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We can analyze fields transforming by the Lorentz group as $(m, n)$ representations, where $m,n$ are the max eigenvalues of two SU(2) operators, which generate the irreducible representation of the Lorentz group as $$ S^{m, n} = S^{m}\times S^{n}, $$ and then to identify these representations with vectors, spinors, scalars etc with spin $m + n$. We can build irreducible spinor representation $\left( \frac{n}{2}, \frac{m}{2} \right)$

$$\psi_{c_{1}...c_{n}\dot {c}_{1}...\dot {c}_{m}}, \quad \psi_{c_{1}...c_{n}\dot {c}_{1}...\dot {c}_{m}} {'} = S^{\quad a_{1}}_{c_{1}}\dots S^{\quad a_{n}}_{c_{n}}S^{\quad \dot {a}_{1}}_{\dot {c}_{1}}\dots S^{\quad \dot {a}_{m}}_{\dot {c}_{m}}\psi_{a_{1}...a_{n}\dot {a}_{1}...\dot {a}_{m}}, $$ which refers to the symmetrical spinor tensor - product of n undotted spinors and m dotted spinors.

Then, the relativistic field is the field transforming by the Poincare group: $$ \psi^{'}(x') = T\psi, \quad x^{\mu }{'} = \Lambda^{\mu}_{\quad \nu}x^{\nu} + a^{\mu}. $$ By using spinor representation of the Lorentz group we can get $$ T = T^{\quad a_{1}}_{c_{1}}\dots T^{\quad a_{n}}_{c_{n}}T^{\quad \dot {a}_{1}}_{\dot {c}_{1}}\dots T^{\quad \dot {a}_{m}}_{\dot {c}_{m}}, \quad \psi = \psi_{a_{1}...a_{n}\dot {a}_{1}...\dot {a}_{m}}. $$ Stupid question: why do we say that the fields classification in case of their Lorentz representations remains unchanged for a case of Poincare representations?

I.e., if we say that the field is antisymmetrical tensor with spin 1 under transformations Lorentz group, why do we say, that field is antisymmetrical tensor with spin 1 under transformations of the Poincare group?

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This isn't a stupid question at all. The representations of the Poincare group are labelled by mass and spin. For each fixed mass the representations look like those for the Lorentz group. Beware - your claim is false for massless representations however! In the massless case the representations are labelled by helicity. So while naively you expect the electromagnetic field to have three spin states, a photon only has 2 polarization states, for example. Let me know if you want more details - I can give a complete answer tomorrow if you like! –  Edward Hughes Aug 31 '13 at 22:29
    
@EdwardHughes . "...For each fixed mass the representations look like those for the Lorentz group...", - it's a little unclear to me. Can you give a complete answer? "...Beware - your claim is false for massless representations however!..", - why? Photon has three states natively, because the helicity can be equal to the spin number. But the gauge nature of the electromagnetic field leads to freezing of 0-state. But anyway EM field is escribing by the second-rank tensor, which natively corresponds to three state of freedom. –  PhysiXxx Sep 1 '13 at 5:01
    
Anti-symmetric tensor with spin-one is confusing, $F_{\mu\nu}$ has Lorentz spin $(1,1)$ or using isomorphism with $sl(2,C)$ it is labelled by $(2,0)+(0,2)$. However, $F_{\mu\nu}$ can (not must) describe a massless spin-one (now this refers to representation of $iso(3,1)$). The same spin-one can be described in terms of $A_\mu$. $A_\mu$ can describe a massive spin-one as well. The same time $F_{\mu\nu}$ can describe a massless scalar in $4d$ (so-called notoph). A lot depends on the equations imposed on the field. The dictionary between particles and fields is not one-to-one. –  John Sep 1 '13 at 6:44
    
@John. You're wrong. It is represented as (0, 1) or (1, 0), because refer to complex 3-vector with $(2*1 + 1)*(1*0 + 1)$ degrees of freedom. –  PhysiXxx Sep 1 '13 at 10:31
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@PhysiXxx - I'll try to address your questions and write a full response this evening (UK time)! –  Edward Hughes Sep 3 '13 at 10:16
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