# Does relativistic QM obey rotational symmetry?

1. Oct 5, 2015

### fxdung

SO(3) is subgroup of Poicare group.Does Relativistic Quantum Mechanics obey rotational symmetry.If it is,why we do not still keep the non-relativistic concept of angular momentum(orbit angular momentum plus spin) for relativistic concept of angular momentum,but we instead replace the concept by Pauli-Plebanski vector for relativistic regime?Because if it is obeyed SO(3) the angular momentum still conserves despite it is non-covariant with Lorentz tranformation.

2. Oct 5, 2015

### haushofer

Because the boosts are different in both cases. The Poincare algebra has different Casimir-operators compared to the Bargmann (the centrally extended Galilei) algebra, as you can readily check. Actually, the central extension becomes one of the 'Casimirs', playing the role of mass.

Physically you can also understand it like this: relativistically, two boosts can give you a rotation, but non-relativistically two boosts commute.

3. Oct 5, 2015

### vanhees71

No, in the case of the Galileo group, mass is a central charge (the only non-trivial one in the Galileo algebra). In the case of the Poincare group mass (squared) is a Casimir operator of the the Poincare group, which does not have non-trivial central charges. All unitary ray representations of the Poincare group can thus be lifted to unitary representations. For the Galileo group you have the central charge (you can look at a quantum Galileo group, which extends the usual Galileo group to an 11 dimensional group with the mass as additional generator).

The distinction between spin and orbital angular momentum does not make too much sense in relativistic quantum theory, because there this split is frame dependent. Only the total angular momentum makes sense here. The Pauli-lubanski vector is defined by the total angular momentum and the momentum of the particle and describes the generators of the little group in the Wigner classification of the irreducible representations of the Poincare group in a manifestly covariant way:

https://en.wikipedia.org/wiki/Pauli–Lubanski_pseudovector

In this sense it describes spin in a covariant way. For massive particles you can define the spin as the representation of the rotation group (as a subgroup of the Poincare group) for the states of the particle with $\vec{p}=0$, i.e., in the restframe of the particle.

The massless case is a bit more complicated, because it involves additional gauge symmetries to make physical sense of the corresponding little group, which is an ISO(2) rather than a compact semisimple Lie group. In general that would lead to continuous spin-like degrees of freedom. The usual resolution is that you demand that the "translations" of the ISO(2) little group must be represented trivially, which leads to the usual massless particles. If they are scalar particles, they have 0 spin. For Spin $s \geq 1/2$ you have only 2 spin-like degrees of freedom, and you can use the helicity, which for spin $s$ takes on the 2 values $\pm 1$.

For details, see

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Last edited: Oct 5, 2015
4. Oct 5, 2015

### haushofer

I'm aware that the central extension is not a casimir because it's not quadratic, hence the '', I'm a bit sloppy :P Nice notes by the way, I put them in my Ibooks for future reference!

5. Oct 8, 2015

### Meir Achuz

Relativistic QM does have rotational symmetry. Some representations of it lose the rotational symmetry, as do some approximations.
A boost matrix and a rotation matrix do not commute, but neither due two rotation matrices.

6. Nov 13, 2015

### fxdung

If we define spin through SL(2,C) symmetry then spin is constant for boots(spin is integer or haft-integer)?If we define spin through the little group SO(3)(massive case) then spin depend on boots?In non-relativistic QM the sum of orbit and spin angular momentum is discrete arbitrary number because quantum number l of orbit angular momentum is only depended on energy quantum number.But Sir van Hess said that in relativistic we consider the total angular momentum,because the split orbit and spin angular momentum is depended on boots.So I have a question:Does the relativistic total angular momentum equal the spin in rest frame or not? If it is why does it is ''arbitrary'' values in non-relativistic QM but is restricted in relativistic QM(equal spin in rest frame)