For generators of the Lorentz group we have the following algebra: $$ [\hat {R}_{i}, \hat {R}_{j} ] = -\varepsilon_{ijk}\hat {R}_{k}, \quad [\hat {R}_{i}, \hat {L}_{j} ] = -\varepsilon_{ijk}\hat {L}_{k}, \quad [\hat {L}_{i}, \hat {L}_{j} ] = \varepsilon_{ijk}\hat {R}_{k}. $$ For the splitting of algebra, we can introduce operators $$ \hat {J}_{k} = \hat {R}_{k} + i\hat {L}_{k}, \quad \hat {K}_{k} = \hat {R}_{k} - i\hat {L}_{k}. $$ So $$ [\hat {J}_{i}, \hat {J}_{j} ] = -\varepsilon_{ijk}\hat {J}_{k}, \quad [\hat {K}_{i}, \hat {K}_{j} ] = -\varepsilon_{ijk}\hat {K}_{k}, \quad [\hat {J}_{i}, \hat {K}_{j}] = 0. $$ So, each irreducible representation of Lie algebra is characterized by $(j_{1}, j_{2})$, where $j_{1}$ is max eigenvalue of $\hat {J}_{3}$ and $j_{2}$ is max eigenvalue of $\hat {K}_{3}$.

Then I can classify objects that transform through the matrices of the irreducible representations, $$ \Psi_{\mu \nu}' = S^{j_{2}}_{\mu \alpha }S^{j_{2}}_{\nu \beta}\Psi_{\alpha \beta}, $$ where $S^{j_{i}}_{\gamma \delta}: (2j_{i} + 1)\times (2j_{i} + 1)$.

For $(0, 0)$ I have scalar field, for $\left(\frac{1}{2}, 0\right); \left(0; \frac{1}{2}\right)$ I have spinor, for $(1, 0); (0, 1)$ I have 3-vectors $\mathbf a, \mathbf b -> \mathbf a + i\mathbf b$ creating antisymmetrical tensor etc.

Also, for scalar $j_{1} + j_{2} = 0$, for spinor - $\frac{1}{2}$, for tensor - $1$. So, the question: is sum $j_{1} + j_{2}$ experimentally observed? Is it connected with a spin?


Yes, the comination $j_1 + j_2$ determines the spin of the particle. Note however, that this is an addition of angular mementum which may be complicated.

Furthermore, you can count the degrees of freedom: In $(j_1, j_2)$, each contribute $2j_1 + 1$ states and we construct a tensor product, so $(j_1, j_2)$ gives $(2j_1 + 1) * (2j_2 + 2)$ degrees of freedom. For the vector we have $(1/2, 1/2) \mapsto 2 * 2 = 4$ degrees of freedom.

If the representation is reducible, i.e. of the form $(j_1, j_2) + (k_1, k_2)$, then you simply add the d.o.f. you get from each pair. The dirac spinor has $(1/2, 0) + (0, 1/2) \mapsto (2) + (2) = 4$ degrees of freedom, the field-strength tensor has $(1, 0) + (0, 1) \mapsto 3 + 3 = 6$ d.o.f.

As a sidenote: the representations $(1, 0)$ do not corresond to vectorlike degrees of freedom, but rather to antisymmetric self-dual tensors. $(0, 1)$ is the antisymmetric anti-self-dual tensor. The vector (and the only way to get a vector out of this) is $(1/2, 1/2)$!

  • $\begingroup$ But how did you realize that the $j_{1} + j_{2}$ is a spin? $\endgroup$ – user8817 Jul 27 '13 at 12:45
  • $\begingroup$ Maybe, it's refer to the fact, that $j_{1} + j_{2}$ corresponds to some representation of $\frac{1}{2}(\hat {R}_{3} + i\hat {L}_{3}) + \frac{1}{2}(\hat {R}_{3} - i\hat {L}_{3})$, which is rotations generator? $\endgroup$ – user8817 Jul 27 '13 at 13:06
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    $\begingroup$ Indeed. Spin in general (not limited to the S = 1/2 case) is nothing but the representation w.r.t. the rotation group, which is a subgroup of the Lorentz group. $\endgroup$ – Neuneck Jul 27 '13 at 14:18
  • $\begingroup$ And we can talk about $j_{1} + j_{2}$ as observable because operator-matrix (as tensor product of two operators $\hat {J}_{i}, \hat {K}_{i}$) of corresponding irreducibles representation is hermite? $\endgroup$ – user8817 Jul 27 '13 at 15:12
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    $\begingroup$ This is a difficult matter, as the $\hat K_i$ themselves are actually not Hermitian. This has to do with the fact that the underlying group's parameter space is non-compact (boosts can go up to any rapidity). The $\hat J_i$ are perfectly observable, though, as is the Pauli-Lubanski pseudovector. $\endgroup$ – Neuneck Jul 27 '13 at 15:38

Yes, a representation labeled by $(j_1,j_2)$ corresponds to the total spin $j_1+j_2$, (rigourously speaking of spin needs that one of $j_1$ or $j_2$ is zero) and if $j_1=j_2$, this is a real representation, but you may have a representation which is a sum or irreductible representations , some examples:

$(\frac{1}{2},0)$ corresponds to a left-handed Weyl spinor

$(0,\frac{1}{2})$ corresponds to a right-handed Weyl spinor

$(\frac{1}{2},0) + (0,\frac{1}{2})$, is the Dirac bi-spinor

$(\frac{1}{2},\frac{1}{2})$ corresponds to a Lorentz vector.

$(1,0) + (0,1)$, is the electromagnetic field representation

More generally, if the complex conjugate of a representation (interverting in all terms $j_1$ and $j_2$) is the same as the representation, then the representation is real.

For instance, the Dirac representation, or the electromagnetic field representation, are real representations.

  • $\begingroup$ Thank you. But I don't understand, how to identity experimentally observed spin and $j_{1} + j_{2}$? $\endgroup$ – user8817 Jul 26 '13 at 14:12
  • $\begingroup$ Good question...Well, I would say, that practically, the observed particles are in a representation $(0,x)$ or $(x,0)$ or a sum $(x,0)+(0,x)$ $\endgroup$ – Trimok Jul 26 '13 at 14:20
  • $\begingroup$ So maybe it is abusive to speak of spin if $j_1$ and $j_2$ are both not zero. $\endgroup$ – Trimok Jul 26 '13 at 14:27
  • $\begingroup$ Why? If one of $j_{i}$ is equal to zero, it only means that one of matrices of irreducible representations is absent. Moreover, only a sum of $j_{1}, j_{2}$ may be associated with hermite operator. $\endgroup$ – user8817 Jul 26 '13 at 14:45
  • $\begingroup$ So, if I understand correctly, you can "observe" only the sum $j_{1} + j_{2}$. $\endgroup$ – user8817 Jul 26 '13 at 14:52

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