What defines the spin of a certain field? (formally)

Question

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?

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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?

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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?

Answer 1

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, 2 ⊗ 2= 3 ⊕ 1); 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.

Answer 2

The question is rather old but maybe this answer is yet useful to someone. It all boils down to the rotation group. The thing is: a field is defined by how it transforms according to the universal cover of the Lorentz group ${\rm SL}(2,\mathbb{C})$. Now in the same way that the Lorentz group contains rotations its universal cover ${\rm SL}(2,\mathbb{C})$ contains the universal cover of the rotation group, ${\rm SU}(2)$.

In that case if $\rho : {\rm SL}(2,\mathbb{C})\to{\rm GL}(V)$ is a representation of ${\rm SL}(2,\mathbb{C})$ on a vector space $V$ restricting $\rho$ to the ${\rm SU}(2)$ subgroup we have a representation $\rho|_{\rm SU(2)}:{\rm SU}(2)\to {\rm GL}(V)$ of the universal cover of the rotation group.

Even if $\rho$ is initially an irrep of ${\rm SL}(2,\mathbb{C})$ it may not be an irreducible representation of the rotation group. Still you can always decompose it into its irreducible parts. Therefore you may decompose $$V=\bigoplus_{i}V_i$$

where $V_i$ is invariant under ${\rho}|_{\rm SU(2)}$ and is an irreducible representation of ${\rm SU}(2)$. Now recall the irreps of ${\rm SU}(2)$ are labelled by spin $j\in \frac{1}{2}\mathbb{Z}_+$: these are the spins the field transforming under $\rho$ contains.

Example: Pick the vector representation of ${\rm SL}(2,\mathbb{C})$, which is actually one representation of ${\rm SO}(1,3)$. This is in fact the defining action of the Lorentz group on $\mathbb{R}^4$ by matrix multiplication. Now recall that the rotations are the transformations of the form $$\Lambda=\begin{pmatrix}1 & 0\\ 0 & R\end{pmatrix},$$

where $R\in {\rm SO}(3)$. Now it is very easy to observe that the subspace spanned by $(1,0,0,0)$ alone is left invariant and that the subspace spanned by $\{(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$ is left invariant. These are the subspaces transforming irreducibly under ${\rm SO}(3)$: the first according to spin $0$ and the second according to spin $1$. Therefore the vector representation of the Lorentz group decomposes under rotations are $0\oplus 1$.

More generally consider the irreps of ${\rm SL}(2,\mathbb{C})$. These are labelled by pairs $(A,B)$ where $A$ and $B$ label irreps of ${\rm SU}(2)$. What you can show (please see Weinberg's The Quantum Theory of Fields, Chapter 5, for the full construction) is that the representation of ${\rm SU}(2)$ associated to this $(A,B)$ irrep is the direct sum of all ${\rm SU}(2)$ irreps where the spin ranges in the interval $A+B,A+B-1,\dots, |A-B|$.

In other words: it is one addition of angular momentum problem!

Observe the vector irrep is $(\frac{1}{2},\frac{1}{2})$ therefore its associated ${\rm SU}(2)$ representation will have spins $\frac{1}{2}+\frac{1}{2}=1$ and $\frac{1}{2}-\frac{1}{2}=0$. Again the vector irrep is $0\oplus 1$. The Weyl irreps are $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$ and you easily see that they can only carry spin $\frac{1}{2}$. And this goes on for all kinds of fields.

So the quick answer is: given a field, it transforms according to some specific representation of ${\rm SL}(2,\mathbb{C})$. The spins it carries are determined by how it transforms under the induced ${\rm SU}(2)$ rotation subgroup representation. In many cases it need not have a well-defined spin, because the ${\rm SU}(2)$ representation can be reducible. In that case the field is able to carries whatever spins appear in the decomposition of the ${\rm SU}(2)$ representation into its irreps. Finally for the $(A,B)$ irreps of ${\rm SL}(2,\mathbb{C})$ finding the spins the field carry amounts to solving one angular momentum addition problem with spins $A$ and $B$.

Answer 3

This is a bit late, but I hope it is helpful to people in the future:

We know that the vector representation of the Lorentz group is $\left(\frac{1}{2},\frac{1}{2}\right)$, where I am labelling irreps by their spin, instead of their dimension.$^1$ We then know a general second rank Lorentz tensor must live by definition in the representation: $\left(\frac{1}{2},\frac{1}{2}\right) \otimes \left(\frac{1}{2},\frac{1}{2}\right)$, so lets expand this and see what we find:

$$\left(\frac{1}{2},\frac{1}{2}\right) \otimes \left(\frac{1}{2},\frac{1}{2}\right) = \left(1\oplus0\right) \otimes \left(0\oplus 1\right) = (1,1)\oplus(1,0)\oplus(0,1)\oplus(0,0)$$ where the first decomposition$^2$ is simply what you and Cosmas call $\mathbf{2}\otimes\mathbf{\bar{2}}=\mathbf{1}\oplus\mathbf{3}$. We can see that a rank-2 Lorentz tensors 16 degrees of freedom breakdown into a 9 dimensional spin-2 piece, a 6 dimensional spin-1 piece, and a 1 dimension spin-0 piece. You can either use more sophisticated group theory, or simply construct these objects to see what each of them look like.

_{1: $dim[(s_1,s_2)] = (2s_1+1)(2s_2+1)$}

_{2: we are free to commute objects around a direct sum, but not around tensor products}