Does the excess-baggage theorem count against CSD?

The short answer

There's a family of results, Hardy's and Montina's sharpening of it, with a deliberately deflating name. They say any model which puts real underlying states beneath quantum mechanics must carry more than looks necessary, so for a system with a fixed finite number of distinguishable outcomes no finite number of underlying states will do, and you need a continuum, a smoothly varying range of states like the points of a line, rather than a set you could count off. The name suggests the extra states are luggage the model would rather not be carrying.

For most hidden-variable programmes this is a real cost, because they set out to recover quantum statistics from something smaller or simpler than the quantum state. For CSD it isn't a cost at all, because the continuum isn't a concession. The constraint surface, the space of underlying states, is a continuous space from the first line of the construction, as that's what it is, and the measures that give rise to probabilities are volumes on it, so the theorem asks for exactly what the programme already assumes.

This entry exists so the question is answered rather than left hanging. A referee meeting a theory shaped like a hidden-variable theory is entitled to raise excess baggage, and the answer is short.

The longer answer

The constraint surface is a measurable space, one on which regions can be assigned sizes, carrying a finite Liouville measure, the natural volume preserved by the dynamics. The sector map, which sends each underlying state to the quantum state it presents, is many-to-one onto the space of quantum states, and nothing in the construction attempts a finite or countable space of underlying states, so the excess-baggage lower bound is satisfied trivially rather than evaded.

Two things this answer doesn't claim. It doesn't claim the theorem supports CSD, as satisfying a lower bound isn't evidence for a model. And it doesn't claim sheer size is what makes the programme work, as the load-bearing structure is the fibre, the set of underlying states that share one quantum state, together with its measure and the cells cut out by the moment map, and not the mere fact that the constraint surface has uncountably many points. The excess-baggage theorems are silent about all of that.

What is actually proved

Hardy (2004) and Montina (2008) show that an ontic model reproducing the statistics of an N-level quantum system requires an ontic state space of continuum cardinality, so a finite set of ontic states can't reproduce the continuum of pure-state predictions.

CSD's Sigma is a continuum by construction (P1 measurable constraint surface, P2 finite Liouville measure), so no bound is approached from below.

Related entries

• Ontic vs epistemicWhat exists is one question and what a measurement context lets you say is another, and most quantum paradoxes are category errors between the two.
• FibreThe arena of CSD is larger than the space of quantum states by exactly one compact factor, and from three dimensions upward the natural ways of doing without that factor stop working.
• Many-to-one projectionBehind every quantum state stands a crowd of world-states, and the projection that maps the crowd to one state loses information on purpose, and that loss is where quantum probability comes from.
• Does the PBR theorem rule out CSD?No. For exactly prepared pure states, CSD satisfies the very conclusion the PBR theorem argues for, as two distinct states correspond to sets of underlying world- states that don't overlap at all, and that's a proved theorem rather than something arranged by hand.
• Is the wavefunction real in CSD?The wavefunction is a genuine but incomplete coordinate of a real underlying world, so it can be read off the world's actual configuration, yet it leaves out the part of that configuration which decides what a measurement will show.