Let \(u\) be a left-handed Weyl spinor transforming in the \(\left(\tfrac12,0\right)\) representation. Under an infinitesimal spatial rotation by a small angle vector \(\delta\boldsymbol{\theta}\),
\[ u \to u' = \left(1-\frac{i}{2}\,\delta\boldsymbol{\theta}\!\cdot\!\boldsymbol{\sigma}\right)u. \]Equivalently, for a finite rotation,
\[ u \to u' = S\,u, \qquad S = \exp\!\left(-\frac{i}{2}\,\boldsymbol{\theta}\!\cdot\!\boldsymbol{\sigma}\right). \]Since \(\sigma_i^\dagger=\sigma_i\), we have
\[ S^\dagger = \exp\!\left(+\frac{i}{2}\,\boldsymbol{\theta}\!\cdot\!\boldsymbol{\sigma}\right)=S^{-1}, \]so \(S\) is unitary.
The statement to prove is:
\[ u^\dagger u \text{ is a scalar under rotations,} \qquad u^\dagger \sigma_i u \text{ transforms as a 3-vector.} \]I will show both.
Under the rotation,
\[ u' = Su, \qquad u'^\dagger = u^\dagger S^\dagger. \]Therefore
\[ u'^\dagger u' = u^\dagger S^\dagger S u. \]But \(S^\dagger S=1\), so
\[ u'^\dagger u' = u^\dagger u. \]So \(u^\dagger u\) is invariant under rotations.
Define
\[ V_i \equiv u^\dagger \sigma_i u. \]Under the rotation,
\[ V_i' = u'^\dagger \sigma_i u' = u^\dagger S^\dagger \sigma_i S u. \]So the whole question reduces to showing that
\[ S^\dagger \sigma_i S = R_{ij}\sigma_j, \]where \(R_{ij}\) is the ordinary \(3\times 3\) rotation matrix. Then it follows immediately that
\[ V_i' = u^\dagger (R_{ij}\sigma_j)u = R_{ij}\,u^\dagger \sigma_j u = R_{ij}V_j, \]which is exactly the transformation law of a 3-vector.
So let us prove
\[ S^\dagger \sigma_i S = R_{ij}\sigma_j. \]Take
\[ S = 1-\frac{i}{2}\delta\theta_k \sigma_k, \qquad S^\dagger = 1+\frac{i}{2}\delta\theta_k \sigma_k. \]Then
\[ S^\dagger \sigma_i S = \left(1+\frac{i}{2}\delta\theta_k \sigma_k\right)\sigma_i \left(1-\frac{i}{2}\delta\theta_\ell \sigma_\ell\right). \]Keeping only first-order terms in \(\delta\boldsymbol{\theta}\),
\[ S^\dagger \sigma_i S = \sigma_i +\frac{i}{2}\delta\theta_k \sigma_k\sigma_i -\frac{i}{2}\delta\theta_\ell \sigma_i\sigma_\ell. \]Rename the dummy index \(\ell\to k\):
\[ S^\dagger \sigma_i S = \sigma_i +\frac{i}{2}\delta\theta_k(\sigma_k\sigma_i-\sigma_i\sigma_k). \]Thus
\[ S^\dagger \sigma_i S = \sigma_i+\frac{i}{2}\delta\theta_k[\sigma_k,\sigma_i]. \]Now use the Pauli commutator relation
\[ [\sigma_k,\sigma_i]=2i\,\epsilon_{kij}\sigma_j. \]So
\[ S^\dagger \sigma_i S = \sigma_i+\frac{i}{2}\delta\theta_k(2i\,\epsilon_{kij}\sigma_j) = \sigma_i-\delta\theta_k\epsilon_{kij}\sigma_j. \]Using antisymmetry of \(\epsilon_{kij}\),
\[ -\epsilon_{kij}=\epsilon_{ikj}, \]hence
\[ S^\dagger \sigma_i S = \sigma_i+\delta\theta_k\epsilon_{ikj}\sigma_j. \]This is exactly the infinitesimal rotation law for a vector:
\[ V_i' = V_i+\delta\theta_k\epsilon_{ikj}V_j. \]Therefore \(u^\dagger \sigma_i u\) transforms as a 3-vector.
Start from
\[ V_i'=u'^\dagger\sigma_i u' = u^\dagger S^\dagger \sigma_i S u. \]Using the result above,
\[ S^\dagger \sigma_i S = \sigma_i+\delta\theta_k\epsilon_{ikj}\sigma_j, \]we get
\[ V_i' = u^\dagger\left(\sigma_i+\delta\theta_k\epsilon_{ikj}\sigma_j\right)u = u^\dagger\sigma_i u+\delta\theta_k\epsilon_{ikj}u^\dagger\sigma_j u. \]So
\[ V_i' = V_i+\delta\theta_k\epsilon_{ikj}V_j. \]That is the standard infinitesimal transformation rule for a 3-vector.
The Pauli matrices themselves carry the vector index \(i=1,2,3\). The spinor indices are hidden in the matrix multiplication, but the object
\[ u^\dagger \sigma_i u \]still has one free spatial index \(i\), so it is the natural candidate for a vector. The calculation above shows that it really does transform correctly.
By contrast,
\[ u^\dagger u \]has no free spatial index, so it is a scalar, and the unitarity of the spinor rotation matrix makes this explicit.
Under \(u\to Su\) with \(S=\exp\!\big(-\frac{i}{2}\theta\cdot\sigma\big)\),
\[ u^\dagger u \to u^\dagger S^\dagger S u = u^\dagger u, \]and
\[ u^\dagger \sigma_i u \to u^\dagger S^\dagger \sigma_i S u = R_{ij}\,u^\dagger \sigma_j u. \]So \(u^\dagger u\) is a rotational scalar, while \(u^\dagger\boldsymbol{\sigma}u\) is a rotational 3-vector.
This is additional material to complement page 459 of Group Theory in a Nutshell by Anthony Zee.
Created with assistance from ChatGPT.