# Spin-2 and tensors

I read that we need scalars, spinors, vectors and rank two tensors to describe spin-0, spin-1/2, spin-1 and spin-2 particles, respectively.

But then I reacall from quantum mechanics courses that the intrinsic spin of a particle is described by different finite representations of so(3), (2j+1)(2j+1) matrices acting on (2j+1) vectors, where j is half-integer or integer.

Why then tensors to describe spin-2 particles? Why not a 5-vector that gets transformed by 5x5 matrices that belong to SO(3)?

(EDIT: Or do all half-integer-j reps of so(3) behave like spinors (particles have spin 1/2) and all integer-j reps of so(3) behave like vectors (particles have spin 1)?

Or has it something to do with when we go from SO(3) to SO(3,1)?

hope my question makes sense

thank you

Last edited:
• Antonio Rivera

## Answers and Replies

DrDu
Science Advisor
A symmetric traceless tensor (in 3 dimensions) is indeed a 5-vector and gets transformed as you said by 5x5 matrices which span a representation of SO(3). E.g. in atomic physics they correspond to the d-orbitals. Google for "irreducible tensors".

arivero
Gold Member
Or has it something to do with when we go from SO(3) to SO(3,1)?

Perhaps a simpler question should be in this line: why a vector field has always spin 1? Last time it was asked, https://www.physicsforums.com/showthread.php?t=421052, nobody did a didactically good answer.

In fact a good answer would involve, I think, SO(n) and SO(n,1) for arbitrary space dimension n. An excelent answer would involve tensor products, and Spin(n) too.

Last edited:
arivero
Gold Member
I am a bit ashamed. I have checked some text books and there is not a general proof of the relationship between tensors and spin. Of course it is straignt that a tensor is a shortname for a sum of tensor products of vectors, so it will contain the max spin of the product. But it does not amount to a proof: you must check irreducibility, count the number of reducible components and the number of degrees of freedom of each component, and see how it fits with the number of components of a SO(n) representation. And to do a separate proof for the initial case, the vector.

It could be an exersice or an appendix, but it should be done somewhere in a textbook.

Huang delays the proof to the chapter about electromagnetism, and then it only gives an argument for the massless photon. In Peskin-Schroeder I haven't found any proof neither. And this is for the 4D case; for the general dimensional case I think that at least Weinberg did an appendix about it, but I need to check a copy.

Thanks DrDu and arivero for the replies.

I found that tensors decompose as follows, tensor = scalar (the trace) + antisymmetric part + symmetric traceless part.

For generic Lorentz rank-two tensor (product of two four-vecs), the number of components is 16 = 1 + 6 + 9, and for a cartesian tensor ( product of two spatial vecs), we have 9 = 1 + 3 + 5.

The symmetric traceless part of the spatial tensor describes a spin-2 particle, the antisymmetric part is a spin-1 particle, a vector.

Does the 9 ( the symmetric traceless part) of the Lorentz tensor correspond to the 9 components of a spatial tensor? When someone hands me a Lorentz tensor, are all the spins in the symmetric traceless part of the Lorentz tensor?

Maggiore (QFT textbook) says every generic tensor with n indices has up to n spins. So a rank-two has spin-1 and spin-2.

dextercioby
Science Advisor
Homework Helper
Maggiore (QFT textbook) says every generic tensor with n indices has up to n spins. So a rank-two has spin-1 and spin-2.

That's true. A Lorentz second order, traceless and symmetric tensor describes purely spin 2 as would a traceless linearized spacetime metric.

arivero
Gold Member
That's true. A Lorentz second order, traceless and symmetric tensor describes purely spin 2 as would a traceless linearized spacetime metric.

Hmm can you explain with more detail the "linearized" condition?

dextercioby
Science Advisor
Homework Helper
Well, as far as I remember from 7 years ago, the traceless Pauli-Fierz tensor field purely describes spin 2. Or ? The bibliography (I think the first 10 items or so ) from a known 2000 article on arxiv by Boulanger, Henneaux and other 2 guys contains, as far as I know, all the relevant info on the linearized gravity field.

I don't remember the details from college, but at least I hope to offer you a place to start searching for the relevant material, that, of course, if you have unrestricted access to every possible journal.

But my initial questions still stands, why using a rank-two vector and not a (much simpler) spatial 5 component vector to describe a spin-2 particle?

Ok, the obvious answer is that there are only three spatial components! So one object with three space coordinates and spin-2 must be described by a tensor. Am I right?

I suppose the j>1 vector representations are all meant for composed systems of many particles where spins are added.

arivero
Gold Member
There is a mix of different lores here. For riemaniann tensors, you have the irreducible pieces mentioned above (symmetric, antisymetric, scalars). For pseudoriemann 3,1 metrics, you can consider the 3d spin, and then a symmetric tensor has one spin 2, one spin 1 and two spin 0, while an antisymmetric has two spin 1 pieces.

But when you add dynamics, some of these pieces get simplified. For instance, a 1-tensor (a vector) must agree with Klein Gordon, and this is a set of extra conditions. For any tensor T, you must ask (p^2 - m) T = 0 or p_i T =0 or similar things, and in this way it reduces to the usual lore.

DrDu
Science Advisor
I just remembered some more old lore, as arivero would put it. The irreducible representations of the general linear group of order n GL(n) can be obtained through the operation of the Young operators from the representations of the symmetric S_m group on the indices of the tensor with m indices. For tensors with maximally two indices, this reduces to symmetrization and anti-symmetrization. For more complex tensors, the irreducible representations can be obtained from the Young tableaux and rules on how to fill them with the indices. These are called Weyl tableaux. This can also be applied for sub-groups of GL(n). Down to U(n) the irreducible reps remain irreducible, but are no longer all distinct. For even lower groups, they start to split up but the classification scheme remains useful. Consider e.g. Sternberg or Hamermesh for the details.

A. Neumaier
Science Advisor
Ok, the obvious answer is that there are only three spatial components! So one object with three space coordinates and spin-2 must be described by a tensor. Am I right?

I suppose the j>1 vector representations are all meant for composed systems of many particles where spins are added.

so(n) acts naturally on vectors of size n, and tensors of size n x n, n x n x n, etc.,
and their decomposition in term of partial symmetry and/or antisymmetry.
This means that the 5-dimensional representation of so(3) on traceless symmetric tensors of order 2 is the shortest way to describe the spin j=2 representation. In contrast, the description in terms of a (2j+1)-dimensional vector requires some nontrivial work.

Of course, both descriptions are equivalent since there is only one irreducible unitary representation of a given spin. It is a nice exercise to find the linear mapping between the two 5-dimensional spaces that explains this equivalence.

samalkhaiat
Science Advisor
I read that we need scalars, spinors, vectors and rank two tensors to describe spin-0, spin-1/2, spin-1 and spin-2 particles, respectively.

But then I reacall from quantum mechanics courses that the intrinsic spin of a particle is described by different finite representations of so(3), (2j+1)(2j+1) matrices acting on (2j+1) vectors, where j is half-integer or integer.

Why then tensors to describe spin-2 particles? Why not a 5-vector that gets transformed by 5x5 matrices that belong to SO(3)?

(EDIT: Or do all half-integer-j reps of so(3) behave like spinors (particles have spin 1/2) and all integer-j reps of so(3) behave like vectors (particles have spin 1)?

Or has it something to do with when we go from SO(3) to SO(3,1)?

hope my question makes sense

thank you

See my posts (only MY posts) in the following threads:

www.physicsforums.com/showthread.php?t=192572
www.physicsforums.com/showthread.php?t=91769
www.physicsforums.com/showthread.php?t=252102
www.physicsforums.com/showthread.php?t=237936

regards

sam