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Science2Dmax

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In summary: Its the same as getting the reduced density matrix - in that case we trace over the extra degrees of freedom, in this case we trace over the environment.ThanksBillSo in summary, Schrodinger's cat is a thought experiment that was used to highlight a problem in the then-current understanding of quantum mechanics. The cat is either alive or dead, and the probability of it being either is determined by the half-life of the radioactive atom inside the box. The cat can be considered to be in a superposition of two states, but this is not the same as being unknown. The difference between a superposition and an improper mixture can be seen by using the density matrix formalism, and

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Science2Dmax

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TESL@

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Unknown, or more technically, in the superposition of two states.

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Nugatory

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Science2Dmax said:

You'll find a bunch of threads on Schrodinger's cat here if you search. The mainstream answer is also the commonsense one: The cat is either alive or dead, just as a tossed coin is either heads or tails even before we look.

I should add that that's pretty much always been the standard answer. When Schrodinger posed his famous though experiment almost a century ago, it wasn't because he or anyone else was seriously suggesting that the cat would somehow be "both alive and dead" or "in a superposition of alive and dead". He was pointing out a problem in the then-current understanding of quantum mechanics: nothing in the mathematical formalism clearly explained why the cat would be either dead or alive but not something in between.

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atyy

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rootone

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The thought experiment involves a radioactive atom, which when it decays triggers the mechanism which kills the cat.

The atom type has a known half life, and after this amount of time has passed the probability of the cat being alive or dead is 1:1.

If the cat is left in the box for a longer time the probability of it being dead increases and given sufficient time it become a near certainty.

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OCR

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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 wouldScience2Dmax 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.

creating fundamental disturbances in the Force, or at the vary least, a minor "tremor"...

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Nugatory

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TESL@ said:Unknown, or more technically, in the superposition of two states.

"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.

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bhobba

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Nugatory said:"Unknown" and "in the superposition of two states" are not the same thing.

Indeed.

Technically its the difference between a superposition and a mixed state. If you don't know the difference look it up.

Thanks

Bill

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Nugatory

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bhobba said:Technically its the difference between a superposition and a mixed state.

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.

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TESL@

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Thanks for the correction.

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atyy

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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.

But since this is the density matrix before the measurement outcome, it could be an improper mixture, which is a superposition.

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bhobba

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atyy said:But since this is the density matrix before the measurement outcome, it could be an improper mixture, which is a superposition.

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.

Thanks

Bill

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atyy

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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.

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.

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bhobba

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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.

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

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atyy

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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

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.

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StevieTNZ

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bhobba

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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.

Of course. But we are talking about inside the system. That's what tracing over the environment does.

Thanks

Bill

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bhobba

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StevieTNZ said:

It can't - by the definition of a mixed state. You have traced over the environment to do it.

Thanks

Bill

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bhobba

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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

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

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harrylin

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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.

- https://en.wikipedia.org/wiki/Schrödinger's_cat#The_thought_experiment

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atyy

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bhobba said:

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

The link you give itself shows that the the partial trace is equivalent to observing ##L_{A} \otimes \bf{1}## on the full system which is in a pure state. So the distinction you are making is rather arbitrary. In particular, with respect to your comment in on Nugatory's 'If toss a coin, it's either heads or tails but I don't know which until I look at it. That's "unknown".' - that refers to a proper mixture, not an improper mixture.

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bhobba

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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 an 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

Suppose we have the following superposition |p> = 1/√2|b1>|a1> + 1/√2|b2>|a2>. This is obviously an 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

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- #23

atyy

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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

How is observing A inside the system any different from observing A ##\otimes## I on the full system?

Anyway, the important point is that the improper mixed state does not correspond to Nugatory's definition of unknown - that is a proper mixed state.

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- #24

Buzz Bloom

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"I think it is safe to say that no one understands Quantum Mechanics."

See http://www.spaceandmotion.com/quantum-mechanics-richard-feynman-quotes.htm .

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Buzz Bloom

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Buzz Bloom

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http://www.spaceandmotion.com/quantum-mechanics-richard-feynman-quotes.htm

I don't understand why I have these dupicate posts but I don't seem to be able to delete all but one. When I try to delete one, all vanish.

- #27

bhobba

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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."

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.

In particular you need to get used to the idea the primitive of the theory is observations.

Thanks

Bill

- #28

atyy

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Let me see if I can use two different definitions of "unknown" to explain more clearly the point of view in which an improper mixture and a superposition are the "same thing".

In one definition of "unknown", the quantum system is in a definite state of reality, but we don't know what it is. All classical randomness is "ignorance interpretable" in this sense. In the quantum mechanical formulation, this most closely (but not exactly) corresponds to a measurement, followed by wave function collapse and the formation of a proper mixture. (I believe Nugatory gave this answer in post #3.)

In the definition of "unknown" that is the typical answer to the OP's question, the quantum system is in a definite state, but we don't know what that means because we don't understand whether the quantum state is real. It is in this sense that the pure state and the improper mixture are the "same thing", in that the randomness produced by each is not "ignorance interpretable". (This was my answer in post #4.)

Although there are two different definitions of "unknown" that are used to reply to the OP, it is known in some cases how to make the second definition into the first without wave function collapse, ie. give the quantum system a definite state of reality - by using nonlocal hidden variables.

At our current stage of technology, if nonlocal hidden variables do exist, we are not able to control them. So the second answer is more currently practical, in the sense that as long as we don't have enemies who are able to control such nonlocal hidden variables, then we can use quantum mechanics to produce "true" randomness for secure codes. Of course, if our enemies are much more technologically advanced than we are and can control the hidden variables, then quantum mechanics will not produce "true" randomness, and the enemies can use the determinism to break our codes.

In one definition of "unknown", the quantum system is in a definite state of reality, but we don't know what it is. All classical randomness is "ignorance interpretable" in this sense. In the quantum mechanical formulation, this most closely (but not exactly) corresponds to a measurement, followed by wave function collapse and the formation of a proper mixture. (I believe Nugatory gave this answer in post #3.)

In the definition of "unknown" that is the typical answer to the OP's question, the quantum system is in a definite state, but we don't know what that means because we don't understand whether the quantum state is real. It is in this sense that the pure state and the improper mixture are the "same thing", in that the randomness produced by each is not "ignorance interpretable". (This was my answer in post #4.)

Although there are two different definitions of "unknown" that are used to reply to the OP, it is known in some cases how to make the second definition into the first without wave function collapse, ie. give the quantum system a definite state of reality - by using nonlocal hidden variables.

At our current stage of technology, if nonlocal hidden variables do exist, we are not able to control them. So the second answer is more currently practical, in the sense that as long as we don't have enemies who are able to control such nonlocal hidden variables, then we can use quantum mechanics to produce "true" randomness for secure codes. Of course, if our enemies are much more technologically advanced than we are and can control the hidden variables, then quantum mechanics will not produce "true" randomness, and the enemies can use the determinism to break our codes.

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- #29

Buzz Bloom

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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.

Hi bhobba:

I am not a physisist.

From other material of Feyman's that I have read, I have come to believe that he had a more profound interpretation in mind than the one you give in your post. Unfortunaely, I can't at this time find any other relevant quotes of his I can post.

I think Feynman meant that the interpretive relationship between the QM math and the real physical world was, at the time of his quote, outside the realms of both math and physics. The relationship was (and may still be) entirely philosophical. That is, all the interpretations about this relationship that had been put forth by the best minds in physics soon after were seen to be

A physisist friend introduced me to an interpretation (one that I have been unable to find on the internet anywhere) that so far seems to me to be free of paradoxes, but it would require a somewhat lengthy exposition. As a brief overview, I offer: The interpretation of probability states involves multiple

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.

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- #30

bhobba

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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.

Ok.

A question then.

In that thought experiment the particle detector clicks or not and that is what determiners if the cat lives or dies.

What has the lid opening got to do with anything?

The reason I ask is there is a lot of confusion about it and what it means. I could explain it, but I think its usually better if you think it through for yourself - and beside I may be wrong.

Thanks

Bill

- #31

carllooper

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The state of the cat before opening the box can be regarded as no different from the state of the cat after opening the box.

Before opening the box we can say the cat is alive or dead.

And after opening the box? Well we find the cat is ... as described ... alive or dead.

C

Before opening the box we can say the cat is alive or dead.

And after opening the box? Well we find the cat is ... as described ... alive or dead.

C

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- #32

write4u

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Would it be correct to say that Schrodingers cat is in a state of superimposed potentials?

- #33

Buzz Bloom

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Hi Bill:bhobba said:What has the lid opening got to do with anything?

(I notice you signed your post "Bill", I so I assume that's your name rather than bhobba.)

As I undesrand the various philosophical views about this ambiguous point, it remains controversial regarding what constitutes an observation. There seem to be a spectrum of views. The two extrremes are:

1. Any physical interaction constitutes an "observation" that (in what I remember as being the old Copenhagen terminolgy)

2. It requires a conscious mind to make an observation that will accomplish the "collapse".

In my example, the opening of the lid is based on the second extreme.

Thanks for your post.

Buzz

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- #34

bhobba

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Buzz Bloom said:In my example, the opening of the lid is based on the second extreme.

That is one view, but a very very backwater view these days because of the severe problems it poses. For example imagine we did a Schroedinger's Cat with a robot opening the lid that recorded the result to computer memory. We then made billions of copies and scattered each copy across the cosmos. A million years later someone reads the contents of one of those copies - it would be a very very weird view of the world that's when it collapsed - and all of those copies collapsed. You could probably formulate a consistent view of the world along those lines - but - like solipsism - most would reject it as unnecessarily contrived.

The point I was trying to make is in the standard Copenhagen interpretation QM is a theory about observations that occur here in an assumed common-sense classical world. In Schroedinger's Cat that observation occurs at the particle detector - everything from that point on in common-sense classical - the cat is alive or dead regardless of if the box is opened or not.

Thanks

Bill

- #35

bhobba

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write4u said:Would it be correct to say that Schrodingers cat is in a state of superimposed potentials?

No. Its entirely classical. The observation occurred at the particle detector.

Thanks

Bill

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