Lorentz Group: Tensor Representation Explained

[chingel]

I've been trying to understand representations of the Lorentz group. So as far as I understand, when an object is in an (m,n) representation, then it has two indices (let's say the object is $\phi^{ij}$), where one index $i$ transforms as $\exp(i(\theta_k-i\beta_k)A_k)$ and the other index as $\exp(i(\theta_k+i\beta_k)B_k)$, where A and B are commuting su(2) generators of dimension (2m+1) and (2n+1) respectively and $\theta$ are the rotation angles and $\beta$ the rapidities.

It is often said (for example I am reading the Schwartz QFT book, where it is mentioned), that the (n/2,n/2) representation corresponds to a tensor with n indices.

How is this the case? How can I see it?

Is the meaning of the tensor representation that each index transforms using the usual 4x4 Lorentz transformation matrix? How is it the same as a two index object where each index transforms under a (n+1) dimensional su(2) representation?

 

[chingel]

At least for the (1/2, 1/2) representation I figured it out by surfing on the internet. I can choose for A representation the $\sigma/2$ matrices and for B take $-\sigma^*/2$, then the two index object $\phi^{ij}$ transforms as $$\phi \rightarrow \exp(i(\theta_k-i\beta_k)\sigma_k/2)\, \phi\, \exp(i(\theta_k-i\beta_k)\sigma_k/2)^\dagger,$$ where we have just matrix multiplication. Note that the element $\exp(i(\theta_k-i\beta_k)\sigma_k/2)$ belongs to SL(2) with complex entries. So the determinant of $\phi$ is constant under the transformation, and the transformation keeps the $\phi$ Hermitian if it was Hermitian to begin with (it has to be Hermitian to be connected to the identity). Now we can identify the matrix $\phi$ as (you might want to double check the signs) $$\phi = \begin{bmatrix} p^0+p^3 & p^1-ip^2 \\ p^1+ip^2 & p^0-p^3 \end{bmatrix},$$ where the determinant gives just $p_\mu p^\mu=m^2$, which is good because it is constant, and also then we can check that under the $\phi$ transformation the $p^\mu$ transform as we would expect under Lorentz transformations. So in the end we can say that the (1/2,1/2) representation indeed can be written using four parameters that transform as a four vector.

But now how do I do it for the (1,1) case to show that it has 4x4 components that transform as a two index Lorentz tensor $T_{\mu\nu}$?

 

[vanhees71]

The point is that $(j_1,j_2)$ describe irreducible representations. It's $(2j_1+1) (2j_2+1)$ dimensional, i.e., for $(1,1)$ you have 9, not 16 components. This is easy to understand, because you can decompose the general 2nd-rank Tensor into a scalar part (1 component), given by its trace, the antisymmetric piece (6 components) and the traceless symmetric piece (9 components).

For a very good and didactical introduction to the group- and representation-theoretical analysis of the Lorentz and Poincare groups, see

Sexl, Urbandtke, Relativity, Groups, Particles, Springer (2001)

 

[chingel]

Thanks, I managed to take a quick look at the book and chapter 8 seems to cover exactly what I was asking about (and that I didn't find in my QFT books).

On a quick skim apparently if we had for a four vector $p_\mu\sigma^\mu$ and transforming this under (1/2,1/2) gave the proper transformation for $p_\mu$, then more generally there is this quantity $T^{ij\dots}\sigma_i\sigma_j\dots$, and using this I think they can show the equivalence to tensors. However I'll have to read the chapter in more detail to figure out the details how exactly this works out.

 

[chingel]

So for anybody who is interested, the argument is quite simple in my opinion.

First we know from angular momentum addition that by adding m copies of spin 1/2 (I mean by doing a tensor product of them) and symmetrizing, we get a m/2 representation of SU(2).

So a (m/2,m/2) representation of the Lorentz group can be written as $\phi_{i_1\dots i_m j_1\dots j_m}$, where it is symmetric over the i indices, and also symmetric over the j indices, and each index transforms as (1/2) representation of SU(2) either LH for i or RH for j.

Now before we showed that (1/2,1/2) is a four vector, where we used $\phi_{ij} = p^\mu \sigma_{\mu,ij}$ (or the other way around it is $\frac12 \sigma_{\mu,ji}\phi_{ij} = p^\mu$) where $\sigma_\mu$ is $(1,\vec{\sigma})$. So now we can just consider each pair of indices $(i_1, j_1)$ etc and each of those is a four vector, so we just have a tensor product of m pairs each of which transforms as a four vector, so the 4 tensor is like this $$T^{\mu_1\dots\mu_m} = \phi_{i_1\dots i_m j_1\dots j_m} \sigma^{\mu_1}_{i_1j_1}\dots\sigma^{\mu_m}_{i_mj_m} .$$ From this we see that the 4 tensor T has to be symmetric, because the right hand side is due to $\phi$ being symmetric.