# Density matrix formalism and Poincaré invariance

1. Sep 23, 2014

### arpharazon

The postulates of quantum theory can be given in the density matrix formalism. States correspond to positive trace class operators with trace 1 on a Hilbert space $\mathcal{H}$. Composition is defined through the tensor product and reduction through partial trace. Operations on the system are represented by a set of operators $\{M_m\}$ verifying $\sum_m M_m^\dagger M_m=1$. This, in my understanding, includes all kinds of operations one can perform on a system, including measurements (see Kraus operators). If we dismiss measurement results or deal with noisy situations, one takes $\sum_m M_m^\dagger M_m\leq 1$.

Now, the only thing left for a complete description is dynamics, or more generally the action of a symmetry group $G$ on the system. In my understanding any element $g\in G$ should be represented by a unitary (or antiunitary) operator $U(g)$ acting on $\mathcal{H}$. The action on a state is represented as $\rho \overset{g}{\rightarrow} U(g)^{-1}\rho U(g)$, therefore $U(g)$'s only need to verify $U(g_1\cdot g_2)=e^{i\theta(g_1,g_2)}U(g_1)U(g_2).$
In other words, we have a projective unitary representation of $G$ on $\mathcal{H}$.

Question 1: Groups that are not simply connected cannot have a simple representation and superselection is necessary. However we can go to the universal covering group to have a simple representation and no superselection. Why do we consider that the Poincaré group $$\mathbb{R}(1,3) \ltimes O(1,3)$$ is THE symmetry group of any relativistic theory? This is, in my understanding, only the $(\frac{1}{2},\frac{1}{2})$ (or spin 1 = 4-vector) representation. We should rather say that the Poincaré algebra is the correct object encoding spacetime geometry constraints on physical systems.

Question 2: A confusion I have is the link between the $(\frac{1}{2},\frac{1}{2})$ representation and Minkowski space. Both are composed of 4-vectors, but the former components are amplitudes whereas for the latter we have coordinates.

Question 3: Back to the initial problem, I want to be more specific about $G$=connected Lie group because this is how we can get dynamics. Section 6 in http://en.wikipedia.org/wiki/Quantum_logic seems to point out that we need some conditions to have weak/strong continuity for the one-parameter group that will represent G, however I cannot find clear statements of the hypothesis and/or results.

Last edited: Sep 23, 2014
2. Sep 23, 2014

### dextercioby

It's not easy to understand what you mean by 'simple representation' and 'superselection'. This is to me some non-standard terminology.

3. Sep 23, 2014

### arpharazon

By superselection I mean that you cannot take a superposition of the state of a boson with that of a fermion, which I think is standard terminology. However I must admit that I don't quite understand at which points going to the covering group allows you to get rid from superselection (maybe because the type of particle is fixed with the choice of an irrep?)

4. Sep 23, 2014

### dextercioby

Superselection is not induced by the symmetry group. It's a constraint on the theory from experimental observation. You can't get rid of it. Ignoring it makes you work with a wrong theory. So going from SO(3) to SU(2) doesn't exclude the hypothesis that a state can be a superposition of a spin 1 and spin 3/2 state. This hypothesis is excluded on the basis that it's not seen in nature, no matter what experiment you'd do.

5. Sep 23, 2014

### arpharazon

In understand your point, but I am pretty convinced that superselection should be related to properties of representations of the Poincaré group. Using irreps of this group and gauge groups we can get all particle physics, with some free parameters, but all the rest is deduced, and I guess superselection should be no exception.

6. Sep 23, 2014

### dextercioby

The only sensitive comment on superselection I remember is the first volume of Weinberg. Do you know it?

7. Sep 23, 2014

### arpharazon

No, I'd be glad if you have page reference.

8. Sep 23, 2014

### dextercioby

Page 90 at the end of a technical discussion about representations.

9. Sep 23, 2014

### arpharazon

Great thanks for the reference. It seems Weinberg confirms my statement about getting rid of superselection when taking the universal cover. However I don't quite understand his exact statement "ordinary rotation invariance forbids transitions".

10. Sep 23, 2014

### dextercioby

No, not the universal cover, but rather enlarging the symmetry algebra with the non-trivial central charges, as it happens in the case of the Galilei group. Mass times unit operator is a non-trivial central charge. It imposes a superselection rule which can be evaded if the Galilei algebra grabs a new generator and has 11, the extended Galilei algebra, with mass times 1 as the 11th generator.

11. Sep 23, 2014

### arpharazon

Sorry I read too quickly, you are right! Though I now understand symmetry does not have the answer, I find it very disturbing not to be able to deduce superselection from first principles! Thanks for your help on this point!

Do you have any further comment, especially on question 2?

A related question is the following:

Wigner treatment associates to particles the irreps of the universal covering of the Poincaré group $\mathbb{R}(1,3)\rtimes SL(2,\mathbb{C}).$ Why don't we consider finite dimensional representations of this group? I understand we ask for unitarity when representing its action on states, so the representation cannot be finite dimensional. However we do consider finite dimensional representations of the Lorentz group $O(1,3)$ and associate them to fields. Why associate the Lorentz group to fields? Why do we look in this case for finite dimensional representations? What do we associate to unitary representations of the Lorentz group?

Thanks.

12. Sep 23, 2014

### dextercioby

Question 2 first: <Question 2: A confusion I have is the link between the (1/2,1/2) representation and Minkowski space. Both are composed of 4-vectors, but the former components are amplitudes whereas for the latter we have coordinates>.

The simple answer: vector fields living in the tangent bundle to Minkowski spacetime bear the same 'index type' as normal coordinates on the Minkowski spacetime, due to the existence of the Lorents transformations. The connection of vectors to (1/2,1/2) spinors is a consequence of formulating a quantum theory of 'objects (technically not really) living in Minkowski spacetime'.

13. Sep 23, 2014

### dextercioby

Unitarity, i.e. probability density preservation upon . That's how you end up with the universal cover in the first place, because you're forced to work with projective representations. See you comment bolded below.

The unitary representations of the Lorentz group are found to have no meaning to physics. They only appear in mathematically oriented books: Gelfand+Shapiro, Rühl, Naimark and Carmeli.

How are fields really defined?: starting with vectors of space-time which transform under the fundamental representation of the Lorentz group and going more generally to tensors and for quantum use to spinors (spinor tensors), by moving from the Lorentz group to the SL(2,C) group.

14. Sep 23, 2014

### arpharazon

OK, if I understand what you are saying I'm asking the wrong question. Basically special relativity fields are (1/2,1/2), once we know this we can look for other irreps of the Lorentz group. What is the intuition for asking for Lorentz invariance and not Poincaré invariance for "fields"? Can't we find more general types of fields by looking at finite dimensional irreps of Poincaré, or these don't provide more information than the Lorentz subgroup does? Thanks a lot for your helpful comments.

15. Sep 23, 2014

### dextercioby

Actually, the symmetry group of QFT is Poincaré, Fields are assumed to transform under the (universal cover of) the restricted Poincaré group, not Lorentz. When doing representation theory of the SL(2,C) semidirect product with (R^4,+), you encounter the finite dimensional matrices of SL(2,C) spinor calculus (as shown in Weinberg, Vol.1). Assuming that the quantum fields transform under restricted Poincare transformation with a finite dimensional spinor matrix (as opppsed to an infinite one) is a necessity: the spin-statistics and TCP theorems wouldn't hold, as mentioned by Bogolubov et al., 1975, page 337 and proven starting with page 564.

16. Sep 23, 2014

### arpharazon

In my understanding $\mathbb{R}(1,3)$ induces infinite dimensional unitary representations, and once a choice of momentum (casimir) operator is made, we further classify unitary irreps of the little group, i.e. SU(2), which are finite-dimensional. However this is only the way we classify particles. I read at some point that this is connected to the $(n,m)$ classification of field types, and that field equations are restrictions that ensure some kind of compatibility between how we classify types of particles and types of fields. However I don't remember where I saw this nor did I find an accessible reference giving more details.

17. Sep 23, 2014

### dextercioby

SU(2) is not the only possible choice of little group. The connection with field equations is a different story, because you already suppose that the 'objects' in the field equations bear SL(2,C) indices at least (i.e. in the absence of parity invariance, which throws you to a different group, hence a different index structure, like in the case of the Dirac equation). In axiomatic quantum field theory, the importance of field equations is marginal.

Last edited: Sep 23, 2014
18. Sep 23, 2014

### arpharazon

I was referring to the massive case, for photons the little group is the universal cover of $E(2)$.