Wanted: Quantum physicist, prepared to talk to person insane

[compilerbitch]

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 2^N. With N qubits in a mutually entagled state, the number of possible combinations is strictly less than 2^N, 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?

Comments

[splodgenoodles.livejournal.com]

Bunnies?

*cries*

[ashley-y.livejournal.com]

OK, you create two particles entangled in the same state. What exactly do you do next?

[compilerbitch.livejournal.com]

I don't know. I am assuming that there is an after-particle-creation mechanism that can impose entanglement. If there isn't, this is something worth knowing too.

[compilerbitch.livejournal.com]

(thinks further)

People have built gate-like quantum devices which presumably require interaction between already-entangled particles, so operations with this kind of flavour are at least generally feasible. If this specific thing isn't possible, then knowing exactly why would be useful, and possibly very important indeed if some of the consequences of this idea pan out.

[fluffymark]

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.

[arkady.livejournal.com]

I'm going with 2, but I'm not a quantum physicist. I just know that doing such things usually involves big machines, incredibly high velocities and a big bang at the end....

[compilerbitch.livejournal.com]

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)?

[compilerbitch.livejournal.com]

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?

[vampwillow]

er ... dunno without having another strong coffee first.

I do note that String Theory was *twenty* years old this week though. eep!

[ashley-y.livejournal.com]

Entanglement is when two particles have the same state function. If you observe the particles, you'll collapse the state function. If you do something to one particle, you'll give it another state function. So the answer is 4.

[ashley-y.livejournal.com]

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.

[compilerbitch.livejournal.com]

Is partial collapse possible?

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

[compilerbitch.livejournal.com]

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.

[compilerbitch.livejournal.com]

(and I also suspect that partial collapse has to be possible too)

[compilerbitch.livejournal.com]

I care more about entanglement where N>2.

[hairyears.livejournal.com]

Entanglement means that an unmeasured property of A has a quantum counterpart in B. When A is measured, the 'wave function' collapses and measurement B will be the known counterpart of A, even if that information cannot possibly have been transmitted to 'B' after measurement 'A'

So you've obtained an entangled pair, AB, out of some black box we'll call action 1.

Your second act of entanglement can, theoretically, be an entirely unrelated property such that we might as well be talking about particles C & D, but I'm ignoring that possibility.

The point is that this second act of entanglement is itself a measurement and Action: you've done something that determines the quantum state of A and collapses the wave function formed in action 1. Then you entangling A and B again.

So you're back where you started, with A and B having an umeasured property which will, when A is measured, effect a spooky action at a distance that constrains B. And you don't have any information that links the second measurement to the first one: the values are independant.

Of course, you might commit Ation 2 in such a way that the result isn't 'measured' except that you know it has *changed* AB. Let's say you know it's reversed the polarisation on both. So the wave function does not collapse, and the particles remain entangled.

Clever, except that you didn't know what the value was in the first place, and you obviously still don't know now, until you measure the polarisation at A. Subsequent measurements of B will, by spooky action at a distance, be the predictable counterpart.

So you've achieved nothing.

[fluffymark]

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.

[fluffymark]

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.

[fluffymark]

Where N=2 the 2 particles can be entangled post-creation in theory (although in practice is hard, but possible). For N > 2 I don't think has been achieved in practice, as the act of entangling the 3rd particle tends to "measure" and break the entanglement of the first 2, although in theory a multi-particle entanglement is perfectly acceptable.

[fatdog.livejournal.com]

as far as i remember, in quantum mechanics you're allowed to make up the answer, and then fudge the maths (in extreme cases inventing a whole new branch of maths) to show that you were right the whole time.

[compilerbitch.livejournal.com]

I'm sorted then, in that case. :-)