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?

  • 2
    $\begingroup$ $\sigma^\mu$ which is supposedly a 4 component vector Yes, but do you understand what each of those four components is? $\endgroup$
    – Ghoster
    Commented Nov 13, 2022 at 22:07
  • $\begingroup$ @Ghoster yes, the spin matrices. However we still have 4 of those multiplied by 2 component spinors. $\endgroup$
    – user310742
    Commented Nov 14, 2022 at 1:03
  • $\begingroup$ 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. $\endgroup$
    – Ghoster
    Commented Nov 14, 2022 at 1:06
  • $\begingroup$ The point of the compact notation $v^{\dagger}_{L}\sigma^{\mu}u_R$ is that it’s a four-vector. $\endgroup$
    – Ghoster
    Commented Nov 14, 2022 at 1:09
  • $\begingroup$ @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). $\endgroup$
    – user310742
    Commented Nov 14, 2022 at 1:11

1 Answer 1


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

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

  • $\begingroup$ 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! $\endgroup$
    – user310742
    Commented Nov 16, 2022 at 12:07
  • $\begingroup$ 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.) $\endgroup$
    – Ghoster
    Commented Nov 16, 2022 at 17:38
  • $\begingroup$ 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. $\endgroup$
    – Ghoster
    Commented Nov 16, 2022 at 17:41

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