3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ it is a combination of (usual) 2-dim pauli matrices (in some representations) $\endgroup$ – Nikos M. Oct 28 '14 at 11:01
  • $\begingroup$ Tanks! but can you describe the combination procedure more explicit? I not very familiar with the Lie group theory,please. $\endgroup$ – alxandernashzhang Oct 28 '14 at 11:07
  • $\begingroup$ take a look here: en.wikipedia.org/wiki/Pauli_matrices, physicsforums.com/threads/…, effectively pauli matrices are the generators of $SU(2)$ $\endgroup$ – Nikos M. Oct 28 '14 at 11:15
  • $\begingroup$ see these notes on unitary groups and representations cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter9.pdf as well $\endgroup$ – Nikos M. Oct 28 '14 at 11:17
  • $\begingroup$ 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 $\sigma_{i}$? $\endgroup$ – alxandernashzhang Oct 28 '14 at 11:23
6
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – alxandernashzhang Oct 28 '14 at 11:13
  • $\begingroup$ 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! $\endgroup$ – alxandernashzhang Oct 29 '14 at 4:21
  • $\begingroup$ 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 ${({\gamma}^{\mu}{T}^{a}\psi)}^{\alpha i}$** $\endgroup$ – alxandernashzhang Oct 29 '14 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.