Spinor product in QED scattering Equation 6.36 in Larkoski’s Introduction to Particle Physics says
$v^{\dagger}_{L}\sigma^{\mu}u_R$ =
$E_{cm}(0, -i)(1, \sigma_1, \sigma_2, \sigma_3)\begin{bmatrix}1 & 0\end{bmatrix}$
Which is one of the currents used in calculating the scattering amplitude for muon production.
My question is how are we supposed to multiply $\sigma^\mu$ which is supposedly a 4 component vector with a 2 component spinor $u_R$?. Isn’t multiplication of mathematical objects with different dimensions not defined?
 A: $v^{\dagger}_{L}\sigma^{\mu}u_R$ is a Lorentz four-vector, as the notation suggests.  Each component is just a number, computed by multiplying together -- in "spin space" -- the 2-element row vector $v^\dagger_L$, the 2$\times$2 spin matrix with index $\mu$, and the 2-element column vector $u_R$.
The Lorentz index $\mu$ has nothing to do with this multiplication in spin space, other than choosing which spin matrix to use. Lorentz indices and spinor indices -- the latter being completely omitted in this compact notation -- are completely separate, like apples and oranges; they index components in entirely different vector spaces.
Using the explicit forms of the spin matrices, the four Lorentz components are found to be
$$v^{\dagger}_{L}\sigma^{0}u_R = E_\text{cm}
\begin{bmatrix}0 & -i\end{bmatrix}
\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix}1 \\ 0\end{bmatrix} = E_\text{cm}(0);$$
$$v^{\dagger}_{L}\sigma^{1}u_R = E_\text{cm}
\begin{bmatrix}0 & -i\end{bmatrix}
\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix}1 \\ 0\end{bmatrix} = E_\text{cm}(-i);$$
$$v^{\dagger}_{L}\sigma^{2}u_R = E_\text{cm}
\begin{bmatrix}0 & -i\end{bmatrix}
\begin{bmatrix}0 & -i \\ i & 0 \end{bmatrix}
\begin{bmatrix}1 \\ 0\end{bmatrix} = E_\text{cm}(1);$$
$$v^{\dagger}_{L}\sigma^{3}u_R = E_\text{cm}
\begin{bmatrix}0 & -i\end{bmatrix}
\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}
\begin{bmatrix}1 \\ 0\end{bmatrix} = E_\text{cm}(0).$$
Thus the Lorentz four-vector is
$$v^{\dagger}_{L}\sigma^{\mu}u_R = E_\text{cm}(0, -i, 1, 0).$$
Its temporal component is zero, leaving just the 3-vector
$$v^{\dagger}_{L}\sigma^i u_R = E_\text{cm}(-i, 1, 0)$$
or, in traditional 3D vector notation,
$$v^{\dagger}_{L}\vec\sigma \, u_R = E_\text{cm}(-i\hat x + \hat y).$$
