1
$\begingroup$

I am sorry for the naivety of this question, I am a mathematician and I am trying to put together different ideas.

I am trying to understand the vocabulary of physics, in particular, I want to know:

We know that the isometry group of Minknowski metric is the Lorentz group $O(3,1)$. Furthermore, it acts on $\mathbb R^{3,1}$ by isometries. The isotropy group of this action is precisely $SU(2)$, so it is interested to understand further representations of $SU(2)$.

$1)$ What does mean the notation $(1/2,0)\oplus (0,1/2)?$

$2)$ How can I express an spinor from the $1/2$-representation of $SU(2)$?

$\endgroup$
  • $\begingroup$ Isn’t the isometry group of Minkowski space the Poincaré group? $\endgroup$ – G. Smith Jun 14 at 2:47
  • 1
    $\begingroup$ Isn’t the isotropy group of a timelike vector actually $SO(3)$, with $SU(2)$ being its double cover? $\endgroup$ – G. Smith Jun 14 at 3:27
  • $\begingroup$ Related: physics.stackexchange.com/q/28505/2451 , physics.stackexchange.com/q/149455/2451 and links therein. $\endgroup$ – Qmechanic Jun 14 at 3:39
  • $\begingroup$ Physicists’ labeling of $SU(2)$ irreps is: $0$ means the $1$-dimensional irrep; $1/2$ means the $2$-dimensional irrep; $s$ means the $2s+1$-dimensional irrep. Physicists think in terms of the “spin quantum number” $s$ rather than the dimensionality. $\endgroup$ – G. Smith Jun 14 at 5:21
  • $\begingroup$ For 2), physicists talk about at least three kinds of spinors: Dirac spinors, Weyl spinors, and Majorana spinors. $\endgroup$ – G. Smith Jun 14 at 5:28
5
$\begingroup$

Reference: Srednicki, Quantum Field Theory ch. 33

The Lie algebra of the Lorentz group $SO(3,1)$ can be described in terms of 3 spacial rotations ($J_i$) and 3 velocity boosts ($K_i$):

$$ [J_i,J_j] = i \epsilon_{ijk} J_k $$ $$ [J_i,K_j] = i \epsilon_{ijk} K_k $$ $$ [K_i,K_j] = -i \epsilon_{ijk} J_k $$

To make the $SU(2) \otimes SU(2)$ structure of the (complexified) algebra explicit we need to change to a different set of generators:

$$ N_i = \frac{1}{2} (J_i - iK_i) $$ $$ N_i^\dagger = \frac{1}{2} (J_i + iK_i) $$

This gives:

$$ [N_i,N_j] = i \epsilon_{ijk} N_k $$ $$ [N_i^\dagger,N_j^\dagger] = i \epsilon_{ijk} N_k^\dagger $$ $$ [N_i,N_j^\dagger] = 0 $$

So given that the irredicible representations of $SU(2)$ can be listed by a single index $j= 0, \frac{1}{2} , 1, \frac{3}{2} ...$ , irreducible representations of $SU(2) \otimes SU(2)$ clearly needs two such indices $(j_1,j_2)$ where $j_1$ and $j_2$ are the eigenvalues of $N_3$ and $N_3^\dagger$ above.

Important representations in physics are:

$(0,\frac{1}{2}) $ = Weyl Spinor

$(\frac{1}{2},\frac{1}{2}) $ = Vector

$(0,\frac{1}{2}) \oplus (\frac{1}{2},0) $ = Dirac Spinor

For each of these representations in turn we can calculate the corresponding SU(2) representations of the isotropy group (physicists usually call this the little group).

Physically the little group (isotropy group) corresponds to the subgroup of the Lorentz group which leave the momentum vector unchanged. This (at least for a massive particle) is the group of spatial rotations i.e. the $J_i$ above.

So we need to write $J_i$ in terms of the $N_i$ and $N_i^\dagger$ :

$$ J_i = N_i + N_i^\dagger $$

All physicists then recognise this as the process of "adding two angular momentum" (mathematically it is the process of finding the irreducible representations of the tensor product of two representations):

$(j_1,j_2)$ decomposes into the direct sum of representations $|j_1-j_2|, |j_1-j_2| + 1 , ... j_1 + j_2 $

Applying this we get:

$(0,\frac{1}{2}) : j = \frac{1}{2}$ so this is a Spin $\frac{1}{2}$ particle.

$(\frac{1}{2},\frac{1}{2}) : j = 0,1$ so this contains a Spin 1 (vector) particle.

$(0,\frac{1}{2}) \oplus (\frac{1}{2},0) : j = \frac{1}{2} $ in both parts of the sum, so this again is a Spin $\frac{1}{2}$ particle.

$\endgroup$
  • 1
    $\begingroup$ This paper: arxiv.org/abs/1312.3824 starts the reader from approximately zero knowledge (well a bit of quantum) and gets as far as the above equations (just about). It will read a bit too slowly for someone already familiar with group theory, but perhaps that is better than too fast. I should add that I wrote it. $\endgroup$ – Andrew Steane Jun 14 at 17:00
  • 1
    $\begingroup$ Note the covering group of SO(3,1) is in fact SL(2,C) not SU(2)xSU(2) as described in many more technical answers to this question elsewhere. However from an educational view it is nicer to think in term of SU(2)xSU(2) - this is justified since both groups have the same Lie aglebra structure over C - that structure is all that matters to understand things at this level. Others may object. $\endgroup$ – Bruce Greetham Jun 15 at 15:26

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.