# Weinberg QFT vol 1, chapter 2. On symmetries in QM and the universal cover

While reading chapter 2 of the book Quantum theory of fields: Volume 1 of Weinberg I got pretty much confused. All the confusion starts in page 89 with equation 2.7.43 and 2.7.44:

$$U(\Lambda) U(\bar\Lambda) = \pm U(\Lambda \bar\Lambda) \quad \quad \quad \quad \quad (2.7.43)$$

Then Weinberg explains that these are the projective representations of the Lorentz group for integer and half integer spin. All good here. Now, in the last three lines of page he says:

Eq. (2.7.43) or Eq. (2.7.44) imposes a superselection rule: we must not mix states of integer and half-integer spin.

What does he mean by that? That we can not have a superposition of states with integer and half-integer spin, or what? How does he see that this mixing is not allowed?

In the next page (p. 90) he says that we are going to use the universal covering of the Lorentz group instead of the projective representations and writes further:

This does not mean that we actually can prepare physical systems in linear combinations of states of integer and half-integer spin, but only that the observed Lorentz invariance of nature cannot be used to show that such superpositions are impossible.

Why does he emphasise that when we go to the Universal Cover we cannot have such superpositions? It makes me think that with the projective representation that was not allowed and now that we go to the universal cover it could be allowed.

And to finish the post, last confusing quote:

In general, we may just as well take the symmetry group as $$C$$ (he means the universal cover) instead of $$G$$, because there is no difference in their consequences, except that $$G$$ implies a superselection rule, while $$C$$ does not.

Isn't that contradictory, if one carries a superselection rule while the other does not, seems like they do indeed have different consequences.

Following Weinberg's discussion, as you state we have $$U(\Lambda) U(\bar{\Lambda}) = \pm U(\Lambda \bar{\Lambda})$$ where the $$\pm$$ depends on if the state is bosonic/fermionic in the case where the loop $$1$$ to $$\Lambda$$ to $$\Lambda \bar{\Lambda}$$ and back to $$1$$ is nontrivial. Now suppose that we were to try and build our total Hilbert space as a direct sum of the fermionic states and the bosonic states $$H = H_f \oplus H_b$$, so that we could write down and work with superpositions such as $$|\psi_f\rangle + |\psi_b\rangle$$. Now, how would you write down the projective representation of the Lorentz group? To be faithful to its fermionic and bosonic representations that you already have, you would need to write down $$U(\Lambda) = U_f(\Lambda) \oplus U_b(\Lambda)$$, however upon multiplication you have problems, because one part of the matrix gets a minus sign and the other doesn't: $$U(\Lambda) U(\bar{\Lambda}) = (U_f(\Lambda) \oplus U_b(\Lambda)) (U_f(\bar{\Lambda}) \oplus U_b(\bar{\Lambda})) = (-U_f(\Lambda \bar{\Lambda})) \oplus U_b(\Lambda \bar{\Lambda})$$ which is not the same as $$U(\Lambda \bar{\Lambda})$$ up to a phase; in other words, we don't actually have a projective representation anymore. To fix this, we need the superselection rule which says that the fermionic part of the Hilbert space and the bosonic part of the Hilbert space are supposed to be thought of as separate, and we shouldn't write down quantum superpositions of fermionic and bosonic states.

Another way to see the same thing, mathematically the superselection rule can be formulated as imposing that $$\langle \psi_f | A |\psi_b\rangle = 0$$ for all physical observables $$A$$. Indeed, suppose instead that we had a physical operator $$A = |\psi_f \rangle \langle \psi_b| + |\psi_b \rangle \langle \psi_f |$$, then we see that under the symmetry $$U(\Lambda)U(\bar{\Lambda})$$: $$A \mapsto (U(\Lambda)U(\bar{\Lambda}))^\dagger (|\psi_f \rangle \langle \psi_b| + |\psi_b \rangle \langle \psi_f |) U(\Lambda)U(\bar{\Lambda})$$ $$= - U(\Lambda \bar{\Lambda})^\dagger (|\psi_f \rangle \langle \psi_b| + |\psi_b \rangle \langle \psi_f|) U(\Lambda\bar{\Lambda})$$ $$\neq U(\Lambda \bar{\Lambda})^\dagger (|\psi_f \rangle \langle \psi_b| + |\psi_b \rangle \langle \psi_f|) U(\Lambda\bar{\Lambda})$$ an inconsistency!

With regards to your other questions, I might add yet another quote from Weinberg on top of the pile:

In short, the issue of superselection rules is a bit of a red herring; it may or it may not be possible to prepare physical systems in arbitrary superpositions of states, but one cannot settle the question by reference to symmetry principles, because whatever one thinks the symmetry group of nature may be, there is always another group whose consequences are identical except for the absence of superselection rules.

In other words, he's stressing that the only difference between the two is the presence of superselection rules; so we shouldn't just faithfully believe that superselection rules are actually exhibited in nature, because without actually testing it (via observations), we can't know.

I was reading through https://arxiv.org/abs/0710.1516 to understand the superselection stuff

• That's exactly what I had in mind from the very beggining! Well but then $SO(3)$ invariance is actually more restrictive than $SU(2)$ invariance, because with $SU(2)$ you could have superpositions of bosons and fermions, which is something that (I guess) has physical observable consequences. Therefore it is not a matter of "taste" to choose SU(2) or SO(3) and if superpositions were possible we should rule out $SO(3)$. You should be allowed to chose between $SU(2)$ and $SO(3)$ just if they have the same physical observable consequences---which seems is not the case here.
– erni
Jan 13 at 11:23
• Let me add about the second question..."Why does he emphasise that when we go to the Universal Cover we cannot have such superpositions?" I don't think that is what he is saying, it is perhaps the opposite. I believe when one goes to the universal cover one gets linear representations (not projective) which can be found for all spin values (integers or half-integers) and there one is allowed to form direct sums with no problems (at the level of representations) and the fact that one can do this procedure always (going to the Universal cover) is the reason he is saying Lorentz is not enough. Jan 13 at 16:52
• Thats precisely the point, Lorentz do not allow you to have superpositions of fermions and bosons. Rather just superpositions of the type fermion-fermion or boson-boson. This seems to be a consequence of representation theory. Instead SL(2,C) do allow you such superpositions. Therefore both groups should not be equivalent from a physical point of view. If we can observe superpositions of the type fermion-boson in nature this should rule out Lorentz. Right?
– erni
Jan 14 at 13:18