Matrix construction for spinors

[shinobi20]

TL;DR Summary

Questions about the motivation of Ryder in his construction of the $2 \times 2$ traceless Hermitian matrix for spinor transformation.

I'm reading the book QFT by Ryder, in the section where $\rm{SU(2)}$ is discussed.

First, he considered the group of $2 \times 2$ unitary matrices $U$ with unit determinant such that it has the form,

$$U =\begin{bmatrix}
a & b \\
-b^* & a^*
\end{bmatrix}, \qquad \xi =
\begin{bmatrix}
\xi_1 \\
\xi_2
\end{bmatrix}
$$

where this is the transformation matrix of a spinor $\xi$. He went on to show that the spinor and its Hermitian conjugate transform in a certain way (eq. 2.39). He constructed the outer product $\xi \xi^\dagger$ (eq. 2.40) and said that is a Hermitian matrix.

From (eq. 2.39), it is obvious that the spinor and its Hermitian conjugate transform in a different way so he proposed that we can use the unitarity of $U$ to show that,

$$
\begin{bmatrix}
\xi_1 \\
\xi_2
\end{bmatrix}
, \qquad
\begin{bmatrix}
-\xi^*_2 \\
\xi^*_1
\end{bmatrix}
$$

transform in the same way under $\rm{SU(2)}.$

In the end, he said that $\xi \sim \zeta\xi^*$ ($\sim$ means transform in the same way and $\zeta$ is the spinor metric) and $\xi^\dagger \sim (\zeta \xi)^T$ so that,

$$\xi \xi^\dagger \sim
\begin{bmatrix}
\xi_1 \\
\xi_2
\end{bmatrix}
\begin{bmatrix}
-\xi_2~ \xi_1\\
\end{bmatrix}
=
\begin{bmatrix}
-\xi_1 \xi_2 & \xi_1^2 \\
-\xi_2^2 & \xi_1 \xi_2
\end{bmatrix} = -\rm{H}
$$

and this is a traceless matrix.

After all this, he said that we can now construct from the position vector $\vec{r}$ a traceless $2 \times 2$ matrix transforming under $\rm{SU(2)}$ like $\rm{H}.$ It is given by,

$$h =\begin{bmatrix}
z & x-i y \\
x+i y & -z
\end{bmatrix}.$$

My questions are,

1. I do not understand what is the motivation and purpose of constructing a Hermitian matrix and then saying that the spinors used to construct it do not transform the same way and then constructing a traceless matrix using a new set of spinors that transform the same way and YET again forgetting about it and in the end just says that "we can now construct a traceless $2 \times 2$ matrix transforming under $\rm{SU(2)}$ like $\rm{H}$". My guess is that the separate Hermitian and traceless construction is to justify that there is a matrix that has both of those properties? I'm not sure...

2. What is the reason for constructing two spinors that transform the same way and how did he get that new spinor?

Please see the attached file for the exact detail. It consists of three pages talking about what I briefly discussed.

 

[samalkhaiat]

My questions are,
1. I do not understand what is the motivation and purpose of constructing a Hermitian matrix ..
2. What is the reason for constructing two spinors that transform the same way and how did he get that new spinor?

He is explaining, in elementary terms, the following two properties of $\mbox{SU}(2)$:

1) There is one, and only one, 2-dimensional irreducible representation.

2) There exists a homomorphism $\mbox{SU}(2) \to \mbox{SO}(3)$ whose kernel, $\mbox{Z}_{2}$, coincides with the centre of $\mbox{SU}(2)$, i.e., a 2 to 1 mapping of the elements of $\mbox{SU}(2)$ onto those of the rotation group $\mbox{SO}(3)$. In plain English, this means that an $\mbox{SU}(2)$ transformation on the 2-component spinor $\psi$ corresponds to a $\mbox{SO}(3)$ rotation of the coordinates $x^{i} = x^{i}(\psi)$.

In order to explain the above, we need the following two “facts” from the theory of representation:

a) Complex conjugation defines a representation: If $U$ is a representation, then the mapping $U \mapsto D(U) = U^{\ast}$ is also a representation. The proof is very easy, $D(UV) = (UV)^{\ast} = U^{\ast}V^{\ast} = D(U)D(V)$. This tells you that Hermitian conjugation, $U \mapsto U^{\dagger}$, does not form a representation. Why?

So, for a general $\mbox{SU}(n)$, there are two kinds of n-component spinor (lower & upper). If the lower spinors $\psi_{a} \in [n]$ transform by the matrices $U \in \mbox{SU}(n)$, the upper spinors $\psi^{a} \equiv (\psi_{a})^{\ast} \in [n^{\ast}]$ transform according to the conjugate representation, i.e., by the matrices $U^{\ast} = (U^{-1})^{t}$.

b) Two representations, $U$ and $V$ are said to be equivalent if they are related by similarity transformation, i.e., if there exists a non-singular matrix $C$ such that $CUC^{-1} = V$.

Now, we will use (a) and (b) to show that, in $\mbox{SU}(2)$, the two fundamental representations $[2]$ and $[2^{\ast}]$ are equivalent.

For $\psi \in [2]$, we have $$\psi^{\prime} = U(\alpha)\psi, \ \ \ U = e^{-\frac{i}{2} \alpha_{i} \sigma_{i}} \in \mbox{SU}(2) .$$

Now, consider the matrix $$C = i \sigma_{2} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} .$$ It is an antisymmetric matrix with the properties $C^{\dagger} = C^{-1} = - C$.

Using the three Pauli matrices $\sigma_{i}$, you can easily show that $$C \sigma_{i} C^{-1} = - \sigma^{\ast}_{i}.$$ These $\mbox{SU}(2)$ relations, together with the identity $$C e^{A}C^{-1} = e^{CAC^{-1}} ,$$ allow you to show that $$C U C^{-1} = U^{\ast}.$$ Thus, according to (b), the representations $U$ and $U^{\ast}$ are equivalent, i.e., the mapping $U \mapsto U^{\ast}$ is just an inner automorphism. Rewriting the above as $UC^{-1} = C^{-1}U^{\ast}$ and operating both sides on the complex conjugated spinor $\psi^{\ast}$, we find $$U \left( C^{-1} \psi^{\ast} \right) = C^{-1} \left( U^{\ast}\psi^{\ast}\right) = C^{-1} \left(\psi^{\ast}\right)^{\prime}.$$ Using the fact that $C$ is the $\mbox{SU}(2)$ invariant tensor $\epsilon_{ab}$, we can rewrite the above as $$U \left( C^{-1}\psi^{\ast} \right) = \left( C^{-1} \psi^{\ast}\right)^{\prime} .$$ If we define the spinor $\varphi \equiv C^{-1}\psi^{\ast}$, we find $$\varphi^{\prime} = U \varphi .$$ Thus, both $\psi = \begin{pmatrix}\psi_{1} \\ \psi_{2} \end{pmatrix}$ and $C^{-1}\psi^{\ast} = \begin{pmatrix} - \psi_{2}^{\ast} \\ \psi_{1}^{\ast} \end{pmatrix}$ transform by the same matrix $U \in \mbox{SU}(2)$, i.e., $$\psi \sim C^{-1}\psi^{\ast}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ meaning that they belong to the same fundamental representation $[2]$.

Now, from Eq(1), it follows that the Hermitian conjugate spinor $\psi^{\dagger}$ transforms like $\left( C^{-1} \psi \right)^{t} = \left( - \psi_{2} , \ \psi_{1} \right)$ and, therefore, for the tensor product $\psi \psi^{\dagger}$, we have $$\psi \psi^{\dagger} \sim - \begin{pmatrix} \psi_{1}\psi_{2} & - \psi_{1}^{2} \\ \psi_{2}^{2} & - \psi_{1}\psi_{2} \end{pmatrix} .$$ Since the traceless matrix on the RHS transforms like the Hermitian matrix $\psi \psi^{\dagger} \to U (\psi \psi^{\dagger}) U^{\dagger}$, then any $2 \times 2$ traceless Hermitian matrix, such as $X = \sigma_{i}x^{i}, \ x^{i} \in \mathbb{R}^{3}$, transforms like $\psi \psi^{\dagger}$: $$X \to X^{\prime} = U X U^{\dagger} .$$ The rest is a textbook material which I don’t repeat in here: you simply use the above transformation law for $X$ to show that the $\mbox{SU}(2)$ transformations on spinors $\psi \to U(\alpha)\psi$ correspond to $\mbox{SO}(3)$ rotations on the coordinates $x \to R(\theta)x$.