Consider for example the (trivial) spin-1/2 representation of the $SU(2)$ group. This representation has dimension two, which is clear from a quantum mechanical perspective since we need to specify the two coefficients $$|\psi\rangle=a_{+}|+\rangle+a_{-}|-\rangle$$ using the vector $$(a_{+},a_{-})$$ The relevant matrices are the Pauli matrices.

Now consider the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group. This representation has dimension four (implying that we are dealing with four components vectors), however the generators are still written in terms of the Pauli matrices and are thus $2\times 2$. Why is this ok?


1 Answer 1


The six generators are $S^{ab}\equiv i[\gamma^a,\gamma^b]/4$, where $(a,b)$ are the three boosts $(0,1), (0,2) (0,3),$ and the three spatial rotations $(1,2), (2,3), (3,1)$. The Dirac gamma matrices are $4$-by-$4$ matrices. It's true that you can write down the gamma's in terms of two-by-two blocks of two-by-two Pauli sigma's, but that is still four-by-four matrices.

The Lie algebra of $SO(3,1)$ is locally isomophic to that of $SU(2)\times SU(2)$ which is what the $1/2,1/2$ representation refers to, but unless you are dealing with massless (Weyl) particles you need both $SU(2)$'s.

We can take $$ \gamma_0 = \left(\matrix{0&1\\ 1&0}\right), \quad \gamma^a= \left(\matrix{0&\sigma_a\\ -\sigma_a &0}\right). $$ We also have $$ \quad \gamma^5= \left(\matrix{1&0\\ 0&-1}\right), $$ Then setting $(\Sigma_1,\Sigma_2,\Sigma_3)=(S_{23},S_{31},S_{12})$ $$ \Sigma_a= \frac 12 \left(\matrix{\sigma_a&0\\ 0&\sigma_a}\right) $$ are the rotation generators, and $$ K_a=\frac i2 \left(\matrix{\sigma_a&0\\ 0&-\sigma_a}\right) $$ are the boost generators (I do not guarantee that the signs are correct). Note that the boost generators are not Hermitian. The non-compact Lorentz group $SO(1,3)$ has no finite dimensional unitary representation.

My representation matrices are adapted to the decomposition of the algebra into the two $S_a\pm iK_a$ mutually commuting $SU(2)$ subalgebras of $SO(1,3)$.

If I got my signs right, the matrices should obey the Lorentz algebra $$ [\Sigma^a,\Sigma^b]= i \epsilon_{abc} \Sigma^c\\ [\Sigma^a,K^b]= i \epsilon_{abc} K^c\\ [K^a,K^b]= - i\epsilon_{abc} \Sigma^c. $$

  • $\begingroup$ How do you know how to construct these generators given the 2x2 representation? $\endgroup$
    – Watw
    Mar 25, 2017 at 18:17
  • $\begingroup$ You construct gamma matrices: I'll amend my original answer to describe. $\endgroup$
    – mike stone
    Mar 25, 2017 at 20:56
  • $\begingroup$ I don't really think this explains how you combine the two $SU(2)$ algebras to get the Lorentz algebra... $\endgroup$
    – Watw
    Mar 26, 2017 at 10:11
  • $\begingroup$ @Watw. Set $J^{(1)}_a = \Sigma_a+iK_a$ and $J^{2)}_a= \Sigma_a-iK_a$, then $[J^{(1)}_a, J^{(1)}_a]=i\epsilon_{abc} J^{(1)}_c$. Same for $J^{(2)}_a$. And $[J^{(1)}_a, J^{(2)}_a]=0$. $\endgroup$
    – mike stone
    Mar 26, 2017 at 12:55
  • $\begingroup$ @Mike Stone: Do you realize that the question was about the $D(\frac{1}{2},\frac{1}{2})$-representation whereas you explained about the $D(\frac{1}{2},0) \oplus D(0,\frac{1}{2})$-representation. $\endgroup$ Dec 19, 2017 at 13:44

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