Why is Decoherence Needed in Quantum Mechanics?

[stevendaryl]

There is another way of explaining the lack of macroscopic superpositions that doesn't involve collapse (though it sort of involves decoherence).

Suppose you set up a system with a single starting state $A$, two possible orthogonal intermediate states $B_1$ and $B_2$ and a final state $C$. The probability of starting in $A$, passing through either $B_1$ or $B_2$ and winding up in state $C$ is given by:

$P_{AC} = P_{AB_1C} + P_{AB_2C} + 2 Re((\psi_{AB_1C})^* \psi_{AB_2C})$

where $\psi_{AB_1C}$ is the amplitude for going from $A$ to $C$ via $B_1$
$\psi_{AB_2C}$ is the amplitude for going from $A$ to $C$ via $B_2$
$P_{AB_1C}$ is the probability for going from $A$ to $C$ via $B_1$, which is $(\psi_{AB_1C})^*\psi_{AB_1C}$
$P_{AB_2C}$ is the probability for going from $A$ to $C$ via $B_2$, which is $(\psi_{AB_2C})^*\psi_{AB_2C}$

The first two terms in the expression for $P_{AC}$ is what you would expect from classical probability. The term that is essentially quantum-mechanical is the term:
$2 Re((\psi_{AB_1C})^* \psi_{AB_2C})$

That's the interference term between the two alternatives, $B_1$ and $B_2$. So observing this term is a kind of evidence of there being an intermediate state that is a superposition (as opposed to a mixture, which is the only kind of alternative possible in classical probability).

So here, to me, is the simplest way to understand the implications of decoherence, and the reason why we never see the effects of macroscopic superpositions: If $B_1$ and $B_2$ are macroscopically distinguishable (say, a dead cat and a live cat), then for any final state $C$ one or the other of the transition amplitudes will be negligible:

$\psi_{AB_1C} \approx 0$ or $\psi_{AB_2C} \approx 0$

If the intermediate states are macroscopically distinguishable, then there will be some evidence in the final state, $C$ of which alternative was chosen. Only one alternative will be compatible with final state $C$ (that is, have a non-negligible amplitude for ending up in that state).

 

[stevendaryl]

So there is a possibility that one would see the superposition if open and folded, and write "both"?

I don't think that's a coherent possibility. Imagine the further implications: Suppose I hand the notebook to someone trying to decide what to eat for dinner. They decide:

1. If the notebook says "open", I'll eat Chinese.
2. If the notebook says "folded", I'll eat Mexican.
3. If the notebook says "both", I'll eat Italian.

So "both" being a superposition of "open" and "folded" would imply that Italian food is a superposition of Chinese food and Mexican food. I think that's pretty silly.

You could substitute any consequence you like. "If open, vote for Rand Paul, if folded vote for Hillary Clinton, if both vote for Ted Cruz". That would imply that Ted Cruz is a superposition of Rand Paul and Hillary Clinton.

Daryl

 

[atyy]

I don't think that's a coherent possibility. Imagine the further implications: Suppose I hand the notebook to someone trying to decide what to eat for dinner. They decide:

1. If the notebook says "open", I'll eat Chinese.
2. If the notebook says "folded", I'll eat Mexican.
3. If the notebook says "both", I'll eat Italian.

So "both" being a superposition of "open" and "folded" would imply that Italian food is a superposition of Chinese food and Mexican food. I think that's pretty silly.

You could substitute any consequence you like. "If open, vote for Rand Paul, if folded vote for Hillary Clinton, if both vote for Ted Cruz". That would imply that Ted Cruz is a superposition of Rand Paul and Hillary Clinton.

Silly, but impossible?

Also, in the other argument you gave. What if I used for the intermediate state D1 = B1 + B2, and D2 = B1 - B2. Wouldn't I still end up with the same conclusion?

 

[stevendaryl]

Silly, but impossible?

Yes, I think it's impossible. For one thing, if I have a plan

1. If open, do A
2. If folded, do B
3. If both, do C

A, B and C are ARBITRARY. So this would imply that any choice I make, C, is a superposition of any other two choices, A and B. That's provably not true. For example, suppose I choose C = A. Then A certainly cannot be equal to a superposition of A and B. (Unless B = A, also)

Also, in the other argument you gave. What if I used for the intermediate state D1 = B1 + B2, and D2 = B1 - B2. Wouldn't I still end up with the same conclusion?

No, because the assumption was that $B_1$ and $B_2$ are macroscopically distinguishable. But your $D_1$ and $D_2$ are not macroscopically distinguishable (or at least, there is no reason to assume that they are). I'm not sure what a "Dead cat + Live cat" would look like, but there is no reason to think it would look any different than a "Dead cat - Live cat".

 

[stevendaryl]

There's another argument that "both" can't be the superposition of "open" and "folded":

If "both" is a superposition of "open" and "folded", then that means that "both" and "open" are NOT orthogonal states. And what that means is that if the word is "both" one day, there is a nonzero chance that it will be discovered to be "open". In the same way, if an electron is put into the state "spin-up in the x-direction", there is a nonzero chance that it will later be detected to be "spin-up in the z-direction". So if "open", "folded" and "both" are permanent records, they can't possibly be superpositions of one another.

This seems ridiculous, arguing over a possibility that is clearly ludicrous. But when it comes to QM, it's really hard to know if ludicrous just means that you're using pre-quantum intuitions that don't apply.

 

[atyy]

This seems ridiculous, arguing over a possibility that is clearly ludicrous. But when it comes to QM, it's really hard to know if ludicrous just means that you're using pre-quantum intuitions that don't apply.

Yes, that's why I'm skeptical about the explanations as alternatives to collapse. Assuming we take collapse within Copenhagen, the interpretation itself has "commonsense" put in by hand, because it has the classical observer. So if one is putting in commonsense by hand, I don't see how it is much different from Copenhagen.

 

[zonde]

The photons from CMBR are everywhere and can't be shielded and said to decohere things to position eigenstates. But how come one can perform double slit experiment or the c60 buckyball... won't the CMBR photons interact with them or are they somehow shielded from this, and how?.

you can read in this paper - https://vcq.quantum.at/fileadmin/Publications/2003-17.pdf
p.324 under "B. Coherence and which-path information" that more serious concern is photons radiated out by buckyball molecules themselves and why that does not destroy interference pattern.

 

[zonde]

There is another way of explaining the lack of macroscopic superpositions that doesn't involve collapse (though it sort of involves decoherence).

Suppose you set up a system with a single starting state $A$, two possible orthogonal intermediate states $B_1$ and $B_2$ and a final state $C$. The probability of starting in $A$, passing through either $B_1$ or $B_2$ and winding up in state $C$ is given by:

$P_{AC} = P_{AB_1C} + P_{AB_2C} + 2 Re((\psi_{AB_1C})^* \psi_{AB_2C})$

where $\psi_{AB_1C}$ is the amplitude for going from $A$ to $C$ via $B_1$
$\psi_{AB_2C}$ is the amplitude for going from $A$ to $C$ via $B_2$
$P_{AB_1C}$ is the probability for going from $A$ to $C$ via $B_1$, which is $(\psi_{AB_1C})^*\psi_{AB_1C}$
$P_{AB_2C}$ is the probability for going from $A$ to $C$ via $B_2$, which is $(\psi_{AB_2C})^*\psi_{AB_2C}$

The first two terms in the expression for $P_{AC}$ is what you would expect from classical probability. The term that is essentially quantum-mechanical is the term:
$2 Re((\psi_{AB_1C})^* \psi_{AB_2C})$

That's the interference term between the two alternatives, $B_1$ and $B_2$. So observing this term is a kind of evidence of there being an intermediate state that is a superposition (as opposed to a mixture, which is the only kind of alternative possible in classical probability).

So here, to me, is the simplest way to understand the implications of decoherence, and the reason why we never see the effects of macroscopic superpositions: If $B_1$ and $B_2$ are macroscopically distinguishable (say, a dead cat and a live cat), then for any final state $C$ one or the other of the transition amplitudes will be negligible:

$\psi_{AB_1C} \approx 0$ or $\psi_{AB_2C} \approx 0$

If the intermediate states are macroscopically distinguishable, then there will be some evidence in the final state, $C$ of which alternative was chosen. Only one alternative will be compatible with final state $C$ (that is, have a non-negligible amplitude for ending up in that state).

In order to produce interference you make a setup where both $B_1$ and $B_2$ can probabilistically transition to $C_1$ or $C_2$. And if you can't come up with such a setup it has little to do with decoherence.

For example, consider two cases:
1. Two beams are heading in different direction. One of them goes out to nowhere. You place in it's path a mirror and make it interfere with the other beam. Only it does not produce interference. We say that two beams are not coherent.
2. Two beams are heading in different direction. One of them goes out to nowhere. We don't have a handy mirror (say they are not invented yet) to place in the path of the beam. So there is no way we can produce interference. We won't call it decoherence, right?

 

[stevendaryl]

In order to produce interference you make a setup where both $B_1$ and $B_2$ can probabilistically transition to $C_1$ or $C_2$. And if you can't come up with such a setup it has little to do with decoherence.

For example, consider two cases:
1. Two beams are heading in different direction. One of them goes out to nowhere. You place in it's path a mirror and make it interfere with the other beam. Only it does not produce interference. We say that two beams are not coherent.
2. Two beams are heading in different direction. One of them goes out to nowhere. We don't have a handy mirror (say they are not invented yet) to place in the path of the beam. So there is no way we can produce interference. We won't call it decoherence, right?

No, as I said, interference is observable when you have two possible alternative intermediate states that both lead to the same final state. For example, in a two-slit experiment, the particle can either go through one slit or the other, but the final state is that the particle reaches a single spot on the screen.

[edit]I'm not claiming that lack of interference implies decoherence, I'm claiming that decoherence implies lack of interference.

 

[stevendaryl]

No, as I said, interference is observable when you have two possible alternative intermediate states that both lead to the same final state. For example, in a two-slit experiment, the particle can either go through one slit or the other, but the final state is that the particle reaches a single spot on the screen.

So decoherence is what makes it impossible to find a common final state that is reachable from both intermediate states.

 

[zonde]

No, as I said, interference is observable when you have two possible alternative intermediate states that both lead to the same final state. For example, in a two-slit experiment, the particle can either go through one slit or the other, but the final state is that the particle reaches a single spot on the screen.

There is always constructive interference outcome and destructive interference outcome. Give an example where it is not so if you do not agree.
And speaking about the idea that two alternative states are necessarily intermediate states of the same initial state, there is an experiment you might consider interesting:
Interference of Independent Photon Beams

[edit]I'm not claiming that lack of interference implies decoherence, I'm claiming that decoherence implies lack of interference.

Right.

So decoherence is what makes it impossible to find a common final state that is reachable from both intermediate states.

No, I don't agree. Decoherence makes probabilities of outcomes add up classically.
Take an example where you can observe interference. Take away coherence. You can still reach final states but probabilities add up classically.

 

[stevendaryl]

No, I don't agree. Decoherence makes probabilities of outcomes add up classically.

Yes, and why is that?

Take an example where you can observe interference. Take away coherence. You can still reach final states but probabilities add up classically.

I think you might be mixing up two different things. When I say "decoherence", I'm talking about the process whereby superpositions evolve into what is, for all practical purposes, mixtures. So it's the process by which alternatives become decoherent. From Wikipedia.

Quantum decoherence gives the appearance of wave function collapse, which is the reduction of the physical possibilities into a single possibility as seen by an observer.
​

It seems to me that you are talking about a case where things start out decoherent.

 

[stevendaryl]

There is always constructive interference outcome and destructive interference outcome. Give an example where it is not so if you do not agree.

There is no interference between classically distinguishable alternatives.

 

[zonde]

I think you might be mixing up two different things. When I say "decoherence", I'm talking about the process whereby superpositions evolve into what is, for all practical purposes, mixtures. So it's the process by which alternatives become decoherent. From Wikipedia.

Quantum decoherence gives the appearance of wave function collapse, which is the reduction of the physical possibilities into a single possibility as seen by an observer.
​

It seems to me that you are talking about a case where things start out decoherent.

I don't see that we are talking about different things.
Look, if you have two ensembles of particles where one ensemble is represented by superposition of coherent states and the other one is represented by mixture. How you can tell apart two ensembles by observation? In first case you have interference and in second you don't.

 

[zonde]

There is always constructive interference outcome and destructive interference outcome. Give an example where it is not so if you do not agree.

There is no interference between classically distinguishable alternatives.

By my comment I wanted to say that you can't meaningfully speak about single outcome when you speak about interference.

 

[stevendaryl]

By my comment I wanted to say that you can't meaningfully speak about single outcome when you speak about interference.

Right. My point was that $P_{AC} = P_{AB_1C}P_{AB_2C} + 2 Re(\psi^*_{AB_1C} \psi_{AB_2C}$, where $B_1$ and $B_2$ are alternative intermediate states. If you repeat the experiment many many times of starting in state A and checking if the final state is C, then you will be able to observe the second interference term. No single experiment can show it, of course (well, unless it makes $P_{AC}$ zero, which sometimes happens).

 

[stevendaryl]

I don't see that we are talking about different things.
Look, if you have two ensembles of particles where one ensemble is represented by superposition of coherent states and the other one is represented by mixture. How you can tell apart two ensembles by observation? In first case you have interference and in second you don't.

That's true. But the evolution operator evolves pure states into pure states. So the issue for decoherence is: why in some circumstances does a pure state seem to become a mixed state?

 

[zonde]

Right. My point was that $P_{AC} = P_{AB_1C}P_{AB_2C} + 2 Re(\psi^*_{AB_1C} \psi_{AB_2C}$, where $B_1$ and $B_2$ are alternative intermediate states. If you repeat the experiment many many times of starting in state A and checking if the final state is C, then you will be able to observe the second interference term. No single experiment can show it, of course (well, unless it makes $P_{AC}$ zero, which sometimes happens).

No, it's not about repeating experiment. Look, Mach–Zehnder interferometer has two outputs and you can't make anything like Mach–Zehnder interferometer with single output.

 

[stevendaryl]

No, it's not about repeating experiment.

I'm sorry, I misunderstood your "single outcome". You mean multiple measurements of a single event, rather than multiple events?

 

[stevendaryl]

No, it's not about repeating experiment. Look, Mach–Zehnder interferometer has two outputs and you can't make anything like Mach–Zehnder interferometer with single output.

Okay, I refreshed my memory about what that device does (by looking it up in Wikipedia). But I don't understand what point you are making about it.

 

[Ilja]

That's true. But the evolution operator evolves pure states into pure states. So the issue for decoherence is: why in some circumstances does a pure state seem to become a mixed state?

This is not the issue of decoherence. For decoherence, a pure state remains a pure state. The mixed state appears if one ignores the environment, and considers the effective state of the subsystem. Purely pragmatical, the question considered by decoherence is, given the system, and the environment (their subdivision defined by some actual situation) how long it takes until the interaction of the subsystem with the environment was strong enough to lead to a mixed states if one considers the projection of the whole system to the subsystem.

 

[stevendaryl]

This is not the issue of decoherence. For decoherence, a pure state remains a pure state.

Yes, but I said "seems to".

The mixed state appears if one ignores the environment, and considers the effective state of the subsystem. Purely pragmatical, the question considered by decoherence is, given the system, and the environment (their subdivision defined by some actual situation) how long it takes until the interaction of the subsystem with the environment was strong enough to lead to a mixed states if one considers the projection of the whole system to the subsystem.

 

[zonde]

Okay, I refreshed my memory about what that device does (by looking it up in Wikipedia). But I don't understand what point you are making about it.

I tried to make a point that in order to observe interference we would have to subject two alternative states to a setup similar to Mach–Zehnder interferometer where there are at least two outputs and both states can appear on either output. Say a dead cat and living cat is subjected to the some procedure that can kill/resurrect or leave intact either cat.

But it was minor point and I will try to get back on subject.
If two intermediate alternative states are made/become distinguishable then they do not produce interference and we say that they are no longer coherent.
This process is called decoherence.
And two states become distinguishable when they are undergoing interactions asymmetrically, right?

 

[lucas_]

This is not the issue of decoherence. For decoherence, a pure state remains a pure state. The mixed state appears if one ignores the environment, and considers the effective state of the subsystem. Purely pragmatical, the question considered by decoherence is, given the system, and the environment (their subdivision defined by some actual situation) how long it takes until the interaction of the subsystem with the environment was strong enough to lead to a mixed states if one considers the projection of the whole system to the subsystem.

Stevendaryl, If one just considers the effect state of the subsystem without the born rule, can you still call it mixed state? Because it seems some people call it mixed state even without born rule by simply considering the state of the subsystem (such as Iija above).

 

[stevendaryl]

If two intermediate alternative states are made/become distinguishable then they do not produce interference and we say that they are no longer coherent.
This process is called decoherence. And two states become distinguishable when they are undergoing interactions asymmetrically, right?

Yes. The clearest case is when there is a permanent record of which choice was made: A dot on a photographic plate, for example.

 

[stevendaryl]

Stevendaryl, If one just considers the effect state of the subsystem without the born rule, can you still call it mixed state? Because it seems some people call it mixed state even without born rule by simply considering the state of the subsystem (such as Iija above).

Well, it seems to me that the mathematical treatment of mixed states already assumes the Born rule, indirectly.

 

[zonde]

I think I got idea about decoherence using example of double-slit experiment.
Say we are performing double slit experiment with buckyballs. Buckyball going trough first or second slit we can describe as states A and B. Then we can write superposition as $|A\rangle\pm|B\rangle$ where + or - depends on position of detector.
If we illuminate buckyball after the slits with light of short enough wavelength combined buckyball and photon superposition can be written as $|A\rangle|P_A\rangle\pm|B\rangle|P_B\rangle$ so that interference is still observable if we somehow observe buckyball and photon together. But if we observe buckyball alone we get no interference (as there is no certain phase in complex plain for two states) and it is represented by product state $|A\rangle\otimes|B\rangle$

 

[zonde]

I would like to return to that stevendaryl's post as it perfectly illustrates my confusion with or even objection to decoherence as measurement explanation:

I have a complaint about the claim "We see that [the wavefunctions of] chairs and tables are collapsed". It seems obvious that it's true, but think about what it would mean to be otherwise.

In quantum mechanics, the behavior of a superposition (or mixture--there is a technical difference which isn't important here) is completely determined by the behavior of the corresponding pure states. Suppose you set things up so that there is a consequence of being in one state or another:

1. If the system is in state $|A\rangle$, then consequence $C_A$ happens.
2. If the system is in state $|B\rangle$, then consequence $C_B$ happens.

Then if the consequence itself is governed by quantum-mechanical laws, then we conclude:

If the system is in a superposition/mixture of states $|A\rangle$ and $|B\rangle$, then the consequence will be a superposition/mixture of $C_A$ and $C_B$​

So how does this apply to tables and chairs? Well, suppose you have a folding chair, and for simplicity, we consider two states, either "open" or "folded". So you take a notebook and walk into the room where the chair is, resolved to record what you see:

1. If it is open, you write "open".
2. If it is folded, you write "folded".
3. If it is in a superposition or mixture of these two states, you write "both"

Well, according to QM if you yourself are governed by quantum mechanics, then you'll never write "both". Instead, what will happen is:

1. If it is open, afterward the notebook will contain the word "open"
2. If it is folded, afterward the notebook will contain the word "folded"
3. If it is in a superposition or mixture, afterward the notebook will be in a superposition or mixture of having the word "open" and having the word "folded"

There is no possibility of your writing the word "both" in the notebook (at least not if we assume that you always write "open" if it's open, and "folded" if it's folded)

Another way to say it is that the three possible consequences: write "open", write "folded", write "both" are contradictory; if the first two happen, then the third will never happen.

Note: this is assuming that you yourself are governed by quantum mechanical laws. Some interpretations of quantum mechanics treat observers as special cases. But in these interpretations, observing the chair causes its wavefunction to "collapse". So you wouldn't write "both" in that interpretation, either.

For example I can read here http://arxiv.org/abs/quant-ph?0312059 motivation for decoherence as measurement explanation:
"A book has never been ever observed to be in a state of being both “here” and “there” (i.e., to be in a superposition of macroscopically distinguishable positions), nor does a Schr¨odinger cat that is a superposition of being alive and dead bear much resemblence to reality as we perceive it. The problem is, then, how to reconcile the vastness of the Hilbert space of possible states with the observation of a comparatively few “classical” macrosopic states, defined by having a small number of determinate and robust properties such as position and momentum. Why does the world appear classical to us, in spite of its supposed underlying quantum nature, which would, in principle, allow for arbitrary superpositions?"

Fine, decoherence explains why I would not observe interference. But even with interference (say in buckyball double-slit) world seems fairly classical.

So the measurement problem as I see it is like this: Superposition describes combination of probability amplitudes, but we do not observe probability amplitudes instead we observe individual clicks or classical amplitudes. And I would say that measurement problem is that conversion from probability amplitude to individual events.

 

[craigi]

I would like to return to that stevendaryl's post as it perfectly illustrates my confusion with or even objection to decoherence as measurement explanation:For example I can read here http://arxiv.org/abs/quant-ph?0312059 motivation for decoherence as measurement explanation:
"A book has never been ever observed to be in a state of being both “here” and “there” (i.e., to be in a superposition of macroscopically distinguishable positions), nor does a Schr¨odinger cat that is a superposition of being alive and dead bear much resemblence to reality as we perceive it. The problem is, then, how to reconcile the vastness of the Hilbert space of possible states with the observation of a comparatively few “classical” macrosopic states, defined by having a small number of determinate and robust properties such as position and momentum. Why does the world appear classical to us, in spite of its supposed underlying quantum nature, which would, in principle, allow for arbitrary superpositions?"

Fine, decoherence explains why I would not observe interference. But even with interference (say in buckyball double-slit) world seems fairly classical.

So the measurement problem as I see it is like this: Superposition describes combination of probability amplitudes, but we do not observe probability amplitudes instead we observe individual clicks or classical amplitudes. And I would say that measurement problem is that conversion from probability amplitude to individual events.

Pretty much. More correctly the Measurement Problem refers to the unanswered question from the CI about the missing mechanism for wavefunction collapse.

Later Everett noticed that wavefunction collapse was a completely unrequired additional axiom and instead proposed that the observed state is relative to the observer state.

 

[zonde]

Pretty much. More correctly the Measurement Problem refers to the unanswered question from the CI about the missing mechanism for wavefunction collapse.

As I understand, wavefunction collapse originally referred to single quantum system and the outcome of collapse was single eigenstate of operator. In that case it mostly covers what I had on mind.
But in context of decoherence wave function collapse have somehow morphed into something else. In particular how superposition of states is converted into density matrix. So it describes ensembles rather than individual systems.

 

[lucas_]

In preferred basis, note the basis is a set of vectors.. and particular basis is chosen based on predictability sieve. Do you have any idea how to do experiments in which you can adjust preferred basis in position such that for example a marble can be made to disappear at a certain position and reappear elsewhere? Shielding from microwave radiation can be done according to Mr. Hobba and the marble can be frozen to remove thermal agitations so repreparat. Then if you can change the preferred basis, then the marble can

I have a complaint about the claim "We see that [the wavefunctions of] chairs and tables are collapsed". It seems obvious that it's true, but think about what it would mean to be otherwise.

In quantum mechanics, the behavior of a superposition (or mixture--there is a technical difference which isn't important here) is completely determined by the behavior of the corresponding pure states. Suppose you set things up so that there is a consequence of being in one state or another:

1. If the system is in state $|A\rangle$, then consequence $C_A$ happens.
2. If the system is in state $|B\rangle$, then consequence $C_B$ happens.

Then if the consequence itself is governed by quantum-mechanical laws, then we conclude:

If the system is in a superposition/mixture of states $|A\rangle$ and $|B\rangle$, then the consequence will be a superposition/mixture of $C_A$ and $C_B$​

So how does this apply to tables and chairs? Well, suppose you have a folding chair, and for simplicity, we consider two states, either "open" or "folded". So you take a notebook and walk into the room where the chair is, resolved to record what you see:

1. If it is open, you write "open".
2. If it is folded, you write "folded".
3. If it is in a superposition or mixture of these two states, you write "both"

Well, according to QM if you yourself are governed by quantum mechanics, then you'll never write "both". Instead, what will happen is:

1. If it is open, afterward the notebook will contain the word "open"
2. If it is folded, afterward the notebook will contain the word "folded"
3. If it is in a superposition or mixture, afterward the notebook will be in a superposition or mixture of having the word "open" and having the word "folded"

There is no possibility of your writing the word "both" in the notebook (at least not if we assume that you always write "open" if it's open, and "folded" if it's folded)

Another way to say it is that the three possible consequences: write "open", write "folded", write "both" are contradictory; if the first two happen, then the third will never happen.

Note: this is assuming that you yourself are governed by quantum mechanical laws. Some interpretations of quantum mechanics treat observers as special cases. But in these interpretations, observing the chair causes its wavefunction to "collapse". So you wouldn't write "both" in that interpretation, either.

In preferred basis, note the basis is a set of vectors.. and particular basis is chosen based on predictability sieve. Do you have any idea how to do experiments in which you can adjust preferred basis in position such that for example a marble can be made to disappear at a certain position and reappear elsewhere? Shielding from microwave radiation can be done according to Mr. Hobba and the marble can be frozen to remove thermal agitations so repreparation of preferred basis can be theoretically possible. Any idea how?

 

[stevendaryl]

In preferred basis, note the basis is a set of vectors.. and particular basis is chosen based on predictability sieve. Do you have any idea how to do experiments in which you can adjust preferred basis in position such that for example a marble can be made to disappear at a certain position and reappear elsewhere? Shielding from microwave radiation can be done according to Mr. Hobba and the marble can be frozen to remove thermal agitations so repreparation of preferred basis can be theoretically possible. Any idea how?

For macroscopic objects, in general, the only basis that is feasible to use is the one in which the objects have definite positions at all times (at least up to some trivial uncertainty).

 

[lucas_]

For macroscopic objects, in general, the only basis that is feasible to use is the one in which the objects have definite positions at all times (at least up to some trivial uncertainty).

If you can shield it from CMBR and make it near absolute zero.. why can't you make the position basis jump from say 0.1 to 0.5 x coordinate and make the macroscopic object jumpy? What prevent it? I'm wondering also because it seems some of my food in my very cold metal freezer seems to change position or disappear when I got home.

 

[stevendaryl]

If you can shield it from CMBR and make it near absolute zero.. why can't you make the position basis jump from say 0.1 to 0.5 x coordinate and make the macroscopic object jumpy? What prevent it? I'm wondering also because it seems some of my food in my very cold metal freezer seems to change position or disappear when I got home.

You're right, the sort of measures you're talking about (isolating the system, reducing its temperature) allow for larger objects to be put into coherent superpositions. I think there is a limit though. I'm not sure what the largest object for which superpositions have been observed, but I would guess it would be the size of a molecule.

 

[Nugatory]

why can't you make the position basis jump from say 0.1 to 0.5 x coordinate and make the macroscopic object jumpy?

You're asking about a marble. A marble is not a point particle described by a wave function $\psi(x)$ such that the probability of finding it at position $x$ is given by $|\psi(x)|^2$. Instead, a marble is made up of about $5\times{10}^{22}$ molecules of silicon dioxide (I get that number by assuming that the marble is solid glass, weighs about ten grams, and rounding off enough to do the arithmetic in my head) and its wave function is the product of the product of the wave functions of each of these individual particles in the position basis.

What is the chance of all of these molecules all randomly jumping in the same direction by the same amount at the same time? You are as likely to see a scrambled egg unscramble itself and separate back into white and yolk as it is stirred.

This situation really isn't that different from the way that in classical mechanics the random movements of gas molecules average out in such a way that (for example) any reasonably-sized volume of gas obeys Boyle's Law. In this case, the randomness that you're considering is quantum mechanical in origin, but it still averages out the same way.

Thus stevendaryl's point that "in general, the only basis that is feasible to use is the one in which the objects have definite positions at all times (at least up to some trivial uncertainty)".

some of my food in my very cold metal freezer seems to change position or disappear when I got home.

Better put a smiley next to that... or you'll find someone taking it seriously.