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Isn't the detector always measuring, and thus always collapsing the state?
How can quantum tunnelling lead to spontaneous decay?Can the absence of information provide which-way knowledge?Does spontaneous emission count as a measurement?Is Schrodinger's Cat a real conceptual problem or just a problem with approximations?Does the reduced density matrix describes a real mixed state?Measurement of quantum stateCollapsing a wave function without hitting the detectorGeiger counter in the Schrodinger's cat experimentWhat is the quantum state of spin 1/2 particle in $y$ direction?Why does superposition principle and Copenhagen interpretation not contradict with themselves?Confusion about mixed states and pure states
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have a radioactive particle in a box, prepared so as to initially be in a pure state
$psi_0 =1 theta_U+ 0 theta_D$
(U is Undecayed, D is Decayed).
I put a Geiger counter in the box.
Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger
$psi_t =a theta_U+ b theta_D$
Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.
However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.
But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.
I want to know where the mistake lies.
In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is
$psi_t =1 theta_U+ 0 theta_D$
which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).
How do I deal with this constant 'passive' measuring?
quantum-mechanics measurement-problem wavefunction-collapse schroedingers-cat
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
$endgroup$
add a comment
|
$begingroup$
I have a radioactive particle in a box, prepared so as to initially be in a pure state
$psi_0 =1 theta_U+ 0 theta_D$
(U is Undecayed, D is Decayed).
I put a Geiger counter in the box.
Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger
$psi_t =a theta_U+ b theta_D$
Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.
However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.
But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.
I want to know where the mistake lies.
In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is
$psi_t =1 theta_U+ 0 theta_D$
which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).
How do I deal with this constant 'passive' measuring?
quantum-mechanics measurement-problem wavefunction-collapse schroedingers-cat
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
$endgroup$
2
$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
1
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago
add a comment
|
$begingroup$
I have a radioactive particle in a box, prepared so as to initially be in a pure state
$psi_0 =1 theta_U+ 0 theta_D$
(U is Undecayed, D is Decayed).
I put a Geiger counter in the box.
Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger
$psi_t =a theta_U+ b theta_D$
Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.
However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.
But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.
I want to know where the mistake lies.
In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is
$psi_t =1 theta_U+ 0 theta_D$
which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).
How do I deal with this constant 'passive' measuring?
quantum-mechanics measurement-problem wavefunction-collapse schroedingers-cat
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
$endgroup$
I have a radioactive particle in a box, prepared so as to initially be in a pure state
$psi_0 =1 theta_U+ 0 theta_D$
(U is Undecayed, D is Decayed).
I put a Geiger counter in the box.
Over time (t), the theory says that the state should evolve into a pure state that is a superposition of Undecayed and Decayed, with the Decayed part getting bigger and bigger
$psi_t =a theta_U+ b theta_D$
Eventually the counter will 'click', indicating that the particle has Decayed. Now I know that the state is 100% Decayed.
However, before this happened, the silence of the counter also indicated that the particle hadn't Decayed yet. So all the time up to that point, I also knew that the state was 100% Undecayed.
But this would be contradicting what the theory suggests (a superposition with a non zero contribution of the Decayed state, after some time), so I'm guessing it's an incorrect way of analysing the experiment.
I want to know where the mistake lies.
In other words, it seems to me the Geiger counter is always measuring the state of the particle. Silence means Undecayed, click means Decayed. So the particle would never actually Decay since I continuously know its state is
$psi_t =1 theta_U+ 0 theta_D$
which means its chance of decaying would be perpetually zero (Zeno's effect, I've heard?).
How do I deal with this constant 'passive' measuring?
quantum-mechanics measurement-problem wavefunction-collapse schroedingers-cat
quantum-mechanics measurement-problem wavefunction-collapse schroedingers-cat
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
edited 6 hours ago
Qmechanic♦
114k13 gold badges225 silver badges1353 bronze badges
114k13 gold badges225 silver badges1353 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
asked 8 hours ago
Juan PerezJuan Perez
1861 silver badge12 bronze badges
1861 silver badge12 bronze badges
2
$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
1
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago
add a comment
|
2
$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
1
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago
2
2
$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
1
1
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
The physical answer to your question, as best answered in the framework of QM, is: The actual wave function, as Macro Ocram said, is more than just a two-state system and has a position wave function. When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you can at least derive a characteristic decay time(*), but the notion of "reaches the detector" is too imprecise to get a probability distribution for the decay time.
(*)Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. But that's a bit of a tangent!
So what about the Zeno effect? Is the chance of decaying perpetually zero? You might argue: "even though it is not a 2-level system, it should be the case that at each moment in time, the detector projects away the part of $psi(x)$ which is inside the detector, because we know it hasn't gotten there yet". And actually, doing this does cause the wave function to never arrive in the region, as you said (I actually just modeled this problem for my thesis). This result is inconsistent with experiment, so the only conclusion to draw from this is: continuously-looking measurement cannot be modeled by projection inside the detector at every instant in time.
So what is the correct way to model this experiment? In particular, could you find a probability density function of time, $rho(t)$, so that when you integrate it from time $t_a$ to $t_b$ it gives you the probability that the particle was found in that time interval? This is actually an open question and there are kind of ad-hoc approaches to answering the question but no agreed-upon approach which works in general & has been experimentally tested. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a major weak point of the theory, which needs work! If you don't want to take my word for it, take a look at Gonzal Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a satisfactory way.
answered 6 hours ago
doublefelixdoublefelix
1,2805 silver badges31 bronze badges
$endgroup$
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
add a comment
|
$begingroup$
My answer without much thought is that the undecayed state is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
add a comment
|
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2 Answers
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2 Answers
2
active
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$begingroup$
Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
The physical answer to your question, as best answered in the framework of QM, is: The actual wave function, as Macro Ocram said, is more than just a two-state system and has a position wave function. When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you can at least derive a characteristic decay time(*), but the notion of "reaches the detector" is too imprecise to get a probability distribution for the decay time.
(*)Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. But that's a bit of a tangent!
So what about the Zeno effect? Is the chance of decaying perpetually zero? You might argue: "even though it is not a 2-level system, it should be the case that at each moment in time, the detector projects away the part of $psi(x)$ which is inside the detector, because we know it hasn't gotten there yet". And actually, doing this does cause the wave function to never arrive in the region, as you said (I actually just modeled this problem for my thesis). This result is inconsistent with experiment, so the only conclusion to draw from this is: continuously-looking measurement cannot be modeled by projection inside the detector at every instant in time.
So what is the correct way to model this experiment? In particular, could you find a probability density function of time, $rho(t)$, so that when you integrate it from time $t_a$ to $t_b$ it gives you the probability that the particle was found in that time interval? This is actually an open question and there are kind of ad-hoc approaches to answering the question but no agreed-upon approach which works in general & has been experimentally tested. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a major weak point of the theory, which needs work! If you don't want to take my word for it, take a look at Gonzal Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a satisfactory way.
answered 6 hours ago
doublefelixdoublefelix
1,2805 silver badges31 bronze badges
$endgroup$
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
add a comment
|
$begingroup$
Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
The physical answer to your question, as best answered in the framework of QM, is: The actual wave function, as Macro Ocram said, is more than just a two-state system and has a position wave function. When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you can at least derive a characteristic decay time(*), but the notion of "reaches the detector" is too imprecise to get a probability distribution for the decay time.
(*)Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. But that's a bit of a tangent!
So what about the Zeno effect? Is the chance of decaying perpetually zero? You might argue: "even though it is not a 2-level system, it should be the case that at each moment in time, the detector projects away the part of $psi(x)$ which is inside the detector, because we know it hasn't gotten there yet". And actually, doing this does cause the wave function to never arrive in the region, as you said (I actually just modeled this problem for my thesis). This result is inconsistent with experiment, so the only conclusion to draw from this is: continuously-looking measurement cannot be modeled by projection inside the detector at every instant in time.
So what is the correct way to model this experiment? In particular, could you find a probability density function of time, $rho(t)$, so that when you integrate it from time $t_a$ to $t_b$ it gives you the probability that the particle was found in that time interval? This is actually an open question and there are kind of ad-hoc approaches to answering the question but no agreed-upon approach which works in general & has been experimentally tested. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a major weak point of the theory, which needs work! If you don't want to take my word for it, take a look at Gonzal Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a satisfactory way.
answered 6 hours ago
doublefelixdoublefelix
1,2805 silver badges31 bronze badges
$endgroup$
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
add a comment
|
$begingroup$
Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
The physical answer to your question, as best answered in the framework of QM, is: The actual wave function, as Macro Ocram said, is more than just a two-state system and has a position wave function. When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you can at least derive a characteristic decay time(*), but the notion of "reaches the detector" is too imprecise to get a probability distribution for the decay time.
(*)Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. But that's a bit of a tangent!
So what about the Zeno effect? Is the chance of decaying perpetually zero? You might argue: "even though it is not a 2-level system, it should be the case that at each moment in time, the detector projects away the part of $psi(x)$ which is inside the detector, because we know it hasn't gotten there yet". And actually, doing this does cause the wave function to never arrive in the region, as you said (I actually just modeled this problem for my thesis). This result is inconsistent with experiment, so the only conclusion to draw from this is: continuously-looking measurement cannot be modeled by projection inside the detector at every instant in time.
So what is the correct way to model this experiment? In particular, could you find a probability density function of time, $rho(t)$, so that when you integrate it from time $t_a$ to $t_b$ it gives you the probability that the particle was found in that time interval? This is actually an open question and there are kind of ad-hoc approaches to answering the question but no agreed-upon approach which works in general & has been experimentally tested. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a major weak point of the theory, which needs work! If you don't want to take my word for it, take a look at Gonzal Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a satisfactory way.
answered 6 hours ago
doublefelixdoublefelix
1,2805 silver badges31 bronze badges
$endgroup$
Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which stray outside of that structure.
The physical answer to your question, as best answered in the framework of QM, is: The actual wave function, as Macro Ocram said, is more than just a two-state system and has a position wave function. When this wave function "reaches the detector" (though it probably has some nonzero value in the detector the entire time) the Geiger counter registers a decay. Using this you can at least derive a characteristic decay time(*), but the notion of "reaches the detector" is too imprecise to get a probability distribution for the decay time.
(*)Sure you can in principle hope to position representation to get a characteristic decay time like this, but it might not be as easy as it sounds because QFT has issues with position space wave functions, and you'd need QFT to describe annihilation/creation of particles. But that's a bit of a tangent!
So what about the Zeno effect? Is the chance of decaying perpetually zero? You might argue: "even though it is not a 2-level system, it should be the case that at each moment in time, the detector projects away the part of $psi(x)$ which is inside the detector, because we know it hasn't gotten there yet". And actually, doing this does cause the wave function to never arrive in the region, as you said (I actually just modeled this problem for my thesis). This result is inconsistent with experiment, so the only conclusion to draw from this is: continuously-looking measurement cannot be modeled by projection inside the detector at every instant in time.
So what is the correct way to model this experiment? In particular, could you find a probability density function of time, $rho(t)$, so that when you integrate it from time $t_a$ to $t_b$ it gives you the probability that the particle was found in that time interval? This is actually an open question and there are kind of ad-hoc approaches to answering the question but no agreed-upon approach which works in general & has been experimentally tested. Predicting "when" something happens in Quantum Mechanics (or the probability density for when it happens) is a major weak point of the theory, which needs work! If you don't want to take my word for it, take a look at Gonzal Muga's textbook Time in Quantum Mechanics which is a good summary of different approaches on time problems in QM which are still open to be solved today in a satisfactory way.
answered 6 hours ago
doublefelixdoublefelix
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edited 19 mins ago
answered 6 hours ago
doublefelixdoublefelix
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answered 6 hours ago
doublefelixdoublefelix
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answered 6 hours ago
doublefelixdoublefelix
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$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
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– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
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– doublefelix
23 mins ago
add a comment
|
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
It's true that I've always thought of QM variables becoming defined only when measured by a detector (an electron "does not have a position" until measured). But as Macro Ocram says, this particle can decay without having to interact with a detector. So the variable "did it decay?" may have a definitive value even if not measured. Perhaps such "variable" is not a proper "QM observable" with a corresponding hermitian operator, and thus the theory can't study it properly? I'm not sure if that's what you were aiming at.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
$begingroup$
I edited the answer to make it clearer, let me know if you still have questions. And yes you can have a particle "decay" without interacting with a detector, in the sense that the "decayed" term is much bigger than the "hasn't decayed" term, but it's almost never true that you'd have exactly 0 for either one of those terms for an unstable particle in a real-life situation.
$endgroup$
– doublefelix
23 mins ago
add a comment
|
$begingroup$
My answer without much thought is that the undecayed state is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
answered 6 hours ago
Marco OcramMarco Ocram
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$endgroup$
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
add a comment
|
$begingroup$
My answer without much thought is that the undecayed state is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
add a comment
|
$begingroup$
My answer without much thought is that the undecayed state is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
My answer without much thought is that the undecayed state is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a significant amplitude in the vicinity of the detector that the counter clicks.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
answered 6 hours ago
Marco OcramMarco Ocram
1775 bronze badges
1775 bronze badges
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marco Ocram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
add a comment
|
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
$begingroup$
So the particle could decay without needing to interact with the detector, and therefore the "variable" (Decayed or Undecayed), can take a definite value even if not measured? Also: I could discard the box and simply shove the particle into the counter, if it makes this easier.
$endgroup$
– Juan Perez
3 hours ago
add a comment
|
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$begingroup$
Great question. Similar to this: physics.stackexchange.com/q/232502 but I look forward to the answer.
$endgroup$
– JPattarini
8 hours ago
1
$begingroup$
Decay is spontaneous. In the way it is semi-classically modelled it is indeed not clear what kind of measurement has to be taken into account. I asked a couple of questions along these lines: physics.stackexchange.com/q/258104/109928 and physics.stackexchange.com/questions/258256/… Anyway, it seems the Geiger counter is not part of the picture, it simply registers the fact that a decay did occur.
$endgroup$
– Stéphane Rollandin
7 hours ago