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Which basis does the wavefunction collapse to?
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Which basis does the wavefunction collapse to?
What are the differences between a $psi$-epistemic ontological model and a $psi$-ontic model of quantum mechanics, exactly?wavefunction collapse and uncertainty principleWhat happens after the collapse of a wavefunction?How does wave function collapse when I measure position?Is something beyond the material needed to solve the Von Neumann Chain?Collapse of wave functionDoes measurement of momentum always collapse the wave function into a plane wave?Wave function. Measurement of the absence
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?
quantum-mechanics wavefunction wavefunction-collapse
$endgroup$
add a comment |
$begingroup$
When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?
quantum-mechanics wavefunction wavefunction-collapse
$endgroup$
add a comment |
$begingroup$
When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?
quantum-mechanics wavefunction wavefunction-collapse
$endgroup$
When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?
quantum-mechanics wavefunction wavefunction-collapse
quantum-mechanics wavefunction wavefunction-collapse
edited 8 hours ago
Qmechanic♦
112k13 gold badges219 silver badges1331 bronze badges
112k13 gold badges219 silver badges1331 bronze badges
asked 8 hours ago
Jeff BassJeff Bass
464 bronze badges
464 bronze badges
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3 Answers
3
active
oldest
votes
$begingroup$
The system doesn't "know" anything.
The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:
Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.
That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.
$endgroup$
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
add a comment |
$begingroup$
The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.
Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction
$endgroup$
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
$begingroup$
I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.
$endgroup$
add a comment |
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3 Answers
3
active
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votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The system doesn't "know" anything.
The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:
Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.
That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.
$endgroup$
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
add a comment |
$begingroup$
The system doesn't "know" anything.
The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:
Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.
That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.
$endgroup$
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
add a comment |
$begingroup$
The system doesn't "know" anything.
The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:
Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.
That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.
$endgroup$
The system doesn't "know" anything.
The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:
Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.
That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.
answered 8 hours ago
ACuriousMind♦ACuriousMind
75.5k18 gold badges139 silver badges352 bronze badges
75.5k18 gold badges139 silver badges352 bronze badges
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
add a comment |
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
$begingroup$
Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
$endgroup$
– Jeff Bass
1 hour ago
add a comment |
$begingroup$
The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.
Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction
$endgroup$
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
$begingroup$
The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.
Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction
$endgroup$
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
$begingroup$
The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.
Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction
$endgroup$
The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.
Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction
edited 8 hours ago
answered 8 hours ago
Aaron StevensAaron Stevens
21k4 gold badges36 silver badges75 bronze badges
21k4 gold badges36 silver badges75 bronze badges
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
This does not answer the question. The question is not about representing a wavefunction in different bases.
$endgroup$
– eigenvalue
4 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
$begingroup$
@eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
$begingroup$
I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.
$endgroup$
add a comment |
$begingroup$
I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.
$endgroup$
add a comment |
$begingroup$
I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.
$endgroup$
I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.
answered 8 hours ago
Andrew SteaneAndrew Steane
9,8461 gold badge12 silver badges52 bronze badges
9,8461 gold badge12 silver badges52 bronze badges
add a comment |
add a comment |
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