Tensor representation of the Lorentz Group

  • #1
300
22

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
300
22
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}##?
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,384
5,976
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)
 
  • #4
300
22
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.
 
  • #5
300
22
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.
 

Related Threads on Tensor representation of the Lorentz Group

Replies
4
Views
487
Replies
6
Views
593
Replies
4
Views
2K
Replies
2
Views
2K
Replies
3
Views
540
  • Last Post
Replies
5
Views
641
  • Last Post
Replies
10
Views
2K
Replies
12
Views
16K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
2K
Top