I) Introduction


The mathematical structure is quite clear: given a spacetime $M$, and a Lie group $G$ (the gauge group), we can construct the Principal bundle $P^{G}_{M}$. The connection $1$-form in $P^{G}_{M}$ introduces the gauge $\mathcal{A}$ field in physics.

Given the spacetime and the principal bundle, matter fields are introduced in another manifold called the associated bundle $A_{P}$. Inside this space, we craft the gauge covariant derivative:

$$D_{\mu} \phi = \partial_{\mu} \phi+\rho(\mathcal{A}_{s})\phi, \tag{1}$$

where, $\rho$ is the representation $\rho: \mathfrak{g} \to GL(V)$ of a given Lie group $G$; $\mathfrak{g}$ is the lie algebra.

Now, the mathematics is imprecise, for example the $\mathcal{A}$ inside $\rho(\mathcal{A})$ in $(1)$, isn't just the connection, but rather the local connection $1$-form $\mathcal{A}_{s}$. But my point is: given the principal bundle and the associated bundle the connection is "just one".


In Weinberg-Salam model, the gauge group is $SU(2)_{L} \times U(1)_{Y}$. This group have this form because we didn't "measure" the right electron neutrino $\nu_{{e}_{R}}$. So, we need a space to acomodate a doublet:

$$\psi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ \end{pmatrix},$$

together with a singlet:

$$\phi = e_{R}.$$

So, the Weinberg-Salam-Glashow model deals with "triplets":

$$\xi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ e_{R} \end{pmatrix} \implies \Psi = \begin{pmatrix} \psi \\ \phi \end{pmatrix} \equiv \begin{pmatrix} L \\ R \end{pmatrix}$$


The covariant derivative for Weinberg-Salam-Glashow model is:

$$D_{\mu} = \partial_{\mu} + \rho(\mathcal{A}) = \partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu} \tag{2}. $$

II) Question

Most texts books says that when you act the covariant derivative in the triplet, $D_{\mu}\Psi$, some internal calculation "breaks" the single $D_{\mu}$ in $(2)$ into two different ones:

$$D_{\mu}\Psi =\Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)\Psi \implies \tag{3}$$

$$ D_{\mu}L = \Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)L \tag{4}$$

$$ D_{\mu}R = \Big(\partial_{\mu} - \frac{ig'}{2}B_{\mu}\Big)R \tag{5}$$

My question is: what is the calculation that breakes $(3)$ into $(4)$ and $(5)$?

III) My attempt

$$D_{\mu}\Psi = \Biggr(\partial_{\mu} \otimes \mathbb{I}_{3\times3} - \frac{ig}{2}\begin{pmatrix} W^{3}_{\mu} & W^{1}_{\mu} - iW^{2}_{\mu} & 0\\ W^{1}_{\mu} + iW^{2}_{\mu} & W^{3}_{\mu} & 0 \\ 0 & 0 & 0 \end{pmatrix} - \frac{ig'}{2}\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & B_{\mu} \end{pmatrix}\Biggr)\Psi $$

this matrix expression breaks the $D_{\mu}$ into different action for left particles and right particles, but is clear that won't give us the right set of equations.

So, I shall ask again: what is the calculation that breakes $(3)$ into $(4)$ and $(5)$? Please, consider to write a step-by-step calculation answer.

  • $\begingroup$ You wrote a bland direct sum; what calculation beyond that are you seeking? $\endgroup$ Jul 18, 2022 at 16:52
  • $\begingroup$ @CosmasZachos I want to know why the covariant derivative acts differently for left particles and right particles. But, I want to know the calculation for that. $\endgroup$
    – M.N.Raia
    Jul 18, 2022 at 17:02
  • $\begingroup$ "Why"? It's a just so statement: a fact. $\endgroup$ Jul 18, 2022 at 17:04
  • $\begingroup$ @CosmasZachos I agree with you: experimental physics just tell us that. But exists a formal mathematical calculation where in some place the action of $W$ field is zero for a right particle. $\endgroup$
    – M.N.Raia
    Jul 18, 2022 at 17:08
  • $\begingroup$ No, the asymmetric isolation of right-handed particles is a mysterious experimental fact. Feynman called the structure "cockeyed" because of it, and ribbed Weinberg about it. He was unreasonable. $\endgroup$ Jul 18, 2022 at 18:26


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