# Polarization vectors for spin-2 particles

1. Oct 24, 2014

### Breo

Who knows where this formula comes?

$$e_i \otimes e_j + e_j \otimes e_i - \frac{2}{3}( \sum_{\substack{k}} e_k \otimes e_k )\delta_{ij}$$

2. Oct 24, 2014

### Staff: Mentor

Where does it appear?
The structure reminds me of the off-diagonal elements of a quadrupol tensor.

3. Oct 24, 2014

### Breo

4. Oct 25, 2014

### Breo

this equation reminds to to group rotations or symmetries. Something of group theory.

5. Oct 25, 2014

### samalkhaiat

Ok, If you couple two spin 1 vectors, you get the following spins
$$\vec{ 1 } + \vec{ 1 } = ( \vec{ 2 } , \vec{ 1 } , \vec{ 0 } ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
In components, this means
$$[ 3 ] \otimes [ 3 ] = [ 5 ] \oplus [ 3 ] \oplus [ 1 ]. \ \ \ \ (1a)$$
What does this equation means? Well, the LHS is the (reducible) tensor product of two 3-vectors, $e_{ i } \otimes e_{ j }$. The RHS, which consists of irreducible tensors, is obtained by the decomposing the tensor product into (direct)sum of independent tensors. In general, you decompose a tensor into irreducible tensors by forming symmetric and antisymmetric combinations and subtracting all possible traces. So, your example is the simplest one:
$$e_{ i } \otimes e_{ j } = \frac{ 1 }{ 2 } G_{ i j } + \frac{ 1 }{ 2 } A_{ i j } + \frac{ 1 }{ 3 } \delta_{ i j } e_{ k } \otimes e_{ k } , \ \ \ (1b)$$
where the tensor
$$G_{ i j } = e_{ i } \otimes e_{ j } + e_{ j } \otimes e_{ i } - \frac{ 2 }{ 3 } \delta_{ i j } e_{ k } \otimes e_{ k } ,$$
is symmetric, $G_{ i j } = G_{ j i }$ and traceless $\delta_{ i j } G_{ i j } = 0$. Therefore, it has $(3/2)(3 + 1) - 1= [5]$ components and can represent a massive spin $\vec{ 2 }$ particle,
$$A_{ i j } = e_{ i } \otimes e_{ j } - e_{ j } \otimes e_{ i } ,$$
is anti-symmetric tensor. In 3-dimension, it has $(3/2)(3 - 1 ) = [3]$ components. Therefore, it is equivalent to spin $\vec{1}$ represented by the 3-vector $v_{ i } \equiv \epsilon_{ i j k} A_{ j k }$, and finally
$$e_{ k } \otimes e_{ k } = \mbox{ Tr } ( e_{ i } \otimes e_{ j } ) = \delta_{ m n} e_{ m } \otimes e_{ n } ,$$
is the invariant trace, i.e., $[ 1 ]$ component scalar representing spin $\vec{ 0 }$.
So, equations eq(1), eq(1a) and eq(1b) all have the same meaning.

6. Oct 25, 2014

### arivero

Guys, you are being very helpful this week :) great!

Has symmetry or antisymmetry some deeper significance, beyond building the irreps? From the notation in the irrep sum it seems that this $A_{ij}$ with two indexes is still the same thing, [3], that the initial vector $e_i$... is it? And we could also get a spin 1 vector from a product 1/2 x 1/2, and then it would be in the symmetric representation, but still be just a spin 1 vector as the others.

7. Oct 25, 2014

### Breo

8. Oct 25, 2014

### samalkhaiat

No, that is deep enough. The point is this, symmetric and anti-symmetric tensors do not mix under the transformations in the question, i.e., they belong to different multiplets (invariant subspaces).
Yes, they are (as I said) equivalent, because of the invariant tensor $\epsilon_{ i j k }$
The above construction has no spin (1/2) object! For that, you need to consider the group $SU(2)$.

9. Oct 25, 2014

### samalkhaiat

Try textbooks on group theory and particle physics.

10. Oct 26, 2014

### dextercioby

Try first H. Georgi's <Lie Algebras and Particle Physics>.

11. Oct 26, 2014

### arivero

uh? "Learn mathematics from books written by mathematicians" :-D

Well, for this topic, it is true that physicist books are usual bussiness. I'd add Slansky's Report, which is scanned in the KEK database and surely also at SLAC, and some modern ones. Ramond? Cvitanovic?

12. Oct 29, 2014

### lpetrich

For 2 and 3 dimensions, there is a connection between spin n and symmetric traceless n-tensors.

Let's consider rotation / angular momentum operators for vectors and tensors. $(L_{ab})_{ij} = \delta_{ai} \delta_{bj} - \delta_{bi} \delta_{aj}$ with
L(total) = L(index 1) + L(index 2) + ... + L(index n)

The square $L^2 = - \frac12 L_{ab} L_{ab}$ where the -1/2 is for identifying it with quantum-mechanical spin. Its general expression for a tensor is
$$(L^2)_{(i)(j)} = n (d - 1) \delta_{(i)(j)} + \sum_{q != p} ( \delta_{p:i,q:j} \delta_{q:i,p:j} - \delta_{p:i,p:j} \delta_{q:i,q:j} ) \delta_{other (i)(j)}$$
for indices p and q and values i and j of them, and also d dimensions of of vector index.

It is evident that the largest L2 is for a symmetric traceless tensor, and in that case, we get L2 = n*(n+d-2). That's the right value for the square of the spin for both d = 2 and d = 3.

13. Oct 29, 2014

### samalkhaiat

What does that garbage mean, exactly?

Last edited: Oct 29, 2014
14. Oct 29, 2014

### lpetrich

The Lab's are generators of the SO(d) algebra for d dimensions, appropriately extended to n-tensors. The L2 is the square Casimir invariant of that algebra, and it gives the total spin. This n-tensor representation of SO(d) is reducible, and one of its parts is for a symmetric traceless n-tensor.