(A,A) representation of Lorentz group-why is it tensor?

  • #1
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Why representation of Lorentz group of shape (A,A) corespond to totally symmetric traceless tensor of rank 2A?

For example (5,5)=9+7+5+3+1 (where + is dirrect sum), but 1+5+3+9+7<>(5,5) implies that (5,5) isn't symmetric ?

See Weinberg QFT Book Vol.1 page 231.
 

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  • #2
vanhees71
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You look at the representation of the rotation group as a subgroup (or more precisely on its covring group SU(2)). This is a unitary transformation of this group, and the angular-momentum operators are the self-adjoint generators. They are given by the "pseudo-rotators" by (using Weinberg's notation) ##\vec{J}=\vec{A}+\vec{B}##, i.e., the irrep. ##(A,B)## of the Lorentz group is a reducible unitary representation of SU(2) which can be decomposed into irreps in the usual way of "angular-momentum addition". The decomposition consists of ##j \in \{A+B,A+B-1,\ldots,|A-B| \}##.

Now the special case ##(A,A)## thus consists of only integer-spin representations ##j \in \{2A,2A-1,\ldots,0 \}##. Now the irreducible integer-spin ##j## representations consist of the symmetric traceless tensors of rank ##j##.

This can be seen by rewriting the spherical harmonics in Cartesian components.
 
  • #3
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Yes that is clearer 😀. But why this representation corespond to tensor field T(traceless) with covariant indices i1 i2...in where indices is in interval from 0 to 3 when representation (n,n) corespond to block diagonal matrix which dimension of each matrix is (2k+1)*(2k+1) for every irrep? 2k+1<>4 what is real interval for indices ik in tensor field.Sory for my English
 
  • #4
vanhees71
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I think it's most easily explained with an example and to use the representation with local quantum fields, which you can heuristically get from "canonical quantization" of the corresponding classical fields. Let's take a (free) massive vector field (describing, e.g., my favorite ##\rho## meson in an effective hadronic model). It's the most simple example, where all the issues become clear.

In the local-field realization you can start with a classical field theory and represent a spin-1 field as a four-vector field ##V^{\mu}(x)##. A four-vector builds of course a representation of the full proper orthochronous Poincare group and is easily extended to also describe spatial reflections (P/parity symmetry), time reversal (T symmetry), and in quantized form also charge-conjugation (C symmetry).

Now let's see, what this means in terms of representations of the rotation group as a subgroup. We expect to get an irreducible representation ##J=1##, i.e., one expects ##2J+1=3## spin-components, but of course ##V^{\mu}## has four field-degrees of freedom! That's understandable, because you can easily build a scalar field, even a four-scalar field, which is ##\partial_{\mu} V^{\mu}##. This piece is indeed a scalar field under proper orthochronous Poincare transformations and thus also under the rotations as a subgroup. So in order to describe only spin-1 particles you need to impose a constraint
$$\partial_{\mu} V^{\mu}=0$$
to get rid of the unphysical scalar degrees of freedom.

Another constraint is that the particles should have mass ##m##, i.e., it should fulfill
$$\partial_{\mu} \partial^{\mu} V^{\nu}=-m^2 V^{\nu}.$$
There I've used the fact that the generator for temporal and spacial translations, ##x^{\mu} \rightarrow \bar{x}^{\mu}=x^{\mu}+\delta a^{\mu}## are given by ##\hat{p}_{\mu}=-\mathrm{i} \partial_{\mu}## since
$$\bar{A}^{\nu}(\bar{x})=A^{\nu}(x)=A^{\nu}(\bar{x}-\delta a) = A^{\nu}(x) -\delta a^{\mu} \partial_{\mu} A^{\nu}(x)=A^{\nu} + \mathrm{i} \hat{p}_{\mu} a^{\mu} A^{\nu}(x).$$
There are different ways to achieve this. The most simple one is to consider the Proca equation. Here the idea is to make the constraint automatically being fulfilled by the equations of motion. Using the 4D curl, ##F^{\mu \nu} = \partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu}##, we can write the eom. with some constant ##\lambda##
$$\partial_{\mu} F^{\mu \nu}=\lambda A^{\nu}.$$
The left-hand side expands to
$$\Box A^{\nu} - \partial^{\nu} \partial_{\mu} A^{\mu}=\lambda A^{\nu}.$$
Assuming that the constraint projecting out the scalar degree of freedom, ##\partial_{\mu} A^{\mu}=0## is valid, leads to ##\lambda=-m^2##:
$$\Box A^{\nu} - \partial_{\nu} \partial^{\mu} A^{\mu}=\partial_{\mu} F^{\mu \nu}=-m^2 A^{\nu}.$$
Now contracting this equation with ##\partial_{\nu}## yields
$$-m^2 \partial_{\nu} A^{\nu}=0,$$
and as long as ##m^2>0## we see that the constraint ##\partial_{\nu} A^{\nu}=0## is automatically fulfilled.

So the only thing to go on with quantizing the free spin-1 field is to write down a Lagrangian fulfilling the equations of motion, and this is achieved by
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{m^2}{2} A_{\mu} A^{\mu},$$
where I have also assumed ##A^{\mu} \in \mathbb{R}## and then do the usual steps for canonical quantization.

The only trouble with this is that (a) it's not easy to get the massless limit ##m^2 \rightarrow 0##, if you want to use ##m^2## as a regulator for IR problems with photons and (b) it's not easy to build renormalizable interacting theories with such massive vector bosons, because the propagator in energy-momentum space goes not like ##1/p^2## for large ##p## but has a term ##\propto p^{\mu} p^{\nu}/(m^2 p^2)##.

The way out of both problems is to describe the spin-1 field as massive Abelian vector field by introducing an additional auxilliary scalar field, the socalled Stueckelberg formalism, but that's another story and boils finally down to the same conclusion concerning the spin: You have to project out the spin-0 part of the four-vector field ##A^{\mu}##.

As you know, you can build all irreps of the rotation group and its covering group SU(2) by just using tensor products of the fundamental representation of SU(2), i.e., the spin-1/2 representation. The most simple way is to use just the space ##1/2 \otimes 1/2##, i.e., two spins ##|\sigma_1,\sigma_2 \rangle## with ##\sigma_{j} \in \{-1/2,1/2 \}##. This product vectors however are not transforming according to irreps. of SU(2). Here the reduction into the irreducible pieces is very simple. Just consider the anti-symmetriced and the symmetrices product states,
$$|S=0,M=0 \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2,1/2 \rangle)$$
and
$$|S=1,M=1 \rangle = |1/2,1/2 \rangle,\\
|S=1,M=0 \rangle = \frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle+|-1/2,1/2 \rangle),\\
|S=1,M=-1 \rangle = |-1/2,-1/2 \rangle.$$
The point is to have local fields you cannot start right away in the irred. subspace for ##S=1## but you need to consider the representation ##(1/2,1/2)## of the proper orthochronous Lorentz group, which concerning the rotations contain not only ##S=1## but also ##S=0##, and you have to project out the unwanted ##S=0## piece. This is achieved by imposing corresponding constraints by the equations of motion.
 
  • #5
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I read this https://math.stackexchange.com/ques...sor-product-of-two-representations-of-a-group !!! In answer if we set rho1 as representation of spin j1, rho2 as spin j2 rep, v as matrix of matrices of generators of Lorentz group represented in j1, and all equivalentli do with w and rho2. This matrices of matrices of generators transformed as 2-covariant tensor under Lorentz transformation. If we set rho1(g1) similarty transformation (U(L)*g1*U^-1(L))under Lorentz transformation L1=g1, corespondly for other rhos and gs and L2. But L1=L2 and this Lorentz transform represented equivalently, but with different indicies.We have that dirrect sum of generators of lorenz group transform as 4 covariant tensor. Is right this ? Question is how direct sum of matrix of generators in different representations transform ?
 
  • #6
vanhees71
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The proper orthochronous Lorentz group, i.e., the subgroup of the full Lorentz group that is smoothly connected to the group identity has as finite-dimensional representations what's given in Weinberg's book as the ##(A,B)## representations for its covering group (the ##\mathrm{SL}(2,\mathbb{C})##), where ##A,B \in \{0,1/2,1,3/2,\ldots \}##. The Lorentz group has 6 independent generators. The most intuitive choice of a basis of the corresponding Lie algebra are the rotations (around three perpendicular space-like axes) and three boosts (along the same three perpendicular space-like axes). From the commutation relations you can easily see that the Lie algebra ##\mathrm{sl}(2,\mathbb{C})## is isomorphic to ##\mathrm{su}(2) \oplus\mathrm{su}(2)##. The two ##\mathrm{su}(2)## algebras give rise to the representations ##A## and ##B##.

The rotations as a compact subgroup of the proper orthochronous Lorentz group can be unitarily represented with the generators given by ##\vec{J}=\vec{A}+\vec{B}##. For the rotation subgroup the representation induced by ##(A,B)## is thus containing all representations which turn up in the corresponding problem addition of two independent angular momenta, and it can be split into a direct sum of irreducible parts, containing the irreps with ##J \in \{A+B,A+B-1,\ldots,|A-B| \}##.

The issue is a bit complicated. Perhaps it helps to read the excellent book

R. U. Sexl, H. K. Urbantke, Relativity, Groups, Particles, Springer, Wien (2001).

in parallel to Weinberg's vol. I. It helped me a lot to understand the unitary representations of the proper orthochonous Poincare group, which indeed is the fundamental basis of relativistic QFT.

Another very good source is the original paper by Wigner:

E. P. Wigner, On Unitary Representations of the Inhomgeneous Lorentz Group, Annals of Mathematics 40
(1939) 149.
http://dx.doi.org/10.1016/0920-5632(89)90402-7
 
  • #7
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Ok, it is complicated. But where exist proof that arbitrary bosonic field ((A,A) representation) corespond to totally symmetric tensor field which is traceless ?
 
  • #8
vanhees71
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What's missing in Weinberg's book?
 
  • #9
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Nothing, I read book studiously. But Weinberg says that (A,A) rep corespond to totaly symmetric tensor, ok this rep is symmetric as you says. But Weinberg say that (A,A) rep can constructed with D1, D2,...,D2A, where D is partial derivative, but he doesn't explain this !!!
 

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