compilerbitch: That's me, that is! (Default)
compilerbitch ([personal profile] compilerbitch) wrote2004-12-17 12:01 pm

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?
fluffymark: (Default)

[personal profile] fluffymark 2004-12-17 04:26 am (UTC)(link)
4, I would think. Or 7, if you were attempting with Fermions (which cannot ever be in the same quantum state almost by definition).

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.

[identity profile] compilerbitch.livejournal.com 2004-12-17 04:34 am (UTC)(link)
Further to the garbled nonsense I just posted and deleted, given a pair of qubits, can I constrain them to never both be in a particular state, but allow all other possibilities (e.g. 00, 01 or 10 but never 11)?

[identity profile] compilerbitch.livejournal.com 2004-12-17 04:36 am (UTC)(link)
Entangling in the different state has no memory of previous entanglement in the same state.

Does this presuppose a disentanglement then a fresh entanglement, or retained entanglement and a changed state?
fluffymark: (Default)

[personal profile] fluffymark 2004-12-17 08:09 am (UTC)(link)
As far as I know, "entanglement" just means that a measurement of a state of one particle depends on the state of another - a bit like conditional probability, although not entirely the same. Particles are only "disentangled" if they are truly independent.

[identity profile] ashley-y.livejournal.com 2004-12-17 04:39 am (UTC)(link)
Fermions can't be in exactly the same state: for instance, two electrons cannot have the same principal, azimuthal, magnetic and spin numbers around the same nucleus, but you can just take the spin aspect of their state and entangle them any which way.

[identity profile] compilerbitch.livejournal.com 2004-12-17 04:40 am (UTC)(link)
Is partial collapse possible?

That is, can an entangled state allowing many combinations collapse to a state allowing fewer (but more than one) combinations?
fluffymark: (Default)

[personal profile] fluffymark 2004-12-17 08:11 am (UTC)(link)
Yes. Say you just measure the spin, then just the spin wavefunction collapses, but the position/momentum wavefunciton remains, unless it is entangled with the spin so that you know a particualar spin state corresponds to a particular position/momentum state.

[identity profile] compilerbitch.livejournal.com 2004-12-17 04:45 am (UTC)(link)
PS: I personally suspect that, given the throwing an exception or core dumping doesn't seem to happen, 6 has to be the answer, or some things seem break mathematically if you start considering more complex entanglement.

[identity profile] compilerbitch.livejournal.com 2004-12-17 04:46 am (UTC)(link)
(and I also suspect that partial collapse has to be possible too)