You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.

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It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.

The transformation matrices would remain the same.

$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \qquad |\alpha|^2 + |\beta|^2 = 1,$$

The spinors would still be expressible as $2\times 2$ matrices.

$$ \left( \begin{array} \ z_1 & - \bar{z_2} \\ z_2 & \bar{z_1} \end{array}\right)$$