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Maybe it is a simple question but I have some difficulty to understand the explicit matrix form of this usual relation:

$$A_\mu=A^a_\mu \tau_a$$

where $A^a_\mu $ is the Lie algebra valued potential vector and $\tau_a$ the generators of $SU(2)$. Any help is appreciated.

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In this case, $\tau_a$'s are Pauli matrices, i.e., $$ \tau_1=\left( \begin{array}{cc} 0&1\\ 1&0 \end{array} \right),\quad\tau_2=\left( \begin{array}{cc} 0&-i\\ i&0 \end{array} \right),\quad\tau_3=\left( \begin{array}{cc} 1&0\\ 0&-1 \end{array} \right). $$ Therefore, \begin{align} A_{\mu}=A_{\mu}^a\tau_a&=A_{\mu}^1\tau_1+A_{\mu}^2\tau_2+A_{\mu}^3\tau_3\\ &=A_{\mu}^1\left( \begin{array}{cc} 0&1\\ 1&0 \end{array} \right)+A_{\mu}^2\left( \begin{array}{cc} 0&-i\\ i&0 \end{array} \right)+A_{\mu}^3\left( \begin{array}{cc} 1&0\\ 0&-1 \end{array} \right), \end{align} where $A_{\mu}^a$'s are scalar-valued functions.

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    $\begingroup$ More explicitly $A_\mu =\left( \begin{array}{cc} A^3_\mu & A^1_\mu -i A^2_\mu \\ A^1_\mu +i A^2_\mu & -A^3_\mu \end{array} \right)$ $\endgroup$ – Saksith Jaksri May 11 '18 at 15:28
  • $\begingroup$ But, how can you cope with the fact that on the left side you have a 4-vector and on the right side you have a 2x2 matrix? Thank you for your help in advanced. $\endgroup$ – John May 11 '18 at 15:32
  • $\begingroup$ @Oscar: That's fully tensorial. For this $A_{\mu}$ part, it deals with two spinors $\left(\psi_1,\psi_2\right)$, each component of which is a Dirac spinor. $\endgroup$ – hypernova May 11 '18 at 15:44
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    $\begingroup$ @Oscar: More precisely, you may put it as$$i\left[\gamma_a^b\right]^{\mu}\left(\partial_{\mu}+ie\left[A_c^d\right]_{\mu}\right)\left[\psi_d^a\right]=m\left[\psi_c^b\right],$$where the indices $a,b,c,d$ indicate the matrix rows and columns (respective matrices are highlighted in the brackets). You could see that in the isospin case, $\psi$ has eight components rather than four, because it contains two Dirac spinors. $\endgroup$ – hypernova May 11 '18 at 15:52

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