- #1
Science2Dmax
- 9
- 0
my girlfriend is wondering if while shroodingers cat was in the box, we should consider it as both unknown rather than both alive and dead. please shed some light on the subject.
Science2Dmax said:my girlfriend is wondering if while shroodingers cat was in the box, we should consider it as both unknown rather than both alive and dead. please shed some light on the subject.
atyy said:Unknown. It "is" in a superposition of two states, but we don't now what that means.
So never, never, put a dead cat in "the box", at the beginning... such an action would have a probability ofScience2Dmax said:my girlfriend is wondering if while shroodingers cat was in the box, we should consider it as both unknown rather than both alive and dead.
TESL@ said:Unknown, or more technically, in the superposition of two states.
Nugatory said:"Unknown" and "in the superposition of two states" are not the same thing.
bhobba said:Technically its the difference between a superposition and a mixed state.
bhobba said:Indeed.
Technically its the difference between a superposition and a mixed state. If you don't know the difference look it up.
Thanks
Bill
Nugatory said:What Bhobba said.
You have to use the density matrix formalism to see the difference. Googling for "quantum density matrix" will find some good links - it's unfortunate that this is not covered by some intro textbooks.
atyy said:But since this is the density matrix before the measurement outcome, it could be an improper mixture, which is a superposition.
bhobba said:Improper mixtures are not superpositions from the very definition of a mixture. Its called improper because its physical origin is different to a proper one - not that its not a mixture - which it obviously is.
atyy said:Improper mixtures are superpositions which is why they are not proper. A superposition refers to a pure state, which is what an improper mixture is.
bhobba said:An improper mixture is NOT a superposition.
Outside the system it remains in superposition - inside it isn't. That is the key difference - see the section 1.2.3:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf
It is removing system B from our control by tracing over the environment that does it.
Thanks
Bill
atyy said:There is no difference because the entire system is still in a superposition. Everything you do on the improper mixture has a counterpart in the full system.
StevieTNZ said:Bernard d'Espagnat makes the distinction between an improper mixture v a proper mixture in his book "On Physics and Philosophy". atyy is correct in stating an improper mixture refers to a superposition (pure) state.
This thread is not about the radioactive atom, but about the cat which will be killed when that atom decays. The question is if the special state that the atom is in according to the theory, necessarily affects the cat because we don't know about it. It was Schrödinger's intention to ridicule that idea by means of his cat example.Nugatory said:"Unknown" and "in the superposition of two states" are not the same thing.
If toss a coin, it's either heads or tails but I don't know which until I look at it. That's "unknown".
A spin-1/2 particle with its spin aligned up along the z axis is not in an unknown state. It's in the up eigenstate of the ##S_z## operator and that's a complete and unambiguous specification of its state that leaves nothing unknown. However, that state is also a superposition of spin-up along the x-axis and spin-down along the x axis.
bhobba said:Just to expand further on the issue, Lubos has done an excellent post giving the detail - if I was to write something it would basically be what he wrote:
http://physics.stackexchange.com/qu...ake-the-partial-trace-to-describe-a-subsystem.
What's going on is you have the whole system A+B - that will be assumed to be a pure state |p><p| - here I have used the operator description of a pure state to avoid confusion later. The mixed state comes about because we are only interested in observations on A. But its entangled with B. Its that entanglement that leads to it being a mixed state because you need to do a partial trace over B due to only A being observed. It is this only observing A when it entangled with B that's the cause ie you are observing inside the system ie the state is (trace over B |p><p|) - which, lo and behold, turns out to be a mixed state - see equation 1.2.3 in the paper I linked to before. Outside the system its a pure state |p><p|, hence a superposition of all sorts of things, and remains in superposition, until, of course, it becomes entangled with something else. But because of the partial trace inside the system, since you are observing only A, its a mixed state.
Thanks
Bill
bhobba said:Ok - maybe what I said was a bit general so I will do a simple specific example and we can see exactly what's going on.
Suppose we have the following superposition |p> = 1/√2|b1>|a1> + 1/√2|b2>|a2>. This is obviously and entangled system where system A is entangled with system B. It's a pure state. It remains in a pure state until observed ie until its interacted with.
But now we will do an observation on just system A with the observable A.
E(A) = <p|A|p> = 1/2 <a1|<b1|A|b1>|a1> + 1/2 <a1|<b1|A|b2>|a2> + 1/2 <a2|<b2|A|b1>|a1> + 1/2 <a2|<b2|A|b2>|a2>
Now here is the kicker - since you are only observing system A the observable A has no effect on the B system or its states. So we have:
<p|A|p> = 1/2 <Aa1|<b1|b1>|a1> + 1/2 <Aa1|<b1|b2>|a2> + 1/2 <Aa2|<b2|b1>|a1> + 1/2 <Aa2|<b2|b2>|a2> = 1/2 <a1|A|a1> + 1/2 <a2|A|a2>
= Trace((1/2|a1><a1| + 1/2|a2><a2|) A) = Trace (p' A)
Here p' is the mixed state 1/2|a1><a1| + 1/2|a2><a2|. Thus observing system A is equivalent to observing a system in the mixed state p' - which by definition is the state from |p> by doing a partial trace over B. The observation will of course give |a1> or |a2> and the entanglement will be broken so that if you get |a1> system B will be in |b1> and conversely. We still have collapse if you like that language - but now it has a different interpretation - you are not observing a pure state - but a mixed one. Its not a proper mixed state because its not prepared the way a proper mixed state is prepared - but the state is exactly the same. Any observable A will not be able to tell the difference. This means we, in a sense, can kid ourselves and say, somehow, its a proper mixed state. If it was a proper mixed state then prior to observation it is in state |a1> or state |a2> with probability of half. Prior to observation its in superposition - after it isnt. Until observed it continues in superposition - its simply because of the entanglement it can now be interpreted differently. By observing 'inside' the system - ie only observing system A - it is in a mixed state - not a proper one - but still a mixed state. Because of that it allows a different and clearer interpretation that avoids a lot of problems.
I really can't explain it better - so if its still unclear then there isn't much more I can do.
Thanks
Bill
Buzz Bloom said:Note the following well known quote by Richard Feinman" "I think it is safe to say that no one understands Quantum Mechanics."
bhobba said:Everyone knows Feynmans quotes. But as you learn more about QM you understand plenty of people understand QM - what he meant was understanding it in usual everyday terms.
Buzz Bloom said:Regarding the cat: When the experiment has been set up, and the particle that will determine fate of the cat is emitted, two contigent universes are created, one in which tha cat will survirve, and the other in which the cat will soon after die. When the obsever opens the lid, one of these two contingent universes becomes the real universe in which the observer continues to exist.
Hi Bill:bhobba said:What has the lid opening got to do with anything?
Buzz Bloom said:In my example, the opening of the lid is based on the second extreme.
write4u said:Would it be correct to say that Schrodingers cat is in a state of superimposed potentials?