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The article is actually correct - the general state of a system of 300 spins is described by 2^300 complex amplitudes. The remaining question is how much memory

by danbruc 12y ago

The article is actually correct - the general state of a system of 300 spins is described by 2^300 complex amplitudes. The remaining question is how much memory you are willing to spend per amplitude but with two doubles it will take 2.6 * 10^92 bits. You only get away with 300 amplitudes if the spins are not entangled and the state of the system therefore is a product state, i.e. you can factor the state of the system into 300 independent states, one per spin. A two spin system for example has the general state a |DD> + b |DU> + c |UD> + d |UU> requiring four complex amplitudes, not two bits only selecting one of the basis states. EDIT: Could not resist. The storage density of a 15 mm thick 2 TB 2.5" hard disk is 152.7 Ebit/m³. With the number from above this yields 2 * 10^27 ly³ for the volume which is not to far from the visible universe with 4 * 10^32 ly³ according to Wikipedia. Okay, it is a factor of 200,000 but that will only buy you 17 more bits plus some space for cable routing.

fennecfoxen 12y ago
Of course, if you're trying to use modern hard drive material to fill the observable universe, you'll reach black-hole density in no time flat (if I recall correctly, a Solar System-sized black hole has a density similar to the Earth's atmosphere.) And even if you're not using modern technology to fill the space, there is a quantum limit on the minimum amount of energy required to represent one bit. Interestingly, a black hole's entropy is proportional the surface area of its event horizon, so the maximum information capacity of a given region of space is too. (See also: holographic universe.)
Retric 12y ago
This seems like a clasic analog issue. Does adding 1 spin to a 300 spin system add exactly 2^300 bits of information and 1 spin to a 2 spin situation exactly 2^2 = 4 bits of information?
- bkcooper 12y ago
  As mentioned elsewhere, bits are probably the wrong way to think about it. Does it really double the size of your Hilbert space? Yes. Perhaps this will help clarify why it is different. In the classical case, adding one more spin is doubling the size of your set. In the quantum mechanical case, adding one more spin instead doubles the dimensionality of the space of possible states.

by danbruc 12y ago

fennecfoxen 12y ago

Of course, if you're trying to use modern hard drive material to fill the observable universe, you'll reach black-hole density in no time flat (if I recall correctly, a Solar System-sized black hole has a density similar to the Earth's atmosphere.) And even if you're not using modern technology to fill the space, there is a quantum limit on the minimum amount of energy required to represent one bit. Interestingly, a black hole's entropy is proportional the surface area of its event horizon, so the maximum information capacity of a given region of space is too. (See also: holographic universe.)

Retric 12y ago

This seems like a clasic analog issue. Does adding 1 spin to a 300 spin system add exactly 2^300 bits of information and 1 spin to a 2 spin situation exactly 2^2 = 4 bits of information?

bkcooper 12y ago
As mentioned elsewhere, bits are probably the wrong way to think about it. Does it really double the size of your Hilbert space? Yes. Perhaps this will help clarify why it is different. In the classical case, adding one more spin is doubling the size of your set. In the quantum mechanical case, adding one more spin instead doubles the dimensionality of the space of possible states.