Preparing Quantum Systems: Understanding State Preparation and Its Importance

[lucas_]

In quantum systems, do you always need to prepare the state (and what does this mean) to make it valid quantum system? But the electrons in the atoms.. no one needs to prepare the state yet they exist as atoms and molecules...

 

[atyy]

I think it is a tricky question. Let me write something to start the discussion.

The quantum state is heuristically some mixture of the "objective properties of the system and our knowledge of it", but we don't know what that really means, which is why there is discussion till this day in what senses the quantum state can be considered real or epistemic or both.

Since the theory only predicts probabilities of measurement outcomes, which can most simply be interpreted as relative frequencies, we need an ensemble of systems that are "independently and identically drawn" from some probability distribution on which we make measurements. The preparation of this ensemble is described by the quantum state.

However, quantum theory also recognizes that although we don't know whether the state is real, one certainly makes no mistake if one assigns a pure state as the complete description of a single system. It seems that one can make some tests of a single system by making repeated measurements and seeing if the system is consistent with a stochastic equation from the theory of repeated or continuous quantum measurements. However, in the quantum formalism, measurement and the selection of a system conditioned on the observation of a result is a means of state preparation because of the state reduction postulate. So it seems that even here, one must prepare a system by measurement in order to test the predictions of quantum theory.

 

[Quantum Defect]

In quantum systems, do you always need to prepare the state (and what does this mean) to make it valid quantum system? But the electrons in the atoms.. no one needs to prepare the state yet they exist as atoms and molecules...

Usually, in most experiments you will prepare a special state of the quantum system to study. In natural systems of atoms or molecules, you will usually describe the system with something called a density matrix. I believe that Feynman volume iii talks about this. Definitely in non-introductory qm books you will see a discussion of the density matrix.

E.g. A bunch of H atoms in isotropic space will have electron spins quantized in any old direction. Shooting a beam of these atoms through a strong inhomogeneous magnetic field will yield two sub-beams (spin up and spin down). Even though the original atoms were not in pure up/down states, they look like they were. With the original sample, you will see the same result as you rotate the magnetic field (quantization) direction. We say that the density matrix is diagonal for this isotropic system, whichever quantization axis you choose.

See e.g. http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/DensityMatrix.htm

 

[lucas_]

I think it is a tricky question. Let me write something to start the discussion.

The quantum state is heuristically some mixture of the "objective properties of the system and our knowledge of it", but we don't know what that really means, which is why there is discussion till this day in what senses the quantum state can be considered real or epistemic or both.

Since the theory only predicts probabilities of measurement outcomes, which can most simply be interpreted as relative frequencies, we need an ensemble of systems that are "independently and identically drawn" from some probability distribution on which we make measurements. The preparation of this ensemble is described by the quantum state.

However, quantum theory also recognizes that although we don't know whether the state is real, one certainly makes no mistake if one assigns a pure state as the complete description of a single system. It seems that one can make some tests of a single system by making repeated measurements and seeing if the system is consistent with a stochastic equation from the theory of repeated or continuous quantum measurements. However, in the quantum formalism, measurement and the selection of a system conditioned on the observation of a result is a means of state preparation because of the state reduction postulate. So it seems that even here, one must prepare a system by measurement in order to test the predictions of quantum theory.

Bottom line is that if the quantum state was real.. or Many Worlds, we no longer need to "prepare the state" because it is already prepared by default because it's real? Is this what you are implying?

 

[atyy]

Bottom line is that if the quantum state was real.. or Many Worlds, we no longer need to "prepare the state" because it is already prepared by default because it's real? Is this what you are implying?

No, that is not what I was saying. It is true that if we use an interpretation like Bohmian Mechanics or Many Worlds, then we get a clear(er?) answer to your question. Maybe it is still not entirely clear in Bohmian Mechanics, since in a sense a system has two quantum states - the wave function and the distribution of initial conditions, and it can be argued that the latter is epistemic in some sense.

Rather I was trying to answer the question from within an orthodox Copenhagen-style interpretation, where on the one hand the quantum state is not necessarily real since we have a Heisenberg cut dividing the universe into classical and quantum realms, with only the classical realm being granted the status of real reality. On the other hand, the Copenhagen interpretation does acknowledge that it is possible to assign quantum states to single systems, so that a pure state is the complete description of a single system for all practical purposes. We need to use this if we believe there is only one reality (no many-worlds), and that quantum theory does describe the universe we see. In this case, it seems that we verify quantum theory by making repeated observations on a single system. So is there a preparation here? No in the sense that we did not prepare the universe, but yes in the sense that by making repeated observations we repeatedly collapse the universe, and collapse is the traditional expression of preparing a system by measurement.

 

[lucas_]

No, that is not what I was saying. It is true that if we use an interpretation like Bohmian Mechanics or Many Worlds, then we get a clear(er?) answer to your question. Maybe it is still not entirely clear in Bohmian Mechanics, since in a sense a system has two quantum states - the wave function and the distribution of initial conditions, and it can be argued that the latter is epistemic in some sense.

Rather I was trying to answer the question from within an orthodox Copenhagen-style interpretation, where on the one hand the quantum state is not necessarily real since we have a Heisenberg cut dividing the universe into classical and quantum realms, with only the classical realm being granted the status of real reality. On the other hand, the Copenhagen interpretation does acknowledge that it is possible to assign quantum states to single systems, so that a pure state is the complete description of a single system for all practical purposes. We need to use this if we believe there is only one reality (no many-worlds), and that quantum theory does describe the universe we see. In this case, it seems that we verify quantum theory by making repeated observations on a single system. So is there a preparation here? No in the sense that we did not prepare the universe, but yes in the sense that by making repeated observations we repeatedly collapse the universe, and collapse is the traditional expression of preparing a system by measurement.

Let's take the case of molecular orbital.. or an example of how an electron can be anywhere in the crystal and be any energy. Here the quantum state of the electron is the possibilities of all positions and all energies. Correct? In Many worlds Int, all the possible positions and energy of a single electron in a crystal has each own world? This sounds ridiculous... don't you think so..

And preparing the electron quantum state means repeating the experiments a million times to check the position and energy of a single electron running loose in a crystal? But how do you tell which electron is that since there is over a trillion electrons too? How do you prepare the state of a single electron in this case?

 

[atyy]

Let's take the case of molecular orbital.. or an example of how an electron can be anywhere in the crystal and be any energy. Here the quantum state of the electron is the possibilities of all positions and all energies. Correct? In Many worlds Int, all the possible positions and energy of a single electron in a crystal has each own world? This sounds ridiculous... don't you think so..

And preparing the electron quantum state means repeating the experiments a million times to check the position and energy of a single electron running loose in a crystal? But how do you tell which electron is that since there is over a trillion electrons too? How do you prepare the state of a single electron in this case?

As long as you do a million experiments on systems that are identically and independently drawn from some distribution, then you can easily conceive the state as describing the preparation procedure. In the case where you are not sure exactly what pure state each single member of the ensemble is in, as long as you draw different members of the ensemble from the same distribution, you can write down a probability distribution for the pure states in the ensemble, reflecting the uncertainty in your preparation of the ensemble. The formalism for this is the density matrix formalism, mentioned by Quantum Defect in post #3.

 

[atyy]

As an example where we don't know the precise pure state that each individual member of the ensemble is in, one can consider quantum statistical mechanics. There the individual members of the ensemble are drawn from a thermal ensemble like the canonical ensemble. In that case the density matrix formalism is used.

Here are the notes on the density matrix formalism http://www.pmaweb.caltech.edu/~mcc/Ph127/a/Lecture_12.pdf from Mike Cross's statistical mechanics course http://www.pmaweb.caltech.edu/~mcc/Ph127/a/index.html.

 

[lucas_]

in classical physics, you also need to prepare a system for measurement? For example. When we throw the dice with 6 faces 6 times, we know we would get probability of getting each unique number. Is this preparation too.. in the sense you want to confirm it would indeed produce the conventional statistical mechanics?

 

[atyy]

in classical physics, you also need to prepare a system for measurement? For example. When we throw the dice with 6 faces 6 times, we know we would get probability of getting each unique number. Is this preparation too.. in the sense you want to confirm it would indeed produce the conventional statistical mechanics?

Generally with probabilistic theories, the easiest interpretation is that probabilities are relative frequencies, so it is easiest to think of drawing i.i.d. samples from some distribution that is the preparation procedure. Even classically there are some options given by Bayesian and Frequentist interpretations of probability. The difficult part in quantum theory is the meaning of wave function collapse. Heuristically, it seems to be something like Bayes's rule, but closer inspection shows that we have difficulty interpreting it as such.

So to go with the classical analogy, yes, throwing a die is a preparation procedure that yields outcomes consistent with all numbers {1,2,3,4,5,6} being equally probable. Clearly a different preparation procedure like looking at the dice and putting it down carefully can yield different probabilities, like the 6 being completely certain.

 

[lucas_]

Generally with probabilistic theories, the easiest interpretation is that probabilities are relative frequencies, so it is easiest to think of drawing i.i.d. samples from some distribution that is the preparation procedure. Even classically there are some options given by Bayesian and Frequentist interpretations of probability. The difficult part in quantum theory is the meaning of wave function collapse. Heuristically, it seems to be something like Bayes's rule, but closer inspection shows that we have difficulty interpreting it as such.

So to go with the classical analogy, yes, throwing a die is a preparation procedure that yields outcomes consistent with all numbers {1,2,3,4,5,6} being equally probable. Clearly a different preparation procedure like looking at the dice and putting it down carefully can yield different probabilities, like the 6 being completely certain.

Ok. If you could replace a better word or synonym than quantum state "preparation", what would you use? How about "confirmation" or other less vague words to describe it? maybe "state validation" instead of "state preparation"?

 

[atyy]

Ok. If you could replace a better word or synonym than quantum state "preparation", what would you use? How about "confirmation" or other less vague words to describe it? maybe "state validation" instead of "state preparation"?

I wouldn't worry about it, as long as one knows the concept. In some cases, there is certainly "state validation" somewhere, in a general state reduction rule, one has to calibrate the instrument to know how the instrument collapses the state after a given measurement outcome.

 

[lucas_]

I wouldn't worry about it, as long as one knows the concept. In some cases, there is certainly "state validation" somewhere, in a general state reduction rule, one has to calibrate the instrument to know how the instrument collapses the state after a given measurement outcome.

So state preparation has a lot to do with the collapse postulate as you stated that "measurement and the selection of a system conditioned on the observation of a result is a means of state preparation because of the state reduction postulate". Supposed 100 years ago quantum physics didn't start with Bohr Copenhagen Interpretation but directly started with Many Worlds Interpretations and not only that supposed the holographic principle were also discovered at same time (and we discovered that the universe was really in some 2D surface or our dimension minus one) and techniques were able to make one navigate (of view) the different worlds of the quantum state. Would we still use the "preparation" word? or no more? Point is.. is "preparation" word due to the peculiarity of the Copenhagen interpretation?

 

[atyy]

So state preparation has a lot to do with the collapse postulate as you stated that "measurement and the selection of a system conditioned on the observation of a result is a means of state preparation because of the state reduction postulate". Supposed 100 years ago quantum physics didn't start with Bohr Copenhagen Interpretation but directly started with Many Worlds Interpretations and not only that supposed the holographic principle were also discovered at same time (and we discovered that the universe was really in some 2D surface or our dimension minus one) and techniques were able to make one navigate (of view) the different worlds of the quantum state. Would we still use the "preparation" word? or no more? Point is.. is "preparation" word due to the peculiarity of the Copenhagen interpretation?

I'm pretty sure the holographic principle doesn't allow you to navigate the different worlds of the quantum state - unless you've been reading Susskind and Tegmark or something.

As I understand it, preparation has not so much to do with collapse, as with the Born rule. It is the Born rule that makes things probabilistic, and if one take probabilities to mean relative frequencies, then preparation of an ensemble is the standard way to understand the predictions of the theory, whether in biology or quantum physics.

The collapse was mentioned mainly because of the question of whether the theory applies to single systems, like the universe.

But yes, maybe if Many-Worlds was used, one might have a different perspective. On the other hand, the single observer in each World would still use the Born rule. But perhaps he can escape it if he used a Bayesian interpretation of probability? I don't know, it seems very tricky and related to whether the Deutsch-Wallace version of Many-Worlds is correct. Maybe this paper by Greaves and Myrvold is relevant:


http://philsci-archive.pitt.edu/4222/
Everett and evidence
Greaves, Hilary and Myrvold, Wayne

 

[lucas_]

I'm pretty sure the holographic principle doesn't allow you to navigate the different worlds of the quantum state - unless you've been reading Susskind and Tegmark or something.

As I understand it, preparation has not so much to do with collapse, as with the Born rule. It is the Born rule that makes things probabilistic, and if one take probabilities to mean relative frequencies, then preparation of an ensemble is the standard way to understand the predictions of the theory, whether in biology or quantum physics.

The collapse was mentioned mainly because of the question of whether the theory applies to single systems, like the universe.

But yes, maybe if Many-Worlds was used, one might have a different perspective. On the other hand, the single observer in each World would still use the Born rule. But perhaps he can escape it if he used a Bayesian interpretation of probability? I don't know, it seems very tricky and related to whether the Deutsch-Wallace version of Many-Worlds is correct. Maybe this paper by Greaves and Myrvold is relevant:


http://philsci-archive.pitt.edu/4222/
Everett and evidence
Greaves, Hilary and Myrvold, Wayne

Let's not talk of many worlds but just plain old Copenhagen. Are you aware any papers that detail how prior to state reduction or collapse, the wave function or Hilbert space is located in a holographic principle surface? And it is only upon collapse that the particle occurs in spacetime? Recall geometry only exists in the bulk and didn't exist in the holographic surface (according to a string theorist Macadamia something). So maybe quantum gravity is about collapse and geometry only existing in the bulk? I am not spreading disinformation but mentioned this just to know if papers about them have been written. If none is it due to the fact we don't know where is the holographic surface located?

 

[vanhees71]

I think, there's pretty lot of confusion among theoreticians, mostly among those close to what's called "philosophy of physics", which just comes from the fact that they tend to forget that physics is about observations and experiments in Nature. An experiment is just a controlled setup of sufficiently isolated systems to make observations under well-known conditions. So it's also not different from any observation of any phenomenon, where you have less "prepared" (e.g., in astronomy you cannot really do experiments, because you just look somehow to objects like stars, planets, galaxies, the microwave background radiation, etc. which are there and which we can't manipulate).

Now, from this point of view, what's a quantum state? It's (an equivalence class) of a "preparation procedure"! This "preparation" can be very direct, e.g., at the LHC, the accelerator physicists set up two beams of proton bunches, which are smashed into each other at one of the great detectors built by the detector people, which identify the particles in various clever ways and measure the interaction vertices where they are created, their energy and/or momentum, sometimes their spin/polarization and so on. Nowadays this is stored electronically in huge computer files, which then are evaluated by the experimentalist with statistical means to understand as many aspects of such proton-proton collisions as possible.

The "preparation procedure" can also consist of observations on given objects, like in my astronomy/cosmology example. You just look into the sky and find some object like our Sun. Then you can deduce its state by measuring its temperature, size, mass, etc.

With the discovery of quantum theory in 1925/26, it turned out that a system cannot be prepared in such a way that all the possible observables can have definite values, as was assumed in classical physics without even thinking about it. While according to classical physics you can, e.g., determine the position and momentum of a "particle" at arbitrary precision, and then these both quantities are known to have a definite value. It's also tacitly assumed that indeed they have definite values, even if haven't determined or observed and thus know them. Then we can use statistics as a tool to describe such a system in terms of probabilities, based on the incomplete knowledge we have due to a preparation/observation procedure. Nobody has a problem with this use of probabilities.

In quantum theory it's different, because according to quantum theory, neither position nor momentum can have a precise value, no matter how accurate I try to prepare or observe these quantities on a single particle. I can determine one of the quantities, say position, very accurately but then necessarily the other (here momentum) is only very inaccurately determined, according to the Heisenberg-Robertson uncertainty relation, $\Delta x \Delta p \geq \hbar/2$.

This implies that we have only a probabilistic description in any case, no matter how accurately we try to determine the values of the observables of a system. This disturbs many thinkers about the "interpretation" of quantum theory, but from a physicist's point of view, that's simply a fact about Nature which has been figured out in a long process of observations and theory building since Galilei's times, where modern science has been discovered. Nature does not care, whether we like how she behaves or not, she just behaves as observed.

Maybe (or even most probably) we don't know enough about Nature to see a deterministic description of these quantum phenomena, but if there's really some theory like this, we simply have no evidence for it yet! In all cases of very accurate tests of quantum theory, nowadays also with not so small objects as elementary particles or photons, but with quite large molecules (double-slit experiment with bucky-ball molecules) or even macroscopic objects (entanglement of vibration modes of two macroscopic-sized diamonds), quantum theory has come out to describe the phenomena correctly. So, if you want it or not, at least at the moment we have all reason to believe that Nature behaves as described by quantum theory, including the irreducibly probabilistic behavior of observables, which is not due to a lack of knowledge about a system's state but is inherent in the very definition of the observables themselves.

 

[atyy]

Let's not talk of many worlds but just plain old Copenhagen. Are you aware any papers that detail how prior to state reduction or collapse, the wave function or Hilbert space is located in a holographic principle surface? And it is only upon collapse that the particle occurs in spacetime? Recall geometry only exists in the bulk and didn't exist in the holographic surface (according to a string theorist Macadamia something). So maybe quantum gravity is about collapse and geometry only existing in the bulk? I am not spreading disinformation but mentioned this just to know if papers about them have been written. If none is it due to the fact we don't know where is the holographic surface located?

In the holographic principle, the theory on the boundary is just a quantum field theory, so before collapse it is the same as any other quantum field theory. In the limit in which the bulk geometry is classical, I think collapse of the wave function should work the same as in ordinary quantum theory, but I'm not sure. There is also a lot of uncertainty about the description of someone falling into the black hole.

The holographic principle is speculative, so it should probably not be discussed in here, but in the Beyond the Standard Model section.

 

[lucas_]

In the holographic principle, the theory on the boundary is just a quantum field theory, so before collapse it is the same as any other quantum field theory. In the limit in which the bulk geometry is classical, I think collapse of the wave function should work the same as in ordinary quantum theory, but I'm not sure. There is also a lot of uncertainty about the description of someone falling into the black hole.

The holographic principle is speculative, so it should probably not be discussed in here, but in the Beyond the Standard Model section.

Ok. Back to QM.

We mostly think electrons are delocalized in an atom.. but is the nucleus delocalized too? Or let's take the case of molecules.. let's say a bowling ball. Can we say the inside of it.. both electrons and nuclei are delocalized so when I roll the ball in the floor, only the outer perimeter of the ball exist as localized particles (and the inside existing as ghostly superposition)? Bohr stated that in the absence of measurement to determine it's properties, there is no electron (at least as wave only).. can we say the same thing to nuclei.. that in the absence of measurement to determine it's properties, there is no nucleus.. or just it just existing in ghostly superposition?

 

[bhobba]

We mostly think electrons are delocalized in an atom.. but is the nucleus delocalized too? Or let's take the case of molecules.. let's say a bowling ball. Can we say the inside of it.. both electrons and nuclei are delocalized so when I roll the ball in the floor, only the outer perimeter of the ball exist as localized particles (and the inside existing as ghostly superposition)? Bohr stated that in the absence of measurement to determine it's properties, there is no electron (at least as wave only).. can we say the same thing to nuclei.. that in the absence of measurement to determine it's properties, there is no nucleus.. or just it just existing in ghostly superposition?

The modern conception of measurement is fully quantum and just after decohence has occurred.

Take the bowling ball. It's decohered by the environment so it has all the usual classical properties of position etc. The nuclei and electrons of the atoms inside are entangled forming this thing called an atom and it forms part of larger molecules, latices etc.

Personally I don't think of electrons as anything inside an atom - it's simply described by a wave-function - that is insofar as it can be considered not entangled with the nucleus.

Thanks`
Bill

 

[lucas_]

The modern conception of measurement is fully quantum and just after decohence has occurred.

Take the bowling ball. It's decohered by the environment so it has all the usual classical properties of position etc. The nuclei and electrons of the atoms inside are entangled forming this thing called an atom and it forms part of larger molecules, latices etc.

Personally I don't think of electrons as anything inside an atom - it's simply described by a wave-function - that is insofar as it can be considered not entangled with the nucleus.

Thanks`
Bill

You mentioned how the bowling ball is decohered by thermal exposure to environment and has usual classical properties of position etc. But an electron forming energy bands in a crystal is delocalized even though it has thermal exposure to environment. So we can say the bowling ball electrons are also delocalized because they (ball and crystal) are both exposed to the same temperature of say 30 C. So far ok here. Now if the electrons can be delocalized in spite of the thermal bath.. why can't the nuclei in the the crystal be delocalized too?

 

[bhobba]

But an electron forming energy bands in a crystal is delocalized

Why do you think it has the property of de-localisation when not observed?

In other words precisely what do you mean by de-localisation?

If you mean it's entangled with the nucleus and or crystal lattice - then yes - but from the definition of entanglement you can't consider it being in a separate state. Its modeled that way to get a handle on it - but in reality it actually isn't in a state.

Thanks
Bill

 

[lucas_]

Why do you think it has the property of de-localisation when not observed?

In other words precisely what do you mean by de-localisation?

If you mean it's entangled with the nucleus and or crystal lattice - then yes - but from the definition of entanglement you can't consider it being in a separate state. Its modeled that way to get a handle on it - but in reality it actually isn't in a state.

Thanks
Bill

An electron in crystal can be anywhere and any energy.. it has no fixed location.. the concept of energy bands means there are no fixed electrons.. just wave functions of electrons in the crystal.. this is the context of delocalization. Do you agree with this?

Maybe nuclei are localized because the thermal bath makes them classical.. while electrons didn't have thermal bath unless you did a photoelectric effect trick on them localizing it?

Delocalized means the particles don't exist as particles but as wave function much like before the particle hit the detector in the double slit experiment..

 

[bhobba]

An electron in crystal can be anywhere and any energy.

Wait a minute here. An electron does not have the property of location until observed to have that property.

Electrons in atoms are entangled with the nucleus to form an atom. That atom in solids is part of latices etc and it is that that has the property of location.

Delocalized means the particles don't exist as particles but as wave function much like before the particle hit the detector in the double slit experiment..

But in an atom it doesn't have a state to be expanded as a wave-function - its entangled and you have to consider the system, nucleus and electrons, as a whole.

Thanks
Bill

 

[lucas_]

Wait a minute here. An electron does not have the property of location until observed to have that property.

Electrons in atoms are entangled with the nucleus to form an atom. That atom in solids is part of latices etc and it is that that has the property of location.
But in an atom it doesn't have a state to be expanded as a wave-function - its entangled and you have to consider the system, nucleus and electrons, as a whole.

Thanks
Bill

What I meant by "An electron in crystal can be anywhere and any energy" was technically that an electron wave function become severely smeared and spread out in a crystal lattice and can occupy any energy they desire.. hence it doesn't have location. But you seemed to be saying that the nucleus have location and these two are entangled. How could the electrons in energy band that have no location become entangled with nuclei that have locations? Can you please clarify?

 

[bhobba]

How could the electrons in energy band

Only some electrons wander about in energy bands in solids and that depends on the solid. Not sure if its the case in a bowling ball. But regardless I still think its entangled with the ball and can't wander outside it - insofar that is it can be considered separate from the ball as a whole. However I would like the input of someone more conversant with solid state physics than I am.

But regardless of that if by de-localised you mean the wave-function is spread out - yes - but that doesn't mean it has the property of location.

Earlier you said:

Here the quantum state of the electron is the possibilities of all positions and all energies. Correct?

That's not correct. I think you are getting a bit confused with what a quantum state is. I could give the exact definition but it's probably best for you to explain your conception of it first.

Thanks
Bill

 

[lucas_]

Only some electrons wander about in energy bands in solids and that depends on the solid. Not sure if its the case in a bowling ball. But regardless I still think its entangled with the ball and can't wander outside it - insofar that is it can be considered separate from the ball as a whole. However I would like the input of someone more conversant with solid state physics than I am.

But regardless of that if by de-localised you mean the wave-function is spread out - yes - but that doesn't mean it has the property of location.

Earlier you said:That's not correct. I think you are getting a bit confused with what a quantum state is. I could give the exact definition but it's probably best for you to explain your conception of it first.

Thanks
Bill

quantum state = all possibilities of a particle including superposition of states.. so in a single electron in a crystal lattice.. the single electron is in a quantum state or composing of superposition of all possible position and energies of the energy band

 

[bhobba]

quantum state = all possibilities of a particle including superposition of states.. so in a single electron in a crystal lattice.. the single electron is in a quantum state or composing of superposition of all possible position and energies of the energy band

That's not quite correct - its more subtle than that.

First superposition in QM is rather abstract since any state is a superposition of all sorts of different states. Technically its because the pure states form a vector space and any vector is the supposition of differing things. To be specific suppose |u> = c1*|u1> + c2*|u2> but also its a superposition of other vectors ie c3*|u3> + c4*|u4> and an infinity of others as well. You can't say what a state is a superposition of without the concept of an observational basis to fix what you are expanding it in. I would like the input of a solid state expert to see if position is a sensible expansion. And even then it doesn't mean its in multiple locations at once in a spread out sort of way. It means simply if you were to somehow observe its position would be as per the Born rule.

We also have the issue of entanglement. What this means is suppose we have a system that can be in state |a> or |b> and another system that can also be in state |a> or |b>. If system 1 is in state |a> and system 2 in state |b> that is written as |a>|b>. Similarly if system 1 is in state |b> and system 2 is in state |a> then that is written as |b>|a>. But from the principle of superposition another possible state is 1/root 2 |a>|b> + 1/root 2 |b>|a>. We can't say what state either system is in - it's entangled with the other system. Now an electron bound to a nucleus is in fact entangled with the nucleus - you can't speak of them as separate systems - although as an approximation is done. Most electrons in solids are like that. But we have the issue of electrons being shared by atoms in bonds and as part of latices. It still means they are entangled with those systems. However for some electrons in some solids they wander about in bands - but even then I think they are entangled with the atoms in the solid. However I would like the input of someone more conversant with solid state physics.

The bottom line here is I think what you said before is not correct - namely:

Let's take the case of molecular orbital.. or an example of how an electron can be anywhere in the crystal and be any energy

I don't think it has the property of being somewhere in the crystal or even if it can be considered to have a state except as an approximation used in solid state physics. And if not then the solid needs to be considered as system itself and decohered as a whole.

Thanks
Bill

 

[lucas_]

That's not quite correct - its more subtle than that.

First superposition in QM is rather abstract since any state is a superposition of all sorts of different states. Technically its because the pure states form a vector space and any vector is the supposition of differing things. To be specific suppose |u> = c1*|u1> + c2*|u2> but also its a superposition of other vectors ie c3*|u3> + c4*|u4> and an infinity of others as well. You can't say what a state is a superposition of without the concept of an observational basis to fix what you are expanding it in. I would like the input of a solid state expert to see if position is a sensible expansion. And even then it doesn't mean its in multiple locations at once in a spread out sort of way. It means simply if you were to somehow observe its position would be as per the Born rule.

We also have the issue of entanglement. What this means is suppose we have a system that can be in state |a> or |b> and another system that can also be in state |a> or |b>. If system 1 is in state |a> and system 2 in state |b> that is written as |a>|b>. Similarly if system 1 is in state |b> and system 2 is in state |a> then that is written as |b>|a>. But from the principle of superposition another possible state is 1/root 2 |a>|b> + 1/root 2 |b>|a>. We can't say what state either system is in - it's entangled with the other system. Now an electron bound to a nucleus is in fact entangled with the nucleus - you can't speak of them as separate systems - although as an approximation is done. Most electrons in solids are like that. But we have the issue of electrons being shared by atoms in bonds and as part of latices. It still means they are entangled with those systems. However for some electrons in some solids they wander about in bands - but even then I think they are entangled with the atoms in the solid. However I would like the input of someone more conversant with solid state physics.

The bottom line here is I think what you said before is not correct - namely:I don't think it has the property of being somewhere in the crystal or even if it can be considered to have a state except as an approximation used in solid state physics. And if not then the solid needs to be considered as system itself and decohered as a whole.

Thanks
Bill

Oh. I think I'm just vague. The double slit would illustrates what I had in mind. Of course the electrons didn't duplicate itself as it emits. Similarly, the single electron didn't become many in the crystals.. that's not what I meant. What I meant is that it exists purely as wave function and you need the born rule to make it appear when measured. So thanks for pointing out not to use the vague words that the electron can be anywhere in the crystal and have any energy.

The energy band or the entire Fermi sea is entangled with the nuclei? But the Fermi sea exists only as wave function and not classical.. but yet the nuclei can be classical? But maybe the nuclei is not classical either.. They are decohered but maybe just like the electrons, they only exist as wave function (you can't localized any particle to a specific position because it would make both position and momentum observable have specific values which is not allowed by HUP). So maybe we must agree that nuclei have less smeared out wavefunction than the electrons and they are all wavefunctions.

But then with the atoms in the solids exposed to thermal bath and decoherence, they are still in superposition? maybe they are still entangled with the environment.. and the superposition is something we can't see with our eyes because it's tiny.. yet the macroscopic object has specific position because it's macroscopic or coarse graining of the microscopic degrees of freedom? Agree with the words I use now? I need to be accurate with words because I want to convey quantum physics to others. Thanks.

 

[bhobba]

The energy band or the entire Fermi sea is entangled with the nuclei?

I am not really up on solid state physics but my understanding of the Fermi sea its the energy level electrons occupy in a solid. That's analogous to energy levels in say a hydrogen atom. However the electron is in fact entangled with and bound to the nuclei and its simply modeled as an electrostatic potential in the Hamiltonian to get a handle on it. Intuitively if you move the nucleus the electron goes with it - they are entangled. You can't speak of a separate state of the electron in an exact sense as is seen by the fact in such problems its solved in coordinates where the nucleus is the origin. Its state is dependent on the state of the nucleus.

Now in the bowling ball its electrons, if they are in lattice bands, are still entangled with the ball - move the ball and they go with it.

Thanks
Bill

 

[lucas_]

Dear Bhobba and Atty...

Is the following possible.
First I know QM is about measurements.. the primitive of QM is measurement or classical outcome.
Bhobba fully believes it that he could state improper mixture in decoherence can be interpretated as proper mixture FAPP. But Maui emphasized to Bill "Experiments prove that the world is quantum and not classical. You should not pretend you are explaining the quantum to classical transition by stating the obvious - that you always observe a classical world." But Bill still hold on to ignorant essemble strongly since he has almost given up hope of knowing what's the behind the scene in the quantum. Or how does Improper mixture really become proper mixture.

Without considering Bohmian or Many worlds or even this thing where the wave functions are real. What would be wrong with this Tegmark like interpretation. Why is this not popular. In Tegmark's we are living in simulations inside a program. Improper mixture becomes proper simply because it is what the program do... that is.. rather than saying the quantum state is real and having to face paradox of Wigner friend. We can say the program only makes improper mixture to proper mixture whenever there are measurements and as dictated by the random generator causing wavefunction collapse in the equations in the program.. so eigenvalues chosen are really random.. this means observations can even differ solving Wigner friend because quantum state is not real instantaneously but relatively to observations.. something like relativity of observations.

Look. There is no way to distinguish interpretations because the program did it that way because in truth there is really no bohmian mechanics or many worlds nor even this ontology where wave functions are real. Although the wave functions could be real because they are part of the program and we are part of the program but observations and measurements can differ solving Wigner friend. Since we only have access it the outputs. It makes Bill hobba ensemble natural because that's the only primitive he (and all of us) have access to.

What is the problem why this Matrix interpretation is not as common as Bohmian or Many worlds? It would solve the measurement problem. It would also solve why there is special relativity.. because in the program or simulations.. everything must be relative because there is no boundary because it is just a program.. it would also solve for quantum gravity because the Planck scale is just the pixel of the program output and effective field theory are due to the program behavior etc. But let's just focus on this thread in the transition from improper mixture to proper mixture.. the answer being it is the program doing.

What are your reasonings why this interpretation is not common or not acceptable? are there any near fatal flaws or obvious objections? Would this totally solve the riddle how improper mixture becomes proper mixture?