# What is the four-dimensional representation of the $SU(2)$ generators?

Recently, I have been learning about non-Abelian gauge field theory by myself. Thanks @ACuriousMind very much, as with his help, I have made some progress.

I am trying to extend the Dirac field equation with a coupling to a $SU(2)$ gauge field: $$(i{\gamma}^{\mu }{D}_{\mu}-m)\psi =0$$ where $${ D }_{ \mu }=\partial _{ \mu }+ig{ A }_{ a }^{ \mu }{ T }_{ a }$$ the ${ T }_{ a }$ is the $SU(2)$ Lie group generator, with $[{ T }_{ a },{ T }_{ b }]=i{ f }^{ abc }{ T }_{ c }$, and the ${\gamma}^{\mu }$ are the Dirac matrices. When I write explicitly the first part of the Dirac equation, with spinor form $\psi=(\phi,\chi)^T$, I get (spatial part): $$\begin{pmatrix} 0 & { \sigma }^{ i } \\ -{ \sigma }^{ i } & 0 \end{pmatrix}\partial _{ i }\begin{pmatrix} \begin{matrix} \phi \\ \chi \end{matrix} \end{pmatrix}+ig\begin{pmatrix} 0 & { \sigma }^{ i } \\ -{ \sigma }^{ i } & 0 \end{pmatrix}{ A }_{ a }^{ i }{ T }_{ a }\begin{pmatrix} \begin{matrix} \phi \\ \chi \end{matrix} \end{pmatrix}$$ My problem is: I only known the linear representation of ${ T }_{ a }$ is Pauli spin matrix from text book, but they are the set of 2-dimension matrixes, In above expression, I need to know the 4-dimension matrix of ${ T }_{ a }$ because of the spinor is 4-dimension, I check some test book, but didn't find the explicitly statement of the 4-D matrix.

So, as mentioned in title, What is the 4-dimension representation of the $SU(2)$ generators, or how can I calculate it?

• it is a combination of (usual) 2-dim pauli matrices (in some representations) – Nikos M. Oct 28 '14 at 11:01
• Tanks! but can you describe the combination procedure more explicit? I not very familiar with the Lie group theory,please. – alxandernashzhang Oct 28 '14 at 11:07
• take a look here: en.wikipedia.org/wiki/Pauli_matrices, physicsforums.com/threads/…, effectively pauli matrices are the generators of $SU(2)$ – Nikos M. Oct 28 '14 at 11:15
• see these notes on unitary groups and representations cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter9.pdf as well – Nikos M. Oct 28 '14 at 11:17
• Can you tell me explicitly, when I do the calculation in above, Which matrix representation I can use, dose it the original 2*2 Puali matrix $\sigma_{i}$? – alxandernashzhang Oct 28 '14 at 11:23

Comment to the question (v4): OP seems to effectively conflate spacetime symmetries and internal gauge symmetries. They act in different representations, or more precisely as a tensor product of representations.

For instance the fermion $\psi$ carries two types of indices, say $\psi^{\alpha i}$, $\alpha=1,2,3,4,$ and $i=1,2$. The fermion acts

1. as a $4$-dimensional Dirac spinor representation under Lorentz transformations.

2. as a $2$-dimensional fundamental representation of the gauge group $SU(2)$ under gauge transformations.

Similarly, the $4\times 4$ Dirac matrices $\gamma^{\mu}$ and the $2\times 2$ $SU(2)$ gauge group generator $T^a$ act on different representations. The product of $\gamma^{\mu}$ and $T^a$ is a tensor product. In particular, the term $\gamma^{\mu}T^a\psi$ in OP's formula again carries two types of indices, and is evaluated as

$$(\gamma^{\mu}T^a\psi)^{\alpha i}~=~(\gamma^{\mu})^{\alpha}{}_{\beta}~ (T^a)^{i}{}_{j}~\psi^{\beta j}.$$

• em, Dose you mean,because the SU(2) symmetry is internal (not involve space-time), the representation of the SU(2) generator always 2*2 puali matrix when the Covariant derivative asct on the 4*1 spinor (for SU(2), it is 2*1)?please continuely focus on this post, thanks! – alxandernashzhang Oct 28 '14 at 11:13
• Great!!thanks your powerful help!I think I understand what you means. I write it down, please check it, if it incorrect, please point out, if it is correct, also please tell me,thanks! – alxandernashzhang Oct 29 '14 at 4:21
• Great!thanks your powerful help!I think I understood. I write it down, please check it, if it incorrect,please point out,if it is correct, also please tell me,thanks!**As your mean,the Dirac field $\psi$ is a tensor product of it's space-time part and internal part as $\psi ={\psi}^{\alpha}\bigotimes {\psi}_{i}$, So the expression ${\gamma}^{\mu}{T}^{a}\psi =({\gamma}^{\mu}{ \bigotimes T }^{a})({\psi}^{\alpha}\bigotimes {\psi}_{i})=({\gamma}^{\mu}{\psi}^{\alpha})\bigotimes {(T}^{a}{\psi}_{i})$, result is a 8*1 column vector,and every element is ${({\gamma}^{\mu}{T}^{a}\psi)}^{\alpha i}$** – alxandernashzhang Oct 29 '14 at 4:45