Double slit experiment and Interaction

[bhobba]

Yet, does anything change if the quantum state of the "few stray photons" are also unresolved? Again, if what we are considering is the quantum state of the cosmos as a whole, and no experimental controls are in place, does that change anything. Is every subatomic particle necessarily an apparatus by which a measurement is made?

Scratching head here. Why exactly do you think the state of the photons will not be changed by the interaction?

And not every object in the universe is not an observational apparatus - only those capable of causing decoherence eg a single photon is not enough to decohere a dust particle - you evidently need a few.

.Thanks
Bill

 

[bhobba]

In Qbism, the interpretation of the state is subjective, right.. then what changes in the macro world?

The outcome of the observation.

Thanks
Bill

 

[zonde]

Things with 0 probability of happening don't make the universe branch out so it does happen, only then to realize it shouldn't and cease to exist.
No branching occurs to make a universe where the particle goes along that 0 probability path. (I would have thought?)

Let me use analogy.

Say you are standing on a street corner and you want to get to the diagonally opposite corner of the block. You can go around the block by going left (L path) or right (R path). And as you go by either path you have to count front doors of the buildings that you pass. So if you choose L path you will count X number of doors but if you choose R path you will count Y number of doors.
But the trick is that I want you to tell me at the end if the difference between X and Y is odd or even number (that will decide which one of the further paths will be 0 probability path and which one - 1 probability path).

Can you do that if you are allowed to make two copies of the world?
Say two worlds - L and R. In L world you take L path but at the end you know X but don't know Y. And vice versa for R world. And if you do not allow worlds to communicate some time after splitting you can't tell me the quantity I am asking, right?

 

[kith]

Newer understood this about MWI. Take Mach–Zehnder interferometer, photons are detected only at one output in case of interference. If you say that all outcomes are realized then you contradict experiment.

No. If the probability to detect the photons at one detector is 100%, then there is only a single possible outcome and therefore only a single world.

The MWI doesn't simply assign worlds to all terms in a superposition. This wouldn't make sense because whether a state is a superposition or not depends on the basis and every state is a superposition of eigenstates of some observable.

In the most basic version, the splitting into different worlds occurs whenever you have different possible outcomes in a measurement. More sophisticated, the splitting occurs whenever a superposition is turned into the corresponding mixed state by decoherence.

 

[Feeble Wonk]

Scratching head here. Why exactly do you think the state of the photons will not be changed by the interaction.

And, again, how does the environment know if the photon(s) interacted with the dust particle? My question was making the assumption that the dust particle is not being observed in anyway other than through its "potential" random environmental interaction, with no system preparation and no experimental controls. Furthermore, the dust particle and the photon(s) are only being considered as part of the entire cosmological quantum state. So, would the quantum state of both the photons and the dust particle be less "determined" in such a way as to delay the environmentally induced collapse secondary to decoherence?
I

 

[Feeble Wonk]

Apologies... I botched that lady posting. I'll try again.

And not every object in the universe is not an observational apparatus - only those capable of causing decoherence eg a single photon is not enough to decohere a dust particle - you evidently need a few.

.Thanks
Bill

I'm sorry to be dense about this, but that's what I'm trying to understand. Which subatomic particles perform as an observational apparatus, and why sometimes rather than others. Is it totally dependent on the preparation state of the system being observed? And if the system is not "prepared", and there are no experimental controls in place, such that both system and "environment" are maximally "open", does this change things? It seems that under certain circumstances quantum superpositions can be maintained for a functional period of time, even in warm and "noisy" environments. At least, the demonstration of quantum effects in photosynthesis would appear to suggest that.
http://m.phys.org/news/2012-01-role-quantum-effects-photosynthesis.html

 

[Feeble Wonk]

Scratching head here. Why exactly do you think the state of the photons will not be changed by the interaction?

But, how does the environment know if the photon(s) interacted with the dust particle? My question was making the assumption that the dust particle is not being observed in anyway other than through its "potential" random environmental interaction, with no system preparation and no experimental controls. Furthermore, the dust particle and the photon(s) are only being considered as part of the entire cosmological quantum state. So, would the quantum state of both the photons and the dust particle be less "determined" in such a way as to delay the environmentally induced collapse secondary to decoherence?

 

[bhobba]

Which subatomic particles perform as an observational apparatus, and why sometimes rather than others.

It depends on the situation - specifically if the object or objects interact with what is being observed and the degree of interaction. You must analyse each situation in detail to determine if the decoherence (ie interaction) is enough to give a definite outcome in some observable. Of course its a bit arbitrary what is counted as a definite outcome eg what precision we count the position to be known before we say it has that position. It turns out to nearly always be the position observable and when you chug through the math the reason is virtually all interactions are of the radial type.

Thanks
Bill

 

[bhobba]

But, how does the environment know if the photon(s) interacted with the dust particle?

The environment has nothing to do with it in that situation ie the few stray photons is the environment.

Thanks
Bill

 

[zonde]

No. If the probability to detect the photons at one detector is 100%, then there is only a single possible outcome and therefore only a single world.

The MWI doesn't simply assign worlds to all terms in a superposition. This wouldn't make sense because whether a state is a superposition or not depends on the basis and every state is a superposition of eigenstates of some observable.

In the most basic version, the splitting into different worlds occurs whenever you have different possible outcomes in a measurement. More sophisticated, the splitting occurs whenever a superposition is turned into the corresponding mixed state by decoherence.

Yes, with a bit of help from Wikipedia I understand what I missed about MWI - world splits only at measurement and not before.

And it makes MWI even more bizarre. Quantum system in MWI actually is a wave. So it is non-local.
And then the split. It is non-local - how else if the quantum system is non local?
And it creates only appearance of particle. So there is no such a thing as particle in MWI, right?

 

[bhobba]

So it is non-local.

That doesn't follow nor is the quantum system in MW actually a wave. Its exactly the same as in the formalism - the only difference is the interpretation of decoherence.

Thanks
Bill

 

[kith]

And it makes MWI even more bizarre. Quantum system in MWI actually is a wave.

The quantum system in the MWI is the whole universe and the universal wavefunction is the state of the universe.

 

[Edward Wij]

I think this conciousness stuff is a very very big crock of the proverbial, but it must be said a fully coherent interpretation can be built from it - just a very very weird one. We even have people that believe in solipsism despite the fact it leads to just as weird a view. Most people, correctly IMHO, reject such - but it can't be proven incorrect.

Thanks
Bill

Bill. Do you have any references or papers detailing or summarizing this consciousness interpretation where you said "a fully coherent interpretation can be built from it - just a very very weird one"? Can you give an example how weird? In physics weirdness must not be taken as arguments not to explore clues.. in 16th century, people can be burnt at stake or feed to the dogs for believing in special relativity or superstrings for example.

 

[bhobba]

Bill. Do you have any references or papers detailing or summarizing this consciousness interpretation where you said "a fully coherent interpretation can be built from it - just a very very weird one"?

Don't know of any. But the kind of things it would need to do is pretty obvious eg do a double slit experiment, record the output to computer memory, make a million copies and send them all except one to different distant parts of the universe. Then a century later view the one you kept. At that time the observation becomes real and all those copies contain real data by some strange unspecified means.

Thanks
Bill

 

[zonde]

That doesn't follow nor is the quantum system in MW actually a wave. Its exactly the same as in the formalism - the only difference is the interpretation of decoherence.

As I see it wave function collapse is a bridge between wave function and particles. If you take away this bridge and try to explain everything using only wave functions then it changes a lot of things. Say quantum interference certainly relays on the idea that single wave function can be spatially discontinuous (can be at two places simultaneously) while particles are localized.

 

[bhobba]

As I see it wave function collapse is a bridge between wave function and particles.

The point is that wave function collapse is not part of QM - only interpretations. Quite a few interpretations don't have it.

Thanks
Bill

 

[craigi]

Well, let's say we want to try Bohmian Mechanics or Many Worlds. There is still no Bohmian Mechanics version of the standard model of particle physics, and it is unclear whether Many-Worlds really works, as even proponents like Deustch http://arxiv.org/abs/0712.0149 and Carroll http://www.preposterousuniverse.com...ion-of-quantum-mechanics-is-probably-correct/ agree. So I don't think it is true that we have at least one interpretation that solves the definite outcomes problem. That leaves us with Copenhagen which does have the problem, and the only way to solve it is to assert it is not a problem, which I think the consciousness ones do quite nicely :p

Also, Bohmian Mechanics is in principle testable, just as string theory is. Many-Worlds, if in fact the theory of everything, will not be falsified. But if it isn't, then it can be falsified. So the problem with BM or MWI for solving the definite outcomes problem is not experimentally deciding between them.

This is a good summary. The MWI is the simplest way to interpret QM, but as Einstein asserted, QM is incomplete. Consciousness can be used to complete other interpretations, but leaves with a subjective reality based upon a poorly defined concept, which is too difficult to reconcile

 

[Feeble Wonk]

It depends on the situation - specifically if the object or objects interact with what is being observed and the degree of interaction. You must analyse each situation in detail to determine if the decoherence (ie interaction) is enough to give a definite outcome in some observable.

Sorry for my ignorance, but it seems that I didn't express my question clearly enough. I'd like to return to the question of how decoherence locally determines the subsystem of "collapse" within the universal quantum state.
Again, I'm trying to conceptually avoid the "physics in a box" controls of a system-environment-apparatus relationship. The question is thus assuming that we are discussing the quantum state vector of the entire universe, such that the environment, system and even the "apparatus" are all part of the same state vector. So when we are considering the "potential" states of a few random photons emitted during decoupling more than 13 billion years ago, it is difficult to fathom that our handful of photons did not have the potential paths that would have interacted with potential dust particles at some previous point in time. Why does the decoherence determine that it occurs "now" rather than then.
I suppose a related question would be pertaining to the number of the potential photon paths required to produce the "collapsed" locality of the dust particle. If, as you say, one photon is not sufficient and it would require a "few" random photons from the CMBR, do all of these photons have to interact with the dust particle at a particular instantaneous moment, or is it a cumulative effect of sequential interactions over time?

 

[craigi]

Sorry for my ignorance, but it seems that I didn't express my question clearly enough. I'd like to return to the question of how decoherence locally determines the subsystem of "collapse" within the universal quantum state.
Again, I'm trying to conceptually avoid the "physics in a box" controls of a system-environment-apparatus relationship. The question is thus assuming that we are discussing the quantum state vector of the entire universe, such that the environment, system and even the "apparatus" are all part of the same state vector. So when we are considering the "potential" states of a few random photons emitted during decoupling more than 13 billion years ago, it is difficult to fathom that our handful of photons did not have the potential paths that would have interacted with potential dust particles at some previous point in time. Why does the decoherence determine that it occurs "now" rather than then.

The truth is, that radiation just didn't interact in any significant way in those 13 billion years. Space is very empty and enough radiation survives to be detected today.

I suppose a related question would be pertaining to the number of the potential photon paths required to produce the "collapsed" locality of the dust particle. If, as you say, one photon is not sufficient and it would require a "few" random photons from the CMBR, do all of these photons have to interact with the dust particle at a particular instantaneous moment, or is it a cumulative effect of sequential interactions over time?

Decoherence occurs at thermodynamically irreverisble events. An example of this would be photon absorption. Examples of reversible events are, the specular reflection of light at a mirror, and the refraction of light through lenses or lensing by gravitational fields.

What makes some interactions reversible and some irreversible? We don't need QM for that. It's just back to statistical mechanics.

 

[Feeble Wonk]

The truth is, that radiation just didn't interact in any significant way in those 13 billion years. Space is very empty and enough radiation survives to be detected today.

I guess I'm still not making the point that confuses me clear enough. Sadly, the embarrassing degree of my mathematical ineptitude makes it impossible for me to utilize the formalized calculations as a means of understanding this issue. That forces me to to try to wrap my head around the general idea in a purely conceptual manner, and the quantum state constraints imposed by decoherence is an idea that persistently eludes me. While I fully expect that this failure is largely the result of my utter ignorance, I'll try again...
My understanding is that the quantum wave (state vector?) of a single photon can be demonstrated to "spread out" sufficiently to pass through both slits of the dual slit experiment in the very limited time and distance involved as it travels from the emitting device through the screen. If I extrapolate that concept to the time and distance involved with a remnant photon of CMB radiation produced by a decoupling event more than 13 billion years ago, I would expect the potential path (and/or state vector?) to become increasingly "spread out".
In the dual slit experiment, by preparing the appropriate controls and measurement devices, you can reduce the quantum state of the photon to the path having passed through one or the other slit. This is generally understandable.
However, in a completely uncontrolled environment, when the quantum system being "observed" is the state vector of the entire universe... when the system "is" the environment... and the apparatus is simply part of that system... how does the process of decoherence determine where the Von Neumann cut is placed?

 

[Feeble Wonk]

I suppose I should clarify my previous question even more. I would think that the path of the photon(s) of CMB radiation would sufficiently uncertain that the entirety of the wave function would include paths that might or might not intersect with the potential location of the dust particle in question. If no specific observation is made, other than that by the photon itself, why does it need to decide if it interacted or not?

 

[StevieTNZ]

I suppose I should clarify my previous question even more. I would think that the path of the photon(s) of CMB radiation would sufficiently uncertain that the entirety of the wave function would include paths that might or might not intersect with the potential location of the dust particle in question. If no specific observation is made, other than that by the photon itself, why does it need to decide if it interacted or not?

Decoherence does not solve the measurement problem. It is merely that the photon is entangled with the dust particle, due to interacting by the wave function including paths that intersect with the potential location (yes, you are correct to say the dust particle is in a potential location) of the dust particle.

 

[bhobba]

I suppose I should clarify my previous question even more. I would think that the path of the photon(s) of CMB radiation would sufficiently uncertain that the entirety of the wave function would include paths that might or might not intersect with the potential location of the dust particle in question.

You are trying to understand something highly technical without math,

Its got nothing to do with wavefunctions spreading out etc etc.

What its got to with is if you write down the system of photons and dust particle, indeed any interacting systems, and you observe one part of the entangled system, it mathematically is exactly the same as a proper mixed state - because its only mathematically the same and not prepared the same way its called an improper mixed state. A proper mixed state consists of a number of states randomly presented for observation. That being the case there is no measurement problem, what you are observing is there before observation and everything is sweet. The reason it doesn't solve the measurement problem is its only mathematically the same - it wasn't prepared the same way. However observationally there is no difference.

Here is the detail:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3. The math is likely beyond you but unfortunately this is an area that can't be explained in English - you have to use the math.

Just as an example of where you are going wrong is that decoherence is a phenomena of entanglement - you can't consider the photons and particle having separate wave-functions until they stop interacting.

Unless you go and learn the math, specifically Linear Algebra and the Dirac notation, you must accept the word of those that have learned it.

Thanks
Bill

 

[Feeble Wonk]

Thanks Bill. You've actually given me that resource in the past. What I could decipher was interesting. But, yes, the math made my eyes roll back in my head a little bit.
Let me ask this then. Is there significant debate within the physics academia regarding how severely and how quickly the process of decoherence limits quantum states, or is it essentially accepted as fact across the board?

 

[bhobba]

Let me ask this then. Is there significant debate within the physics academia regarding how severely and how quickly the process of decoherence limits quantum states, or is it essentially accepted as fact across the board?

What do you mean by limits quantum states?

It generally happens so fast you can't observe it. The off-diagonal terms are way below detectability very quickly;y. It is an issue, as a matter of principle, that the models show while those off-diagonal elements go to zero very very quickly they never actually get to zero. FAPP they are zero - but the models do not generally have them as zero. Sometimes you find debate on that point. But the consensus is decoherence does explain apparent collapse - but not actual collapse - which is an entirely different matter. For that you need further interpretive assumptions such as BM or GRW.

Thanks
Bill

 

[rootone]

Please explain 'off diagonal', i don't know what that is.

 

[bhobba]

Please explain 'off diagonal', i don't know what that is.

A mixed state after decoherence is of the form ∑pi |bi><bi| where the |bi> are eigenvectors of what you are observing. If you write it as a matrix then it is a diagonal matrix. However the decoherence models actually have non zero off diagonal terms in the matrix that very quickly go to way below detectability and continue getting smaller - but never are exactly zero. Its like exponential decay - it very quickly goes to zero but never quite reaches it.

Unfortunately explaining this stuff without math is impossible - at least I can't do it.

Thanks
Bill

 

[rootone]

I'll see if I can get my head around that math then.

 

[craigi]

I'll see if I can get my head around that math then.

There are prerequistes to understanding the math. If you explain how far you've studied already we can certainly offer advice. Typically, QM is taught to physics undergrads. A requirement being a strong aptitude for mathematics and a complete understanding of high school mathematics.

The term "off-diagonal" pertains to matrices. Since it's unfamiliar to you, it's likely that there's a lot of groundwork that you should do first. In the meantime, I see no reason why you shouldn't pursue QM from a conceptual perspective.

 

[rootone]

I know what a matrix is, (not the movie version), I might pm you for more detailed advice.