↑ comment by Manfred ·
2013-08-05T10:11:08.417Z · LW(p) · GW(p)
First: check this out.
Second: Suppose I want to demonstrate decoherence. I start out with an entangled state - two electrons that will always be magnetically aligned, but don't have a chosen collective alignment. This state is written like |up, up> + |down, down> (the electrons are both "both up" and "both down" at the same time; the |> notation here just indicates that it's a quantum state).
Now, before introducing decoherence, I just want to check that I can entangle my two electrons. How do I do that? I repeat what's called a "Bell measurement," which has four possible indications: (|up,up>+|down,down>) , (|up,up>-|down,down>) , (|up,down>+|down,up>) , (|up,down>-|down,up>).
Because my state is made of 100% Bell state 1, every time I make some entangled electrons and then measure them, I'll get back result #1. This consistency means they're entangled. If the quantum state of my particles had to be expressed as a mixture of Bell States, there might not be any entanglement - for example state 1 + state 2 just looks like |up,up>, which is boring and unentangled.
To create decoherence, I send the second electron to you. You measure whether it's up or down, then re-magnetize it and send it back with spin up if you measured up, and spin down if you measured down. But since you remember the state of the electron, you have now become entangled with it, and must be included. The relevant state is now |up, up, saw up> + |down, down, saw down>.
This state is weird, because now you, a human, are in a superposition of "saw up" and "saw down." But we'll ignore that for the moment - we can always replace you with with a third electron if it causes philosophical problems :) The question at hand is: what happens when we try to test if our electrons are still entangled?
Again, we do this a bunch of times and do a repeated Bell measurement. If we get result #1 every time, they're entangled just like before. To predict the outcome ahead of time, we can factor our state into Bell States, and see how much of each Bell State we have.
So we factor |up, up, saw up> into |(Bell state 1) + (Bell state 2), saw up>, and we factor |down, down, saw down> into |(Bell state 1) - (Bell state 2), saw down>.
Now, if that extra label about what you saw wasn't here, the ups and the downs would be physically/mathematically equivalent and we could cancel terms to just get Bell state 1. But if any of the labels are different, you can't subtract them to get 0 anymore. That is, they no longer interfere. And so you are just left with equal numbers of Bell state 1 and Bell state 2 terms. And so when we do the Bell measurement, we get results #1 and #2 with equal frequency, just like we would if the electrons were completely unentangled.
This is not to say they're not entangled - they still are. But they can no longer be shown to be entangled by a two-particle test. They're no longer usefully entangled. You need to collect all the pieces together before you can show that they're entangled, now. And that gets awful hard once a macroscopic system like a human gets entangled with the electrons and starts radiating off still-entangled photons into the environment.
This is decoherence. I can have a nice entangled system, but if I let you peek at one of my electrons, you turn the state into into |(Bell state 1) + (Bell state 2), saw up> + |(Bell state 1) - (Bell state 2), saw down>, and they don't behave in the entangled way they did anymore.
Replies from: Luke_A_Somers
↑ comment by Luke_A_Somers ·
2013-08-05T14:56:44.195Z · LW(p) · GW(p)
Not to undermine your point, but |up, up> + |down, down> is perfectly oriented in the X direction.
What works better for this is that you indicate that the state is A |up, up> + B|down, down>, and you don't know A and B.
Replies from: Manfred