Decoherence and standard formalism

[fanieh]

No, the entanglement does not work in that way. Entanglement always involves a superposition, which is precisely what you want to avoid to get a "classical" state.

I know. Without the born rule. So the hat is still in superposition of position but only perhaps about 1 inch (or is it mere millimetres?) from it's original position instead of meters away from the person head. How does one compute the range.. have you done it?

 

[Demystifier]

how's there two meanings of "classical state".. kindly elaborate.. thank you.

There is a classical deterministic state and a classical probability state. In the case of a coin, one classical deterministic state is head, another classical deterministic state is tail. A classical probability state is $p(head)=1/2, \; p(tail)=1/2$. Decoherence, i.e. entanglement with the environment, explains the origin of classical probability states, but it does not explain the origin of classical deterministic states.

 

[fanieh]

I would suggest forgetting Copenhagen to begin with. It really is old hat:
http://scitation.aip.org/content/aip/magazine/physicstoday/article/58/11/10.1063/1.2155755

We now know both Bohr and Einstein were wrong so its a rather hard issue to debate.

Thanks
Bil

Let's have a last word from Peterdonis about whether we should forget Copenhagen to begin with (as Bhobba suggested above). I think Peterdonis was saying even if the environment was treated classically.. you would still get the same experimental result or it has the same mathematical formalism? Or perhaps is it correct to say in front of audience that Copenhagen has limited power or more limited in scope (what is the right thing to say)?

 

[PeterDonis]

whether we should forget Copenhagen to begin with

I agree with Demystifier's point that the term "Copenhagen interpretation" doesn't actually name a specific well-defined interpretation. But as far as the mathematical formalism of QM is concerned, there is no such thing as "interpretation"; the formalism is what it is.

I think Peterdonis was saying even if the environment was treated classically.. you would still get the same experimental result or it has the same mathematical formalism?

I was saying that "if the environment was treated classically" is not something that is done in the mathematical formalism. In many simple cases the environment doesn't need to be treated at all in the formalism; but in cases where it is (such as when you are trying to model decoherence for a macroscopic object), the environment is treated quantum mechanically.

 

[fanieh]

I agree with Demystifier's point that the term "Copenhagen interpretation" doesn't actually name a specific well-defined interpretation. But as far as the mathematical formalism of QM is concerned, there is no such thing as "interpretation"; the formalism is what it is.
I was saying that "if the environment was treated classically" is not something that is done in the mathematical formalism. In many simple cases the environment doesn't need to be treated at all in the formalism; but in cases where it is (such as when you are trying to model decoherence for a macroscopic object), the environment is treated quantum mechanically.

Ok, so it's the domain of applicability. When you want to compute the range of your hat superposition in the position basis as it interacts with all objects in the universe, you need the mathematical formalism which has this ability. Is it right to say Zeh and Zurek make use of the mathematical formalism that was not there in the old mathematical formalism when Bohr developed Copenhagen?

 

[PeterDonis]

Is it right to say Zeh and Zurek make use of the mathematical formalism that was not there in the old mathematical formalism when Bohr developed Copenhagen?

I'm not sure. The basic formalism hasn't changed: you can either use wave functions and differential operators or state vectors and matrices. Both of those formalisms (Schrodinger and Heisenberg) were developed in the mid 1920s. What has changed is the amount of work that has been done to explore how to apply these formalisms to different cases.

 

[fanieh]

I'm not sure. The basic formalism hasn't changed: you can either use wave functions and differential operators or state vectors and matrices. Both of those formalisms (Schrodinger and Heisenberg) were developed in the mid 1920s. What has changed is the amount of work that has been done to explore how to apply these formalisms to different cases.

So decoherence can be described by just wave functions. Yet the density matrix has its natural home in decoherence.. is the reason due to the density matrix probability being more natural in the formalism.. can you give 3 reasons why the density matrix must go hand in hand with decoherence or they are married compatibility optimal? Or can one teach decoherence by totally bypassing the density matrix.. like instead of saying tracing out the environment... just say collapse or born rule engaged.

 

[PeterDonis]

decoherence can be described by just wave functions

Actually I should have added density matrices to that; I'm not sure wave functions (or state vectors) by themselves are sufficient, because a key point about decoherence is that you don't know the state of the environment, so you have to trace over it. That requires using density matrices. AFAIK those weren't introduced until Von Neumann's classic textbook in the early 1930s; so perhaps it is true that the original formalism wasn't enough by itself.

can one teach decoherence by totally bypassing the density matrix

I don't think so, but I am not an expert in the field.

 

[fanieh]

There is a classical deterministic state and a classical probability state. In the case of a coin, one classical deterministic state is head, another classical deterministic state is tail. A classical probability state is $p(head)=1/2, \; p(tail)=1/2$. Decoherence, i.e. entanglement with the environment, explains the origin of classical probability states, but it does not explain the origin of classical deterministic states.

For you. Proper mixed state is classical probability state and not classical deterministic state? The reason I thought proper mixed state was classical deterministic state was due to Bhobba writing the following in another thread to me. Is he wrong when he stated?:

"There are a number of ways of preparing mixed states. One way is simply to take a state and randomly present it with probability pi. Such are called proper mixed states. With proper mixed states everything is sweet - objective reality exists before observation - much of quantum wierdness disappears. But that is just one way of going it. Another way is to take a state and subject it to the process of decoherence - you get exactly the same mixed state and their is no way to tell the difference - no way at all. But because its prepared differently than a proper mixed state its called an improper one. The trouble is you can't say its in the state prior to opservation - there is simply no way to tell. If not quantum wierdness remains. This is the modern version of the measurement problem. What causes an improper mixture to become a proper one. Colloquially its why we get any outcomes at all. With different interpretations like MW and BM its trivial, with others like ensemble its much more controversial - even to the point of its a problem at all."

Bhobba stated that in proper mixed states, objective reality exists before observation. But in your terminology. It's not in classical determistic state, but only in classical probability state. So Bhobba is really wrong about it?

 

[bhobba]

Bhobba stated that in proper mixed states, objective reality exists before observation. But in your terminology. It's not in classical determistic state, but only in classical probability state. So Bhobba is really wrong about it?

You are confused about basic things.

Before going any further, in your own words, define a pure state and a mixed state. Then apply the Born rule to the mixed state and see what happens.

Once that is done we can proceed,

Thanks
Bill

 

[bhobba]

I don't think so, but I am not an expert in the field.

Its impossible. Decoherence converts a superposition to a mixed state. It harks back to entanglement where if you observe one part of an entangled system its in a mixed state.

Thanks
Bill

 

[fanieh]

You are confused about basic things.

Before going any further, in your own words, define a pure state and a mixed state. Then apply the Born rule to the mixed state and see what happens.

Once that is done we can proceed,

Thanks
Bill

You shared this example yourself, the double slit without detector screen is in pure state, then the screen caused it to be in mixed state (to be in left or right slit), then born rule applied that make the electron appear in an eigenposition. But note mixed state by itself doesn't mean there is born rule applied. I thought prior to Demystifier message that proper mixed state is classical state.. this is due to your description that "With proper mixed states everything is sweet - objective reality exists before observation - much of quantum wierdness disappears." I think you must add or correct it to "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?

 

[Simon Phoenix]

Or can one teach decoherence by totally bypassing the density matrix

Hi fanieh, let me first off apologize in advance because I'm struggling to understand where the source of your difficulty is and so my comments may not hit the mark properly.

It looks to me like your mixing up (sorry for the unavoidable pun) 'density matrix' and 'decoherence' a bit. The first thing you've got to get clear in your head is the answer to the question "why do we need density matrices in QM at all?" Forget environments and 'classicality' for the moment - just focus on QM.

When we write down a pure state description for some entity, say $| \psi \rangle$, then whatever we might interpret this state to mean (interpretation dependent) it is the most complete description available to us that describes the properties of that entity consistent with the label $\psi$. There's no other mathematical object that does any 'better'.

But how does QM cope when we are less certain about the state? Suppose we're trying to prepare 2-level atoms in the excited state $|e \rangle$, but our preparation procedure is imperfect and we get atoms in the ground state $|g \rangle$ 10% of the time - and for any given atom we just don't know without doing a measurement whether we have $|e \rangle$ or $|g \rangle$.

We can't now write down a pure state for any given atom - even though we know our preparation procedure has produced pure states every time - we simply don't know which of those pure states has been prepared. How are we to treat such an object, or set of objects, within the formalism of QM? In this situation the best we can do is to say something like "for each atom it is in the state $| e \rangle$ with probability 0.9 and in the state $|g \rangle$ with probability 0.1"

The best way to do that is to write down a density operator for each atom that looks like $$\rho = 0.1 |g \rangle \langle g | + 0.9 |e \rangle \langle e |$$ we can think of this as a way to deal with our 'ignorance' of the initial conditions. If you have this ignorance of the preparation conditions then this is the neatest way to describe the properties of your atoms.

You could also calculate expectation values for measurements assuming pure states $| e \rangle$ and then pure states $| g \rangle$ and then add them with the appropriate weightings 0.9 and 0.1 to get your final expectation value. When you're making a measurement you still need to apply the rules of QM (for example, the Born rule) to predict the results whether you use a density matrix approach or this latter approach of just weighting pure state results.

Now let's imagine a different situation involving 2-level atoms. This time I'm going to imagine Alice having a perfect preparation procedure that prepares every atom in its excited state $|e \rangle$. She's now going to fire that atom through a high-Q (lossless) cavity which has nothing in it (the field inside is in its vacuum state, which is a pure state). If we've got the atomic transition frequency matching the cavity mode frequency then there's going to be an interaction between the atom and the field. We can tailor the cavity transit time such that there's going to be $\frac 1 {2}$ probability of finding the atom in its ground state after interaction with the cavity and $\frac 1 {2}$ probability of finding the atom in its excited state after it leaves the cavity.

Suppose Alice now sends these atoms that have been through the cavity on to Bob, but she doesn't make a measurement. What is the 'state' of the atoms that get to Bob? The combined state of the atom-field system is actually given by the pure state $$| \psi \rangle = \frac 1 { \sqrt 2 } ( |e,0 \rangle + |g, 1 \rangle)$$Conceptually now we cannot say that Bob's atoms are in pure states - there's no way that we could legitimately make the statement that they're actually in the pure state $|e \rangle$ with probability $\frac 1 {2}$ and the pure state $|g \rangle$ with probability $\frac 1 {2}$

What we can say is that, if we're interested in the properties of the atoms alone, then we can treat them as if we had them in the pure state $|e \rangle$ with probability $\frac 1 {2}$ and the pure state $|g \rangle$ with probability $\frac 1 {2}$

The beauty of the density matrix approach is that we can use the same formalism for either the first situation in which we had an imperfect preparation procedure AND the second situation. Now in the second situation Bob's 'ignorance' is coming about because he doesn't have access to the full atom-field system, he only has a subset, or a component, to work with. So if you think about it enough it's clear that we're going to have to have a formalism that looks the same (as far as Bob is concerned) for both types of situation above. So physically we can see why 'proper' and 'improper' mixtures have to be equivalent when we're only interested in the properties of subsets of larger systems.

Now - what's the substantive difference here between the fields and an environment? In the second situation above we ignore or 'trace out' over the field to get a description that works for the atoms alone. But that's exactly what we do when we trace out over environmental variables in the decoherence approach. The only difference is really that the field is a very, very 'small' environment. The entanglement between the atom and field is shared only between 2 objects. An 'environment' does not consist of just one other single entity but squillions of them and so the entanglement is shared out. We could think of the entanglement as a lump of butter - in the atom-field case it's just like spreading a big chunk of butter on a tiny drop scone. In the atom-environment case it would be like spreading the same chunk over many, many slices of toast - we'd get to the point with enough toast that we didn't even know it had been buttered!

 

[fanieh]

There is a classical deterministic state and a classical probability state. In the case of a coin, one classical deterministic state is head, another classical deterministic state is tail. A classical probability state is $p(head)=1/2, \; p(tail)=1/2$. Decoherence, i.e. entanglement with the environment, explains the origin of classical probability states, but it does not explain the origin of classical deterministic states.

Maybe what you meant is the following (so as to be compatible with Bhobba statements):

Decoherence = classical probability state
Proper mixed state = classical probability state + classical deterministic state

meaning proper mixed state already has born rule applied..

prior to your message I wrote: "Then why worry how improper mixed state of the hat becomes proper mixed state as per Maximillian.. the billions of entanglement is enough to put it in proper mixed state (position that is classical)?"

What I was asking if the projections (or eigenpositions) from born rule can be imitated by surrounding the hat with all kinds of entanglement such that it would be lock in position.. I was well aware of the born rule and prior to it. And your reply was:

"You are mixing two different meanings of the word "classical state". You refer to one meaning of classical state as "head", "tail" and second meaning of classical state as the probability of 1/2. You are saying that without the born rule.. the hat won't exist in the first place even if decoherence has produced the probability or classical probability state.. please confirm if this is what you mean. Thanks.

 

[Demystifier]

Maybe what you meant is the following (so as to be compatible with Bhobba statements):

Decoherence = classical probability state
Proper mixed state = classical probability state + classical deterministic state

meaning proper mixed state already has born rule applied..

Yes, that's it. Very concise and elegant explanation.

 

[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.