Update: see the restatement of the question below!

I've seen this question over and over through the archive of questions, but so far the closer to an answer was this. But I still don't understand.

First attempt

For tensor fields, the spin should be related with the number of indices (and their symmetries). This works OK for scalars and vectors.

A rank two tensor decomposes into: (1) symmetric-traceless, (2) skew-symmetric, and (3) trace.

  • The first one is a nine-dimensional irrep of $SO(3,1)$ should be a spin-two field ---in terms of $SU(2)\times SU(2)$ it is a $(3,3)$---, and there is no other possibility.
  • The second irrep. is a $(2,1) \oplus (1,2)$ of $SU(2)\times SU(2)$ which has spin one.
  • Finally, the trace is a $(1,1)$ irrep of $SU(2)\times SU(2)$, which correspond to spin zero.

Using this analysis, Is there an easy method to known what the spin of a certain field is?

Restatement of the question

After the comments by @ACuriousMind, I'd like to restate my question.

First, I understand that there is no isomorphism between $SO(3,1)$ and $SU(2)\times SU(2)$, that is why I emphasised the word. However, I really appreciate the concern (and the link!), because to many times doubts are all about the clarity in the language.

Second, the spin is part of the definition of the field (if we have a field transforming under an irrep of $SO(3,1)$). As stated above, a rank two tensor has no defined spin because it does not transform under an irrep of the Lorentz group.

So, finally... to my question!

Assume you have a field transforming under an irrep of the Lorentz group. For the sake of definiteness, say a rank three tensor which is symmetric and traceless in the first and second indices, but antisymmetric among the second and third.

Is there an easy, natural, or standard way to know the spin of this field?

The point is: We know that the irreps of $SO(3,1)$ can be classified in terms of the irreps of $SU(2)\times SU(2)$ (although they are not isomophic as groups). If someone tells me how this field is classified as irrep of $SU(2) \times SU(2)$... say $(a,b)\oplus (b,a)$ we would for sure say the spin of the field is $s=a+b$.

So, how could this info be obtained (with ease) from the symmetries of the field in the $SO(3,1)$ irrep?

In the above I used the isomorphism between $SU(2)\times SU(2)$ and $SO(3,1)$ to get some information of the irreps through Young diagrams.

How could the process be applied in dimensions with no such morphisms?

  • 2
    $\begingroup$ 1. There is no isomorphism between $\mathrm{SU}(2)\times\mathrm{SU}(2)$ and $\mathrm{SO}(1,3)$ although their representations are closely related, see this answer by Qmechanic. $\endgroup$ – ACuriousMind Oct 9 '15 at 16:03
  • 1
    $\begingroup$ 2. The spin of a field is part of its definition. If you have a "field" but don't know how it transforms under the global natural $\mathrm{SO}(d,1)$ symmetry, then you don't have a field. What data do you want to take as given for a "field" to derive its spin from? (One can e.g. derive that a field taking values in the tangent vectors is a spin-1 field, because differential geometry tells you how tangent vectors transform, but, well, then you defined the field to be a tangent vector field, hence you defined it to have spin-1) $\endgroup$ – ACuriousMind Oct 9 '15 at 16:04
  • $\begingroup$ 3. Fields don't need to have a well-defined spin. As you noticed with the general rank-2 tensor, it decomposes into three distinct representations (and this decomposition of the rank-2 tensors into symmetric, antisymmetric and trace is very general for $\mathrm{SO}(n)$ representations). I'm not really sure what this question is asking for. $\endgroup$ – ACuriousMind Oct 9 '15 at 16:05
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    $\begingroup$ @ACuriousMind complexified Lie algebras agree, it should be enough to classify representations of $SO(3,1)$. Just emphasizing for OP. $\endgroup$ – Prof. Legolasov Oct 9 '15 at 20:11

The point of Weyl's unitarity trick is precisely that (lacking a complete isomorphism, nevertheless, through complexification) the relevant Lie-Algebraic calculations go through, so the non-unitary finite-diensional irreps of the Lorentz group are classified this way! They are detailed in this Wikipedia article, except note that article uses spins s for the labels, whereas you use the dimensionality of spin multiplets, 2 s+1, instead.

In your notation, which has the arguable advantage that the arithmetic of numbers of states in the relevant vector subspace checks easily, (3,3) is, indeed, the symmetric tensor spin 2 field; but your second case is very wrong: what you wrote is a bispinor, i.e. a Dirac spinor; the antisymmetric 2-tensor, so spin 1, is (3,1)⊕ (1,3), instead. (Recall the angular momentum composition law you used, 22= 31); finally, of course, (1,1) is a singlet.

The reason this confuses you is because you are Kronecker-multiplying ⊗ (spin-adding) tensor-product × (so virtually ⊕ for the algebra with 6 angles!) representations, but the rules are self evident, and you may count your states to ensure you have not lost any.

So, (in your peculiar notation!) you are expanding the combination (spin-addition) of two Lorentz vectors, (2,2)⊗(2,2)=(2⊗2,2⊗2) =(3,3)⊕(3,1)⊕(1,3)⊕(1,1), distributing the answer, and with your error corrected. Traceless symmetric tensor, obviously spin 2 (summing of two 3s symmetrically contains the 5---but note the singlet present as well!); antisymmetric tensor (like the electromagnetic one!), spin one; and trace, scalar.

No pat formula like your proposed one above holds, but normally the answer is evident from context. Further see the Pauli-Lubanski eigenvalues involved.


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