# Does Bekenstein bound imply that the number of possible states of a bounded system is finite?

Bekenstein bound limits the amount of information that can be stored in a system of bounded size and mass. Does that imply that the number of possible states is finite? Does that imply that the number of possible values of measured observables is finite? If so, does that imply mass gap (because there must be a finite gap between any two energy values when the number of possible values is finite)? Does that set a limit on decoherence of a bounded system?

• Your link says it depends on energy (not mass) so doesn't it just mean that if you have more information that either it requires more energy or more space? So if space is a premium you can add more energy. – Timaeus Oct 22 '15 at 17:10

This is not my field (so I may be wrong), but since no-one else has tried to answer I will have a go...

Bekenstein bound limits the amount of information that can be stored in a system of bounded size and mass. Does that imply that the number of possible states is finite?

Since we are talking about a finite-volume system, its energy spectrum must necessarily be discrete, and therefore the number of states below a certain energy will always be finite. None of this is a consequence of the Bekenstein bound. However, the Berkenstein bound does put a specific limit on the number of quantum states below a fixed energy.

Does that imply that the number of possible values of measured observables is finite?

The spectrum will have a finite number of states below any finite energy, but will still have an infinite number of states over the whole energy spectrum. You could argue that the energy should be cut off at the energy required to turn the system in to a black hole (as it will then no longer be observable), which could then give you a finite number of possible values for measured observables below this energy... I'm not sure how 'proper' this argument is... :-P

If so, does that imply mass gap (because there must be a finite gap between any two energy values when the number of possible values is finite)?

No... The discrete energy spectrum is a consequence of the finite size of the system. Photons in a box are still massless even though their energy spectrum is discrete.

Does that set a limit on decoherence of a bounded system?

I don't see why it should... what is your reasoning?

• About decoherence limit: I wonder if decoherence increases the number of possibile measurements, which cannot continue infinitely if the number of possible states is finite... – Michael Nov 3 '15 at 17:18
• @Michael I don't think so... decoherence means that the system becomes entangled with the environment, so that different eigenstates of the system are coupled to orthogonal states of the environment. This removes the interference effects that are usually seen when a quantum system is measured (or more accurately, when the same measurement is repeatedly performed on an ensemble of identically prepared systems), but it doesn't change the possible outcomes of measurement or the number of measurements that can be performed. – Mark A Nov 4 '15 at 2:33
• I probably don't understand what decoherence is. I though that if entangle states could be measured as 00> or 11> after decoherence they could be measured 00> or 01> or 10> or 11>. That is, I thought that decoherence was a loss of entanglement. Am I wrong here? – Michael Nov 5 '15 at 16:52
• @Michael Yes, you are wrong :-P... Decoherence is not the same as loss of entanglement. Rather, it is loss of coherence in the quantum state through entanglement with the environment. The 'quantum decoherence' wikipedia page explains it quite well... – Mark A Nov 6 '15 at 2:35