I Decoherence and standard formalism

[bhobba]

"With proper mixed states PLUS BORN RULE APPLIED, everything is sweet - objective reality exists before observation - much of quantum wierdness disappears."! Am I right Demystifier?

Please write out a mixed state then apply the Born Rule.

Here is a start. M = ∑pi |bi><bi| where the |bi> are orthonormal is a mixed state. The Born rule is given the mixed state M the expectation of observing it with the observable O= |b1><b1| is Trace (MO). This will give a result. Please write down the math and show what that result is. Its crucial - it must be understood before proceeding any further with decoherence.

Please, understanding decoherence is an advanced not a beginner topic. The textbook I mentioned and you quoted is way beyond the beginner level. I and others can tell you the results but can't explain them at the beginner level.

Thanks
Bill

 

[DrClaude]

Please, understanding decoherence is an advanced not a beginner topic. The textbook I mentioned and you quoted is way beyond the beginner level. I and others can tell you the results but can't explain them at the beginner level.

@fanieh, what is your level? Can we upgrade this thread to "undergraduate"?

 

[fanieh]

Please write out a mixed state then apply the Born Rule.

Here is a start. M = ∑pi |bi><bi| where the |bi> are orthonormal is a mixed state. The Born rule is given the mixed state M the expectation of observing it with the observable O= |b1><b1| is Trace (MO). This will give a result. Please write down the math and show what that result is. Its crucial - it must be understood before proceeding any further with decoherence.

Please, understanding decoherence is an advanced not a beginner topic. The textbook I mentioned and you quoted is way beyond the beginner level. I and others can tell you the results but can't explain them at the beginner level.

Thanks
Bill

Thanks for the keys that will finally make me understand the Maximilian Schlosshauer book "Decoherence and the Quantum-to-Classical Transition". When I first read it.. I admit they don't make a lot of sense.. but now second reading it. It does. I'll just reread it before asking further questions.

@fanieh, what is your level? Can we upgrade this thread to "undergraduate"?

I'm a layman. But since I'm getting the hang of Maximillian book. You can change it to "undergraduate".. I can no longer see the thread tool so you just edit it.

 

[bhobba]

I admit they don't make a lot of sense.. but now second reading it. It does..

Then you should be able to do the exercise I asked. Its dead simple.

Please post your solution. It will take less than a line.

If you can't do it there is no shame - we all will work with you to bring you up to the correct level, but the number one sin in both physics and especially in math is to think you understand something when you don't really. That's why in physics and math books you work at least some of the exercises - you must ensure your understanding is correct. Its a long hard process only the very greatest like Von-Neumann and Feynman can skip through (and even they didn't) but there is no other way to get understanding.

Math is a language that just like English must be practised.

Thanks
Bill

 

[zonde]

In the standard mathematical formalism, the environment were treated classically, this is because observers (being macroscopic recording mechanisms) are treated classically, so the system is isolated. Decoherence is about open system, so how is decoherence compatible with Copenhagen or the standard formalism at all?
How can you make the standard formalism have an open system-environment too? Or perhaps is it correct to think the standard mathematical formalism is already updated and Copenhagen is already outdated? How do you link the two if you were to give a lecture about this in class (which I'll do)?

Key feature of Copenhagen interpretation is that wave function is maximum description of quantum system or in other words two quantum systems described by the same wave function are fundamentally identical (no hidden variables).
On the other hand decoherence speaks about composite system being in entangled state. Obviously in that case separate quantum system can not be described by it's own wave function.
So in a sense they speak about two distinct scenarios that do not overlap. In one case we provide interpretation for individual system being in pure state and in other case we provide interpretation for individual system being in mixed state.

 

[martinbn]

Here is a start. M = ∑pi |bi><bi| where the |bi> are orthonormal is a mixed state. The Born rule is given the mixed state M the expectation of observing it with the observable O= |b1><b1| is Trace (MO). This will give a result. Please write down the math and show what that result is. Its crucial - it must be understood before proceeding any further with decoherence.

I don't get it! What is there to calculate! Isn't the answer obviously $p_1$? Or am I confusing myself with the physics notations?

 

[Demystifier]

I don't get it! What is there to calculate! Isn't the answer obviously $p_1$?

It is, but note that the post was addressed to a beginner, for whom it may not look so obvious.

 

[fanieh]

It is, but note that the post was addressed to a beginner, for whom it may not look so obvious.

My original answer to his question about what was the result was "mixed state".. but then I thought he was looking for a mathematical formula akin to e=mc^2 so I spent many weeks reading the Maximilian decoherence book, etc. and getting the hang of it so you guys won't say I didn't read the references suggested. So the answer is $p_1$? just probability value? But the trace has hidden collapse inside it.. and you know collapse is ad hoc. So it's like you just ignore the collapse and just give a probability figure?

Anyway. I need clarification. Is the equivalent of Many Worlds branch an eigenstate?
If our branch is a classical eigenstate. How come we can perform any quantum computing within our branch. Classical branch being a branch made of nut and bolts (Newtonian-like) so everything must be classical yet we can still perform quantum experiment. Remember an eigenstate has already born rule applied and not in superposition.. yet we can still do quantum experiment by isolating quantum system.

 

[bhobba]

Anyway. I need clarification. Is the equivalent of Many Worlds branch an eigenstate?
If our branch is a classical eigenstate. How come we can perform any quantum computing within our branch. Classical branch being a branch made of nut and bolts (Newtonian-like) so everything must be classical yet we can still perform quantum experiment. Remember an eigenstate has already born rule applied and not in superposition.. yet we can still do quantum experiment by isolating quantum system.

The sbove shows many misconceptions eq an eigenstate is not in a superposition is simply wrong. You must be precice in what you are saying..

Now here is your task. You claimed to have read Maximilian's book. He disscusses MW on page 336.

How about giving us a precis of what he said then a well formulated question based on that precis. As it stands you query makes no sensse.

Thanks
Bill

 

[bhobba]

I don't get it! What is there to calculate! Isn't the answer obviously $p_1$? Or am I confusing myself with the physics notations?

Of course pi is the probability of outcome i - but you know that by applying the Born Rule. Use the indicator observable |b1><b1| and apply it to M. This gives 1 if |b1> is the outcome and zero otherwise. The expected value will give the probability of getting |b1>.

So let's do it. The Born Rule is the expected value is Trace (MO) = ∑∑pi<bj|bi><bi|b1><b1|bj>. No only when i=j=1 do we get something non zero so we get p1<pi. ie the pi are the probability of getting |bi> as the outcome.

Thanks
Bill

 

[fanieh]

The sbove shows many misconceptions eq an eigenstate is not in a superposition is simply wrong. You must be precice in what you are saying..

Now here is your task. You claimed to have read Maximilian's book. He disscusses MW on page 336.

How about giving us a precis of what he said then a well formulated question based on that precis. As it stands you query makes no sensse.

Thanks
Bill

In the double slit experiment. The eigenstate or eigenpositions are the spots in the detectors. So they are no longer in superposition because they are collapsed already. So why do you say eigenstate is still in superposition. Born rule is already applied when a system is in an eigenstate.

I already read page 336 a week ago.. It's mostly relative-state interpretation.. But in modern day. Most are into many worlds. I'll quote Maximilians:

"Each of these (physical) states can be understood as relative (a) to the state of the other part in the composite system (as in Everett's original proposal; see also [141,143]), (b) to a particular "branch" of a constantly "splitting" universe (the many-worlds interpretation, popularized by DeWitt (386) and Deutsch (387[), or (c) to a particular "mind" in the set of minds of the conscious observer (the many-minds interpretation; see, for example, [388])"

In (b), Dewitt many worlds are classical nut and bolt world, is it not? and most people refer to Dewitt now in Many worlds. The original Everette relative proposal is less popular.

 

[bhobba]

In the double slit experiment. The eigenstate or eigenpositions are the spots in the detectors'

States are destroyed in the double slit so the question is meaningless.

If not, and this is basic QM, all states are in superposition and in an infinite number of ways. An eigenstate of an observable is still in superposition - its just that an eigenstate will give a definite result if observed using the observable its an eigenstate of. Its in superposition as any state is.

Thanks
Bill

 

[bhobba]

I already read page 336 a week ago..

Then read 337 - its the modern version.

It should be easy not to quote, which I asked you not to, but give us a precis. Whats going on in decoherence terms is very very simple - but I would prefer you to nut it out. If you want to study books at this level you must be prepared to THINK.

Thanks
Bill

 

[fanieh]

States are destroyed in the double slit so the question is meaningless.

If not, and this is basic QM, all states are in superposition and in an infinite number of ways. An eigenstate of an observable is still in superposition - its just that an eigenstate will give a definite result if observed using the observable its an eigenstate of. Its in superposition as any state is.

Thanks
Bill

Is this Bohr's original formalism? Because I was referring to Bohr. His Eigenstate is classical already because the apparatus is classical.. so please specify if it is Bohr formalism.. or other formalism.

 

[bhobba]

Is this Bohr's original formalism?

No.

Thanks
Bill

 

[fanieh]

Then read 337 - its the modern version.

It should be easy not to quote, which I asked you not to, but give us a precis. Whats going on in decoherence terms is very very simple - but I would prefer you to nut it out. If you want to study books at this level you must be prepared to THINK.

Thanks
Bill

But most physicists still think apparatus is classical and there is a division between quantum and classical world. Only few realized even the apparatus is quantum. Hence for those who believe in many worlds. They refer to Dewitt Many World instead of the Everett original formalism. Don't you agree with it? Many physicists who study the mathematical formalism assumes apparatus is classical. How many percentage do you estimate think it is quantum? Therefore I need to consider the mindset of these people.

 

[bhobba]

But most physicists still think apparatus is classical and there is a division between quantum and classical world.

Really.

You know this exactly how?

There is no division and I am not aware of mainstream modern practicing physicists that thinks so. And please don't quote thus or that they are either out of date, hold fringe ideas or you have the wrong context.

The modern view is Wienberg's:
http://scitation.aip.org/content/aip/magazine/physicstoday/article/58/11/10.1063/1.2155755

Both Bohr and Einstein were wrong, there is no classical quantum division and you will never learn QM from reading ancient texts.

Thanks
Bill

 

[PeterDonis]

most physicists still think apparatus is classical and there is a division between quantum and classical world

AFAIK this statement is egregiously false.

I need to consider the mindset of these people.

I don't think you should. I think you should focus on understanding the basic math of QM, how it makes experimental predictions, and how those predictions actually compare with experiments. In other words, start with the "shut up and calculate" interpretation. Until you've mastered that, trying to understand what the other interpretations are saying is going to be beyond you.

 

[PeterDonis]

And with that, this thread is closed.