# 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?

• $\sigma^\mu$ which is supposedly a 4 component vector Yes, but do you understand what each of those four components is? Nov 13, 2022 at 22:07
• @Ghoster yes, the spin matrices. However we still have 4 of those multiplied by 2 component spinors. Nov 14, 2022 at 1:03
• No, the multiplications are in “spin space”. The final result still has a Lorentz index $\mu$… it’s four numbers. Lorentz indices and spinor indices (suppressed here) are apples and oranges. Try writing all the spinor indices to see how the spinors and spin matrices multiply. Nov 14, 2022 at 1:06
• The point of the compact notation $v^{\dagger}_{L}\sigma^{\mu}u_R$ is that it’s a four-vector. Nov 14, 2022 at 1:09
• @Ghoster ok, then how are we supposed to get the answer of (-ix + y) where x and y are the x and y direction unit vectors? I don’t see how we are supposed to multiply them (the textbook doesn’t specify). Nov 14, 2022 at 1:11

$$v^{\dagger}_{L}\sigma^{\mu}u_R$$ 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,

$$v^{\dagger}_{L}\vec\sigma \, u_R = E_\text{cm}(-i\hat x + \hat y).$$

• Ohh, now I understand. So it’s just a 4 vector and each component represents a multiplication. I probably need to look more into spinor indices. As I’m not completely comfortable with them yet. Thanks! Nov 16, 2022 at 12:07
• If you make the spinor indices explicit, $v^{\dagger}_{L}\sigma^{\mu}u_R$ would be $\sum_{a=1}^2\sum_{b=1}^2(v^{\dagger}_{L})_a(\sigma^{\mu})_{ab}(u_R)_b$ or, using the Einstein summation convention on the spinor indices, $(v^{\dagger}_{L})_a(\sigma^{\mu})_{ab}(u_R)_b$. (I’m writing spinor indices as consistently “down”; I think there are various conventions for how to do them. But these indices are usually just suppressed, so I never really learned the details.) Nov 16, 2022 at 17:38
• The multiplications between the spinors and the spin matrices “contract” these indices in pairs, so there is no free spinor index left; the multiplications leave just a number in spin space. But the Lorentz index is uncontracted (free) because it doesn’t take part in the spinor multiplication. Nov 16, 2022 at 17:41