quantum mechanics - Non coherence of Fermions and Bosons through $U(1)$

I "know" the textbook answer why we cannot write,

$$ |\psi\rangle = a|j=\tfrac{1}{2}\rangle + b|j=1\rangle $$ as "each term in the quantum superposition transforms differently under $U(1)$", $$ U(2\pi)\ |\psi\rangle = -a|j=\tfrac{1}{2}\rangle + b|j=1\rangle $$ and "hence the phase difference is unobservable"...

However, I do not understand what is meant by the italicised parts.

Why is it that we cannot observe the phase difference?

The question has arisen from studying notes on a course in Symmetries of Quantum Mechanics focusing on the group theoretical basis of QM.

Answer

This is the prototypical example of a superselection rule.

The operator $U(2\pi)$ commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation of the algebra of observables is reducible by Schur's lemma, and the invariant subspaces are precisely the bosonic and fermionic subspaces.

Because the bosonic and fermionic subspaces (let's call them $H_b,H_f$) are invariant, all cross matrix elements for all observables vanish: $\langle b\vert A \vert f\rangle = 0$ for all $\lvert b\rangle\in H_b,\lvert f \rangle \in H_f$ and all observables $A$.

From this we can directly see that relative phases between bosonic and fermionic states are unobservable: For any $\lvert \psi \rangle = \lvert b\rangle + \mathrm{e}^{\mathrm{i}\phi}\lvert f\rangle$, $\langle \psi \vert A \vert \psi\rangle$ is independent of $\phi$, so there is no possible way to determine such relative phases experimentally - they are "unobservable".