Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's have Lorentz group with generators of 3-rotations, $\hat {R}_{i}$, and Lorentz boosts, $\hat {L}_{i}$. By introducing operators

$\hat {J}_{i} = \frac{1}{2}\left(\hat {R}_{i} + i\hat {L}_{i}\right), \quad \hat {K}_{i} = \frac{1}{2}\left(\hat {R}_{i} - i\hat {L}_{i}\right)$

we makes algebra of the Lorentz group the same as SU(2) (or SO(3)) group. So each irreducible representation of the Lorentz group can be built as $$ \hat {\mathbf S}^{(j_{1}, j_{2})} = \hat {\mathbf S}^{j_{1}}\times \hat {\mathbf S}^{j_{2}}, $$ where $j_{1}, j_{2}$ are the max eigenvalues of $\hat {J}_{i}, \hat {K}_{i}$,

and it has dimention $(2j_{1} + 1)\times (2j_{2} + 1)$. The type of object, transforming via boosts and 3-rotations, is depend on $(j_{1}, j_{2})$: $$ \Psi_{\alpha \beta} = S^{j_{1}}_{\alpha \mu}S^{j_{2}}_{\beta \nu}\Psi_{\mu \nu}. $$ For $(0, 0)$ we have scalar, for $\left(\frac{1}{2}, 0 \right), \left(0, \frac{1}{2}\right)$ we have spinor (left- and right-handled) etc. The value $j_{1} + j_{2}$ corresponds to the maximum value of $\hat {J}_{i} + \hat {K}_{i} = \hat {R}_{i}$, so it is an eigenvalue of irreducible rep of 3-rotation operator and corresponds to the spin number.

But the irreducible rep of Lorentz group isn't unitary.

So, the question: how can we classify the objects via transformations by using non-unitary reps?

share|improve this question
2  
Whether the representation of Lorentz group on space of fields is unitary or not is not of any physical significance. One requires that representation of Lorentz group on space of states be unitary. Space of states is a Fock space generated by Fourier modes of fields and even though the fields themselves are under finite dimensional (hence nonunitary) representation of Lorentz group the Fock space generated by their Fourier modes give a unitary representation of the Lorentz group. –  user10001 Aug 5 '13 at 19:19
1  
@user10001 . How exactly does Fock space give a unitary representation of the Lorentz group? –  PhysiXxx Aug 5 '13 at 23:06
    
@PhysiXxx That's a good question, but one you should ask in its own post, rather than buried in the comments here. –  user1504 Aug 6 '13 at 11:49
add comment

1 Answer

up vote 1 down vote accepted

Note that particles correspond to irreductible unitary representations of the Poincaré group (alias inhomogeneous Lorentz group), not the Lorentz group alone.

In these Poincaré representations, states are represented by $|p, \lambda \rangle$. $p$ is the momentum.

Let's consider positive massive representations ($p^2 = m^2, p^o >0$) Let $\pi=(m,\vec 0)$ . We see that we have a freedom to choose polarization, which corresponds to a $S0(3)$ symmetry. Looking at unitary representations of $SO(3)$ is the same thing that looking at representations of $SU(2)$

Here, $\lambda$ is a state basis for a little group $SU(2)$ representation $s$.

For a translation, we have :

$$U(a)|p, \lambda \rangle = e^{ iP.a}|p, \lambda \rangle$$

For a member $R$ of the little group $SU(2)$ , we have :

$$U(R)|\pi, \lambda \rangle = \sum_{\lambda'} D^{(s)}_{\lambda' \lambda}(R)|\pi, \lambda' \rangle$$

For any $SL(2,C)$ matrix $A$ , and for any $p$, it is possible to write an expression :

$$U(A)|p, \lambda \rangle = \sum_{\lambda'} D^{(s)}_{\lambda' \lambda}(W(p, A))| \Lambda_ap, \lambda' \rangle$$ where $W(p,A)$ is a $SU(2)$ little group element (see formula $18$ in the reference cited below for details)

With all this, you get an unitary representation of the Poincaré group.

The "Fock space" is the quantum version of these representations, that is it allows several-particles states.

See Reference pages 4 and 5

[EDIT] "For fields isn't important to have lorentz-invariant positive definite norm?"

No. Take for instance the Dirac equations for the bi-spinor field. The representation is $(1/2,0) + (0,1/2)$. This is not a unitary representation. There is a left and a right spinor. The transformation could be written :

$$\psi_{L,R} =\rightarrow e^{1/2(i\vec \sigma. \vec \theta \mp \vec \sigma. \vec \phi)}\psi_{L,R},$$

The parameters $\vec \theta$ correspond to rotations, the parameters $\vec \phi$ correspond to boosts.

Because the boost part is not unitary, we see clearly that the representation is not unitary.

So, this means that the bispinor bilinear expression $\psi^* \psi = \psi^*_{L}\psi_{L} + \psi^*_{R}\psi_{R}$ is not conserved in a Lorentz transformation [in fact, separarely, the spinor bilinear expressions $\psi^{*}_{L} \psi_{L}$ or $\psi^{*}_{R} \psi_{R}$ are not conserved too]. Remember here that the $ \psi,\psi_{L}, \psi_{R}$ are fields, not "wave function".

Is this a problem ? No.

What is $\psi^*(x) \psi(x)$ ? It is just (mutliplied by $e$) the charge density of fields, that is $j^0(x)$

So, of corse, $j^0(x)$ is not an invariant for a Lorentz transformation, because it is the time component of a Lorentz vector.

The real Lorentz invariant is here : $\overline \psi(x) \psi(x)= \psi^*(x) \gamma^0 \psi(x)$

share|improve this answer
    
"...Note that particles correspond to irreductible unitary representations of the Poincaré group (alias inhomogeneous Lorentz group), not the Lorentz group alone...", - but we discussing about the fields, not about the wave-functions. For fields isn't important to have lorentz-invariant positive definite norm. –  PhysiXxx Aug 6 '13 at 16:05
    
@PhysiXxx : I have made an edit to the answer –  Trimok Aug 7 '13 at 6:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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