# Lie algebra valued potential vector [closed]

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.

## closed as unclear what you're asking by AccidentalFourierTransform, ZeroTheHero, sammy gerbil, Pulsar, Cosmas ZachosMay 14 '18 at 2:44

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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.
• More explicitly $A_\mu =\left( \begin{array}{cc} A^3_\mu & A^1_\mu -i A^2_\mu \\ A^1_\mu +i A^2_\mu & -A^3_\mu \end{array} \right)$ – Saksith Jaksri May 11 '18 at 15:28
• @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. – hypernova May 11 '18 at 15:44
• @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. – hypernova May 11 '18 at 15:52