Why is Decoherence Needed in Quantum Mechanics?

[bhobba]

The photons from CMBR are everywhere and can't be shielded

They can be shielded.

Thanks
Bill

 

[stevendaryl]

When you see chairs and tables. They are collapsed. Yet when you see atoms and molecules. They are not collapsed. So how do you reconcile the two? Can we say the position eigenstates are collapsed while the Hamiltonians are uncollapsed.. but how could the atoms/molecules in position collapsed eigenvalues still respond to the environmental hamiltonians?

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.

 

[atyy]

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)

How do we know from QM that the notebook being in a superposition of "open" and "folded" is not "both"?

For example, for spins, $|x\rangle = |u\rangle+|d\rangle$

 

[stevendaryl]

How do we know from QM that the notebook being in a superposition of "open" and "folded" is not "both"?

For example, for spins, $|x\rangle = |u\rangle+|d\rangle$

I think it's pretty obvious that the word "both" is not a superposition of the words "open" and "folded". But you could redo the experiment, with the plan of writing "Mixed" instead of "both". If the word "both" is a superposition of "open" and "folded", then the word "Mixed" can't also be the same superposition.

 

[atyy]

I think it's pretty obvious that the word "both" is not a superposition of the words "open" and "folded". But you could redo the experiment, with the plan of writing "Mixed" instead of "both". If the word "both" is a superposition of "open" and "folded", then the word "Mixed" can't also be the same superposition.

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

 

[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.