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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.

Alexander McFarlane
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  • If we define the phase difference $\Delta$ then under a rotation of $2\pi$ I would deduce that we obtain a new phase difference $\Delta'$. Is the answer based in this? – Alexander McFarlane Apr 18 '16 at 07:19
  • Think I have a resolution: Set up a superposition of two normalised wave functions $|\psi\rangle = |\phi_f\rangle + |\phi_b\rangle$ where $|\phi_b\rangle$ is bosonic and $|\phi_f\rangle$ is fermionic. Taking the inner product (up to a constant), $\langle\psi| \psi\rangle \propto \langle\phi_b|\phi_b\rangle + \langle\phi_f|\phi_b\rangle + \langle\phi_b|\phi_f\rangle + \langle\phi_f|\phi_f\rangle$ since the wave functions are normalised, $\langle\phi_b|\phi_b\rangle = \langle\phi_f|\phi_f\rangle = 1$ and they are orthogonal so $\langle\phi_b|\phi_f\rangle = \langle\phi_f|\phi_b\rangle = 0$. – Alexander McFarlane Apr 18 '16 at 07:37
  • Hence, $\langle\psi| \psi\rangle \propto 1-1+0+0 = 0$. Would welcome any better answers though! – Alexander McFarlane Apr 18 '16 at 07:37

1 Answers1

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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".

ACuriousMind
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  • Eloquently phrased. This links in everything I have been studying relevant to the question as well. – Alexander McFarlane Apr 18 '16 at 11:06
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    I particularly like the comment "they are 'unobservable'." ... It seems as though the throwaway comments in many notes on this subject stating that these states "cannot exist" or that they correspond to "nonsense" is, in fact, not exact. In contrast, what you state is more precise: That they simply cannot correspond to real observable quantities. – Alexander McFarlane Apr 18 '16 at 11:17