compilerbitch (
compilerbitch) wrote2004-12-17 12:01 pm
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Wanted: Quantum physicist, prepared to talk to person insane
If I entangle two particles A and B in such a way that they are in the same quantum state, then entangle them such that they are in different quantum states, what happens?
1. The universe ends with the biggest core dump in the history of the multiverse
2. Both particles cease to exist in some kind of impressive explosively thrown exception
3. Both particles end up entangled to be in the same state
4. Both particles end up entangled to be in different states
5. Both particles end up entangled, but are randomly entangled as in 4. or 5. above
6. Entanglement is broken, with both particles left in random states
7. Doing what I said is impossible for some clever but not immediately obvious reason
8. I am asking a stupid question (though I'd argue that knowing exactly why it is stupid would be valuable)
(note that 5 and 6 are not equivalent if either particle ends up further entangled)
The same problem, from a quantum computing point of view, could be restated a bit more simply. Given N classical bits, the number of possible combinations of values is 2N. With N qubits in a mutually entagled state, the number of possible combinations is strictly less than 2N, such that if an observation (and collapse to a classical, non-superposed state) is forced, the resulting classical value is constrained by the original entangled state. My question is, what happens if the entangled state allows exactly zero possible classical values?
1. The universe ends with the biggest core dump in the history of the multiverse
2. Both particles cease to exist in some kind of impressive explosively thrown exception
3. Both particles end up entangled to be in the same state
4. Both particles end up entangled to be in different states
5. Both particles end up entangled, but are randomly entangled as in 4. or 5. above
6. Entanglement is broken, with both particles left in random states
7. Doing what I said is impossible for some clever but not immediately obvious reason
8. I am asking a stupid question (though I'd argue that knowing exactly why it is stupid would be valuable)
(note that 5 and 6 are not equivalent if either particle ends up further entangled)
The same problem, from a quantum computing point of view, could be restated a bit more simply. Given N classical bits, the number of possible combinations of values is 2N. With N qubits in a mutually entagled state, the number of possible combinations is strictly less than 2N, such that if an observation (and collapse to a classical, non-superposed state) is forced, the resulting classical value is constrained by the original entangled state. My question is, what happens if the entangled state allows exactly zero possible classical values?
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Assuming Bosons then, you get 2 Bosons and put them in the same quamtumn state. Fair enough, at this point they are not only identical but indistinguishable. Then you take one of them (doesn't matter which) and alter the quantum state, say, flip the spin. Now they are in different quantum states, and have entangled spins (you know whatever spin one has, the other must have the opposite spin). Entangling in the different state has no memory of previous entanglement in the same state. I think.
As to your last question, I think every quantum state must allow at least one possible classical vlaue - after all, an "observation" collapses the quantum wavefunction. Statistically the quamntum state may not behave like any classical statistics, but each reduced state must collapse to a classical one.
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Does this presuppose a disentanglement then a fresh entanglement, or retained entanglement and a changed state?
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That is, can an entangled state allowing many combinations collapse to a state allowing fewer (but more than one) combinations?
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