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The interaction occurs not through their entanglement.

Their entanglement arises as a result of their interaction.

The quantum state of the first particle may be changed through its interaction with a second particle.

However, if a pair of particles do not interact, then there is nothing you can do to one that will change the state of the other.

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I'm thinking in terms of Feynman's path integral for the wavefunction of a particle propagating through some region. That path integral takes into account every possible path. And I'm trying to distinguish the difference between considering no other particles in the path integral and having another particle to consider in the path integral. If there is another particle to consider in the path integral for the first particle, then some of those paths (an infinite number of paths) will be near and through this second particle we now have to consider. How does the mere existence of the second particle change the calculation of the path integral for the first particle? Do the two particles become entangled in this fashion? And is there a degree as to how strongly two particles can be entangled? And can any number of particles all be entangled with each other? Thanks.

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When you calculate the evolution of a particle, the complete behavior may include self-interaction.

As an example of this, to calculate the behavior of high energy photons:

one must also consider the amplitude that the photon will convert into a particle-antiparticle pair and back again (and all their constituent trajectories)

One must also consider that that particle-antiparticle pair may exchange a virtual photon (and its consituent trajectories)

One must also consider that the photon might convert into a particle-antiparticle pair, recombine, then convert, and recombine again (and their consitiuent trajectories)

and the list of successively complex interactions grows without limit.

For quantum electrodynamics, the contribution to the total amplitude of each interaction decreases with its order of sophistication, by about 1/137, the fine structure constant. Going a few orders in, you can get very accurate calculations, but whether or not you need to depends on how accurate a result you're looking for.

Again, I'm not an expert, so I'd like to see this how this works exactly too.

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bhobba

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I cant think of any interaction that wont entangle the systems interacting - there may be some but I cant think of any. For example consider the hydrogen atom. The wave-function of the electron is expressed in terms of a coordinate system where the nucleus is the centre - move the nucleus and the wave-function of the electron goes with it. They are entangled. Strictly speaking when systems are entangled you cant speak of the state of one system without specifying the state of the other - which is pretty much the very definition of entanglement.

I am not conversant enough with the many particle path integral approach to comment on its application to entangled particles. It is however extensively used in QFT where are particles of the same type are excitations of the same underlying field.

Thanks

Bill

I am not conversant enough with the many particle path integral approach to comment on its application to entangled particles. It is however extensively used in QFT where are particles of the same type are excitations of the same underlying field.

Thanks

Bill

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Nugatory

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How is the wavefunction of one particle with mass, m, and velocity v affected by a nearby particle that has its own wavefunction with mass, M and velocity V. Can the interaction of these wavefunctions be through any other means than entanglement? Can the wavefunction of a single non-entangled free particle be compared to that of the same particle when it is entangled with another free particle? Can the wavefunction of the first particle in some sense seem space-contracted or time-stretched by the mere presence of the second particle (when compared to no entanglement )? Thanks.

There's a basic problem in the way that you're thinking about this situation. In QM, the wave function belongs to the entire system, not to the individual particles that make up the system. A system of two particles is described by a single Hamiltonian that includes contributions from the interaction between the particles, and that Hamiltonian in Schrodinger's equation gives you a single wave function. Neither particle has "its own wavefunction".

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You say, "wave function belongs to the entire system". Is this another way of saying all the particles of that "entire system" are entangled with each other? If so, the wikipedia.org article defines entanglement in terms of whether individual wave functions are added in a separable or inseparable way, if I'm reading the article correctly. If these individual wave functions are combined in an inseparable way, then they are entangled. But still, there is reference to wave functions of subsystems in that definition. It would seem from this, again if I'm understand it correctly, that these "individual" wave functions in themselves are not affected by entanglement; it is only when they are combined (in an inseparable way) that there is entanglement (which describes the larger wavefunction of the larger system). Feel free to correct any misconceptions I might have. Thank you.There's a basic problem in the way that you're thinking about this situation. In QM, the wave function belongs to the entire system, not to the individual particles that make up the system. A system of two particles is described by a single Hamiltonian that includes contributions from the interaction between the particles, and that Hamiltonian in Schrodinger's equation gives you a single wave function. Neither particle has "its own wavefunction".

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bhobba

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But still, there is reference to wave functions of subsystems in that definition.

Here is the correct definition.

Suppose system A can be in two states |a> and |b>. Suppose system B can be in the same two states |a> and |b>. If system A is in state |a> and system B in state |b> this is written as |a>|b>. Similarly if system A is in state |b> and system B in state |a> this is written as |b>|a>. But one of the basic principles of QM is the principle of superposition so that a possible state of the combined system, system A and system B, is 1/root 2 |a>|b> + 1/root 2 |b>|a>. Such a state is called entangled. You cant speak of either system being in a state because its entangled with the other.

In fact this is a purely quantum phenomena and is what separates QM from classical probability theory:

http://arxiv.org/abs/0911.0695

Thanks

Bill

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Here is the correct definition.

Suppose system A can be in two states |a> and |b>. Suppose system B can be in the same two states |a> and |b>. If system A is in state |a> and system B in state |b> this is written as |a>|b>. Similarly if system A is in state |b> and system B in state |a> this is written as |b>|a>. But one of the basic principles of QM is the principle of superposition so that a possible state of the combined system, system A and system B, is 1/root 2 |a>|b> + 1/root 2 |b>|a>. Such a state is called entangled. You cant speak of either system being in a state because its entangled with the other.

I'm sorry, but you just spoke of system A as being different from system B. And then you spoke of the entanglement of them as an entirely different "combined system". If you define entanglement in terms of seemingly independent subsystems, then perhaps my question is answered by saying these subsystems are unaffected by entanglement. Is entanglement just a consideration of the two subsystems taken as a whole for the purposes of making measurements on the whole system? In other words, you can't know if the particle you are measuring happens to be entangled or not with another particle unless you also measure the other particle, right? All you know is that the particle you are measuring has it own seemingly independent wave function that collapses on measurement. It might be correlated to another particle that it is entangled with. But you can't know that unless you make another measurement on the second particle and compare results to see that they were correlated. Otherwise, how is the wave function for the entangled pair even at all useful if TWO measurements are not made, one on each of the entangled particles?

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bhobba

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Is entanglement just a consideration of the two subsystems taken as a whole for the purposes of making measurements on the whole system?

Entanglement is as I defined it. The reason I explained it mathematically is its not amenable to an English type explanation.

In other words, you can't know if the particle you are measuring happens to be entangled or not with another particle unless you also measure the other particle, right?

Why would you say that? In fact decoherence is a form of entanglement and it, in a fundamental way, alters measurements done on the system - instead of being in a superposition it now in a mixed state:

http://www.ipod.org.uk/reality/reality_decoherence.asp [Broken]

Thanks

Bill

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Why would you say that? In fact decoherence is a form of entanglement and it, in a fundamental way, alters measurements done on the system - instead of being in a superposition it now in a mixed state:

http://www.ipod.org.uk/reality/reality_decoherence.asp [Broken]

Well, practically speaking, just as you can't confirm that particles identically prepared will result in measurements according to a simple wave function unless you measure that ensemble, you can't confirm an entanglement between two particles unless you make two measurements on that ensemble. Or can you show me one measurement on a single particle that is affected by the fact that it happens to be entangled? Can we ever know from one measuring device that the particles it measures are entangled? In other words, you have particles moving along and then being captured by your one and only measuring device. Each said particle comes in with about the same energy and momentum. And your one measuring device measures some effect with some probability distribution. How would the probabilities change (and thus the wave function change) at your one measuring device if the ensemble were the result of the propagation of an entangled pair? Or for that matter, how could we ever know with only one measuring device at our end that the entangled pairs have already been measured in a far off galaxy?

So what I think is going on is that the wave function calculated for an entangled pair is used to measure the probability of both measurements in conjunction with each other - what's the probability of one side being "up" and the other side being "down"...

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bhobba

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In other words, you have particles moving along and then being captured by your one and only measuring device.

Actually what's going on is it becomes entangled with the measuring device because that's what measurement is.

So what I think is going on is that the wave function calculated for an entangled pair is used to measure the probability of both measurements in conjunction with each other - what's the probability of one side being "up" and the other side being "down"...

'the wave function calculated for an entangled pair is used to measure'. That doesn't make any sense. Wave functions are not used to measure - they are the expansion of the state in terms of position and are used to help calculate the probability of position if you were to measure it.

By up and down I presume you are referring to an EPR type entanglement which I will take, for definiteness, as spin up and spin down. You measure the spin of one particle it becomes entangled with the measurement apparatus and breaks the entanglement with the other particle. No need to do a measurement on the other particle - although of course you can. Only one measurement is required.

Thanks

Bill

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bhobba

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You can treat improper mixed states as proper ones - in fact that's the assumption of the ignorance ensemble interpretation.

But it isn't really - which is the problem of outcomes - in other words why do we get any outcomes at all. It not an issue axiomatically because observation is the primitive of QM so from it's very foundations you must get an outcome - but the question is - why?

That said viewed in that light its like a point particle in classical physics, events in relativity etc - all theories have primitives not explained by the theory - as the theory is developed it's meaning is elucidated but cant actually be explained by the theory. One must go outside the theory to do that - and we have a number of interpretations that do just that - but experimentally distinguishing them is the issue.

I have said it before, and I will say it again, the usual stuff you read about problems in QM is not IMHO its central problem. Its central problem is it doesn't matter what issue worries you there is an interpretation that fixes it - but not all at once - and, most importantly of all, not in a way that's experimentally testable.

Thanks

Bill

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You can treat improper mixed states as proper ones - in fact that's the assumption of the ignorance ensemble interpretation.

But it isn't really - which is the problem of outcomes - in other words why do we get any outcomes at all. It not an issue axiomatically because observation the primitive of QM so from it's very foundations you must get an outcome - but the question is - why?

That said viewed in that light its like a point particle in classical physics, events in relativity etc - all theories have primitives not explained by the theory - as the theory is developed it's meaning is elucidated but cant actually be explained by the theory. One must go outside the theory to do that - and we have a number of interpretations that do just that - but experimentally distinguishing them is the issue.

I have said it before, and I will say it again, the usual stuff you read about problems in QM is not IMHO its central problem. Its central problem is it doesn't matter what issue worries you there is an interpretation that fixes it - but not all at once - and, most importantly of all, not in a way that's experimentally testable.

Thanks

Bill

You mentioned proper mixed state are there prior to observation.. but improper states are not there prior to observation. But if you measure the environment and system, the improper would become proper.. and since it becomes proper it is there prior to observation. So this means improper mixed state are there prior to observation. Did you get the logic (and the contradiction)? Your knowledge of it (preparing the state of the entire environment-system) shouldn't affect whether it is there or not prior to observation.

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bhobba

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You mentioned proper mixed state are there prior to observation.. but improper states are not there prior to observation. But if you measure the environment and system, the improper would become proper..

Measure the environment and system? The environment is usually way beyond actual measurement being things like the atmosphere, the CMBR etc. And how it would resolve the issue beats me.

We don't know if improper mixed states that result from decoherence are pure states there prior to observation - there is no way to tell. If interpretations like BM and collapse theories like GRW are correct then its a proper mixed state, but until someone can experimentally tell the difference its pure speculation.

Thanks

Bill

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Measure the environment and system? The environment is usually way beyond actual measurement being things like the atmosphere, the CMBR etc. And how it would resolve the issue beats me.

We don't know if improper mixed states that result from decoherence are pure states there prior to observation - there is no way to tell. If interpretations like BM and collapse theories like GRW are correct then its a proper mixed state, but until someone can experimentally tell the difference its pure speculation.

Thanks

Bill

Let's ignore for now decoherence to avoid confusing the issue and just focus on pure entanglement like what "friend" was describing. Are there no entanglement where it is proper mixed states when measured? In the case of nucleus and electron.. can't it be not complete proper mixed states in any measurement setting?

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bhobba

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Let's ignore for now decoherence to avoid confusing the issue and just focus on pure entanglement like what "friend" was describing. Are there no entanglement where it is proper mixed states when measured? In the case of nucleus and electron.. can't it be not complete proper mixed states in any measurement setting?

You get a mixed state whenever you have an entangled system and you observe just one part of the entangled system - for the detail see chapter 7 of Susskind's book:

https://www.amazon.com/dp/0465036678/?tag=pfamazon01-20

That's one way to form a mixed state - such are called improper.

Another way is to take a number of systems prepared in different quantum states and randomly present them for observation. Such mixed states are called proper.

Mathematically and observationally they are exactly the same. Without knowing how they were prepared you can't tell the difference.

If it was a proper mixed state, and the states are eigenvectors of what you are observing it with, QM says everything is common-sense. The observation reveals the state and, if its not an observation that destroys what its observing, it doesn't change the state. If that was the case the measurement problem is solved. You know why you get an outcome and you have not altered anything by observation.

But improper mixed states are not prepared that way, there is no way to tell the difference, but its not the same thing. That is the critical and crucial issue of the measurement problem in QM. I resolve it simply. I, as an interpretive assumption, assume improper mixed states are proper - everything is sweet. There are a number of ways that could be such as if BM or GRW is true - I form no hypothesis. That's why the word ignorance is used. Its basically simply a slight variation of the ensemble interpretation applied to mixed states. That's why the word ensemble is used hence ignorance ensemble.

Thanks

Bill

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You get a mixed state whenever you have an entangled system and you observe just one part of the entangled system - for the detail see chapter 7 of Susskind's book:

https://www.amazon.com/dp/0465036678/?tag=pfamazon01-20

When you said "observe" do you also mean measure? What if the entangled system are not being measured, and you just do non-demolition observation (or just visualizing it) of a subsystem or a part of the entangled pair, would it turn to mixed state?

That's one way to form a mixed state - such are called improper.

Another way is to take a number of systems prepared in different quantum states and randomly present them for observation. Such mixed states are called proper.

Mathematically and observationally they are exactly the same. Without knowing how they were prepared you can't tell the difference.

If it was a proper mixed state, and the states are eigenvectors of what you are observing it with, QM says everything is common-sense. The observation reveals the state and, if its not an observation that destroys what its observing, it doesn't change the state. If that was the case the measurement problem is solved. You know why you get an outcome and you have not altered anything by observation.

But improper mixed states are not prepared that way, there is no way to tell the difference, but its not the same thing. That is the critical and crucial issue of the measurement problem in QM. I resolve it simply. I, as an interpretive assumption, assume improper mixed states are proper - everything is sweet. There are a number of ways that could be such as if BM or GRW is true - I form no hypothesis. That's why the word ignorance is used. Its basically simply a slight variation of the ensemble interpretation applied to mixed states. That's why the word ensemble is used hence ignorance ensemble.

Thanks

Bill

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You get a mixed state whenever you have an entangled system and you observe just one part of the entangled system - for the detail see chapter 7 of Susskind's book:

https://www.amazon.com/dp/0465036678/?tag=pfamazon01-20

That's one way to form a mixed state - such are called improper.

Another way is to take a number of systems prepared in different quantum states and randomly present them for observation. Such mixed states are called proper.

Mathematically and observationally they are exactly the same. Without knowing how they were prepared you can't tell the difference.

If it was a proper mixed state, and the states are eigenvectors of what you are observing it with, QM says everything is common-sense. The observation reveals the state and, if its not an observation that destroys what its observing, it doesn't change the state. If that was the case the measurement problem is solved. You know why you get an outcome and you have not altered anything by observation.

Bill. Can you give an example where an observation would destroy what its observing which change the state that you mentioned above? thanks.

Were you talking about changing a mixed state back to pure state by mere observation? Is this a separate case than observation that destroy what its observing? Please share examples of each case so I can differentiate them. Thanks.

But improper mixed states are not prepared that way, there is no way to tell the difference, but its not the same thing. That is the critical and crucial issue of the measurement problem in QM. I resolve it simply. I, as an interpretive assumption, assume improper mixed states are proper - everything is sweet. There are a number of ways that could be such as if BM or GRW is true - I form no hypothesis. That's why the word ignorance is used. Its basically simply a slight variation of the ensemble interpretation applied to mixed states. That's why the word ensemble is used hence ignorance ensemble.

Thanks

Bill

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- #21

bhobba

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Bill. Can you give an example where an observation would destroy what its observing which change the state that you mentioned above?

Photons impinging on the screen at the back of the double slit.

Only some observations, called filtering type observations, do not destroy the system being observed. They are in fact state preparation procedures. That would be for example electrons going through a single slit - its a position observation and just behind the slit it has a specific position.

Thanks

Bill

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- #22

bhobba

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Were you talking about changing a mixed state back to pure state by mere observation?

No.

But the delayed choice experiment has an interesting take on that

https://www.physicsforums.com/threa...and-the-delayed-choice-quantum-eraser.623648/

'Decoherence is irreversible only when caused by a LARGE number of degrees of freedom. A quantum eraser involves a small number of degrees of freedom, which is why it is reversible.'

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Bill

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No.

But the delayed choice experiment has an interesting take on that

https://www.physicsforums.com/threa...and-the-delayed-choice-quantum-eraser.623648/

'Decoherence is irreversible only when caused by a LARGE number of degrees of freedom. A quantum eraser involves a small number of degrees of freedom, which is why it is reversible.'

Thanks

Bill

Bill. In the search for answers to the measurement problem, why is spacetime ignored? what theorem makes spacetime unnecessary. When you have entangled pair or wavefunction miles away, doesn't spacetime matter esp as the particle has to coupled to spacetime?

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bhobba

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Bill. In the search for answers to the measurement problem, why is spacetime ignored? what theorem makes spacetime unnecessary. When you have entangled pair or wavefunction miles away, doesn't spacetime matter esp as the particle has to coupled to spacetime?

You mean a theory using the space-time formalism of relativity?

We have that - its called QFT.

Heuristically normal QM has a problem with relativity - position is treated as an observable, but time is a parameter. But relativity says they should be treated on the same footing. So what QFT does is treat both time and position as a parameter ie you have a field.

In fact, and you can find the details in chapter 3 of Ballentine, usual QM is based on time being an absolute. That's fixed in QFT.

It still has exactly the same measurement problem issues as standard QM.

But some things are heuristically easier to understand in QFT eg EPR. Since the particles are excitations of exactly the same field that observing one is correlated to another intuitively seems easier to swallow.

Thanks

Bill

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You mean a theory using the space-time formalism of relativity?

We have that - its called QFT.

Heuristically normal QM has a problem with relativity - position is treated as an observable, but time is a parameter. But relativity says they should be treated on the same footing. So what QFT does is treat both time and position as a parameter ie you have a filed.

In fact, and you can find the details in chapter 3 of Ballentine, usual QM is in fact based on time being an absolute. That's fixed in QFT.

It still has exactly the same measurement problem issues as standard QM.

Thanks

Bill

I meant general relativity... or how quantum particles are coupled or connected to spacetime. Is this not a necessity to know to understand the problem better?

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