How does the Uncertainty Principle work? [duplicate]Can the Heisenberg Uncertainty Principle be explained intuitively?Is the uncertainty principle a property of elementary particles or a result of our measurement tools?EPR paradox and uncertainty principleWhat is the meaning of uncertainty in Heisenberg's uncertainty principle?Application of Heisenberg's uncertainty principleIs uncertainty principle a technical difficulty in measurement?Heisenberg uncertainty principle clarificationReason for Uncertainty principleProblem understanding Heisenberg's uncertainty principleIf uncertainty principle is explained by wave function, then doesn't wave function collapse when we measure position or momentum?Does the Uncertainty Principle really rule out the existence of definite trajectory of electrons?A question on Heisenberg's uncertainty principle
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How does the Uncertainty Principle work? [duplicate]
Can the Heisenberg Uncertainty Principle be explained intuitively?Is the uncertainty principle a property of elementary particles or a result of our measurement tools?EPR paradox and uncertainty principleWhat is the meaning of uncertainty in Heisenberg's uncertainty principle?Application of Heisenberg's uncertainty principleIs uncertainty principle a technical difficulty in measurement?Heisenberg uncertainty principle clarificationReason for Uncertainty principleProblem understanding Heisenberg's uncertainty principleIf uncertainty principle is explained by wave function, then doesn't wave function collapse when we measure position or momentum?Does the Uncertainty Principle really rule out the existence of definite trajectory of electrons?A question on Heisenberg's uncertainty principle
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$begingroup$
This question already has an answer here:
Is the uncertainty principle a property of elementary particles or a result of our measurement tools?
9 answers
Can the Heisenberg Uncertainty Principle be explained intuitively?
17 answers
In my textbook they have explained the Heisenberg's uncertainty principle using the example where you cannot measure the position of the electron using the photon because the wavelength of the photon needs to be small which then causes it to have high energy which would knock off the electron by collision. So my first question is isn't this the observer effect?
My second question is, if I were to measure the position (the point with highest probability) how does that cause an increase in the uncertainty in momentum? How does that work?
heisenberg-uncertainty-principle
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
marked as duplicate by Qmechanic♦ Jun 11 at 18:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is the uncertainty principle a property of elementary particles or a result of our measurement tools?
9 answers
Can the Heisenberg Uncertainty Principle be explained intuitively?
17 answers
In my textbook they have explained the Heisenberg's uncertainty principle using the example where you cannot measure the position of the electron using the photon because the wavelength of the photon needs to be small which then causes it to have high energy which would knock off the electron by collision. So my first question is isn't this the observer effect?
My second question is, if I were to measure the position (the point with highest probability) how does that cause an increase in the uncertainty in momentum? How does that work?
heisenberg-uncertainty-principle
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
marked as duplicate by Qmechanic♦ Jun 11 at 18:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
$endgroup$
– John Forkosh
Jun 11 at 9:40
$begingroup$
Observer effect.
$endgroup$
– Cosmas Zachos
Jun 11 at 10:13
$begingroup$
Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
$endgroup$
– Alchimista
Jun 11 at 12:44
add a comment |
$begingroup$
This question already has an answer here:
Is the uncertainty principle a property of elementary particles or a result of our measurement tools?
9 answers
Can the Heisenberg Uncertainty Principle be explained intuitively?
17 answers
In my textbook they have explained the Heisenberg's uncertainty principle using the example where you cannot measure the position of the electron using the photon because the wavelength of the photon needs to be small which then causes it to have high energy which would knock off the electron by collision. So my first question is isn't this the observer effect?
My second question is, if I were to measure the position (the point with highest probability) how does that cause an increase in the uncertainty in momentum? How does that work?
heisenberg-uncertainty-principle
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This question already has an answer here:
Is the uncertainty principle a property of elementary particles or a result of our measurement tools?
9 answers
Can the Heisenberg Uncertainty Principle be explained intuitively?
17 answers
In my textbook they have explained the Heisenberg's uncertainty principle using the example where you cannot measure the position of the electron using the photon because the wavelength of the photon needs to be small which then causes it to have high energy which would knock off the electron by collision. So my first question is isn't this the observer effect?
My second question is, if I were to measure the position (the point with highest probability) how does that cause an increase in the uncertainty in momentum? How does that work?
This question already has an answer here:
Is the uncertainty principle a property of elementary particles or a result of our measurement tools?
9 answers
Can the Heisenberg Uncertainty Principle be explained intuitively?
17 answers
heisenberg-uncertainty-principle
heisenberg-uncertainty-principle
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
edited Jun 11 at 10:05
Qmechanic♦
110k122101289
110k122101289
asked Jun 11 at 9:15
user662650user662650
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 11 at 9:15
user662650user662650
211
asked Jun 11 at 9:15
user662650user662650
211
211
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user662650 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
marked as duplicate by Qmechanic♦ Jun 11 at 18:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Qmechanic♦ Jun 11 at 18:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
$endgroup$
– John Forkosh
Jun 11 at 9:40
$begingroup$
Observer effect.
$endgroup$
– Cosmas Zachos
Jun 11 at 10:13
$begingroup$
Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
$endgroup$
– Alchimista
Jun 11 at 12:44
add a comment |
2
$begingroup$
Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
$endgroup$
– John Forkosh
Jun 11 at 9:40
$begingroup$
Observer effect.
$endgroup$
– Cosmas Zachos
Jun 11 at 10:13
$begingroup$
Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
$endgroup$
– Alchimista
Jun 11 at 12:44
2
2
$begingroup$
Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
$endgroup$
– John Forkosh
Jun 11 at 9:40
$begingroup$
Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
$endgroup$
– John Forkosh
Jun 11 at 9:40
$begingroup$
Observer effect.
$endgroup$
– Cosmas Zachos
Jun 11 at 10:13
$begingroup$
Observer effect.
$endgroup$
– Cosmas Zachos
Jun 11 at 10:13
$begingroup$
Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
$endgroup$
– Alchimista
Jun 11 at 12:44
$begingroup$
Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
$endgroup$
– Alchimista
Jun 11 at 12:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The content of the Uncertainty Principle (UP) is apparently simple. Nevertheless, after its statement by Werner Heisenberg, it underwent an importand mutation after Robertson's derivation of a general inequality for the product of the variances of the statistical distributions of the velues of two non-commuting operators.
After Robertson's work UP relations are presented as a statement about the distribution of the values of two no-commuting observables, when measured in a given quantum state. As such, there is no reference to experimental disturbs and even the observer effect plays a minor role. Actually, the statistical intepretation of UP implies only that, if an ensemble of quantum systems has been prepared in a given state $left| 0 right>$, independent measurements of observable represented by operators $A$ and $B$ would imply that the distribution of measurements of $A$ and $B$ should be subject to the inequality
$$
sigma^2_A sigma^2_B geq big| dfrac12i langle 0[A,B] 0rangle big|^2.
$$
As such, the UP would say something quite different from the original content of Heisenberg's microscope. Therefore, it is usual to find, even here on PhysicsSE, vigorous denial that UP are directly connected with experimental errors.
However, Heisenbergs pont of view should not be confused with the unavoidable presence of uncertainty in any practical measurement. It was stressing something else, which has a common origin in the non-commutation of some pairs of operators representing observables, but does not coincides with Robertson's statistical result.
This last point has emerged quite clear by a revival of interest, in the last couples of decades, for the physical content of UP with respect to the prblem of (almost) simultaneous mesurements of non-communting observables.
Indeed, the non commuting of two operators, according to the basic postulates of QM implies that it is not possible to measure at the same time the two quantites. The reason is that one of the basic postulates of QM says that the effect of a measurement of a quantity A is to bring the quantum system into one of the eigenstates of the corresponding operator. However two non-commuting operators do not have a set of common eigenvectors, then the theoretical impossibility of a simultaneous measurement.
In recent years, people have started to analyze quantitatively such impossibility, asking questions about how good could be, on theoretical basis, a joint measurement of two non-commuting observables. See for instance the paper by Cyril Branciard on PNAS and references therein contained.
Under such new viewpoint, it is possible to recover in a semiquantitative form the original Heisenberg's formulation, although the exact value of the "uncertainty" may be slightly different.
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
$endgroup$
add a comment |
$begingroup$
Such example is misleading (though very common in all quantum mechanics textbooks) as if focused the attention on the experimental details, seemingly addressing the uncertainty to practical measurements practices.
The uncertainty principle is solely due to the fact that two observables do not commute; whenever that is the case you may see that the product of their standard deviations must be at least equal to their non-zero commutator: whenever their non-zero commutator happens to be a constant (like in the case of position and momentum) this implies that the smaller the uncertainty on one, the bigger the uncertainty on the other (in order for the product to be a constant).
The uncertainty in quantum mechanics is theoretical, by definition, and not due to experimental errors.
answered Jun 11 at 9:41
gentedgented
4,959920
$endgroup$
1
$begingroup$
I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
$endgroup$
– Umaxo
Jun 11 at 11:22
3
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
$endgroup$
– gented
Jun 11 at 11:50
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The content of the Uncertainty Principle (UP) is apparently simple. Nevertheless, after its statement by Werner Heisenberg, it underwent an importand mutation after Robertson's derivation of a general inequality for the product of the variances of the statistical distributions of the velues of two non-commuting operators.
After Robertson's work UP relations are presented as a statement about the distribution of the values of two no-commuting observables, when measured in a given quantum state. As such, there is no reference to experimental disturbs and even the observer effect plays a minor role. Actually, the statistical intepretation of UP implies only that, if an ensemble of quantum systems has been prepared in a given state $left| 0 right>$, independent measurements of observable represented by operators $A$ and $B$ would imply that the distribution of measurements of $A$ and $B$ should be subject to the inequality
$$
sigma^2_A sigma^2_B geq big| dfrac12i langle 0[A,B] 0rangle big|^2.
$$
As such, the UP would say something quite different from the original content of Heisenberg's microscope. Therefore, it is usual to find, even here on PhysicsSE, vigorous denial that UP are directly connected with experimental errors.
However, Heisenbergs pont of view should not be confused with the unavoidable presence of uncertainty in any practical measurement. It was stressing something else, which has a common origin in the non-commutation of some pairs of operators representing observables, but does not coincides with Robertson's statistical result.
This last point has emerged quite clear by a revival of interest, in the last couples of decades, for the physical content of UP with respect to the prblem of (almost) simultaneous mesurements of non-communting observables.
Indeed, the non commuting of two operators, according to the basic postulates of QM implies that it is not possible to measure at the same time the two quantites. The reason is that one of the basic postulates of QM says that the effect of a measurement of a quantity A is to bring the quantum system into one of the eigenstates of the corresponding operator. However two non-commuting operators do not have a set of common eigenvectors, then the theoretical impossibility of a simultaneous measurement.
In recent years, people have started to analyze quantitatively such impossibility, asking questions about how good could be, on theoretical basis, a joint measurement of two non-commuting observables. See for instance the paper by Cyril Branciard on PNAS and references therein contained.
Under such new viewpoint, it is possible to recover in a semiquantitative form the original Heisenberg's formulation, although the exact value of the "uncertainty" may be slightly different.
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
$endgroup$
add a comment |
$begingroup$
The content of the Uncertainty Principle (UP) is apparently simple. Nevertheless, after its statement by Werner Heisenberg, it underwent an importand mutation after Robertson's derivation of a general inequality for the product of the variances of the statistical distributions of the velues of two non-commuting operators.
After Robertson's work UP relations are presented as a statement about the distribution of the values of two no-commuting observables, when measured in a given quantum state. As such, there is no reference to experimental disturbs and even the observer effect plays a minor role. Actually, the statistical intepretation of UP implies only that, if an ensemble of quantum systems has been prepared in a given state $left| 0 right>$, independent measurements of observable represented by operators $A$ and $B$ would imply that the distribution of measurements of $A$ and $B$ should be subject to the inequality
$$
sigma^2_A sigma^2_B geq big| dfrac12i langle 0[A,B] 0rangle big|^2.
$$
As such, the UP would say something quite different from the original content of Heisenberg's microscope. Therefore, it is usual to find, even here on PhysicsSE, vigorous denial that UP are directly connected with experimental errors.
However, Heisenbergs pont of view should not be confused with the unavoidable presence of uncertainty in any practical measurement. It was stressing something else, which has a common origin in the non-commutation of some pairs of operators representing observables, but does not coincides with Robertson's statistical result.
This last point has emerged quite clear by a revival of interest, in the last couples of decades, for the physical content of UP with respect to the prblem of (almost) simultaneous mesurements of non-communting observables.
Indeed, the non commuting of two operators, according to the basic postulates of QM implies that it is not possible to measure at the same time the two quantites. The reason is that one of the basic postulates of QM says that the effect of a measurement of a quantity A is to bring the quantum system into one of the eigenstates of the corresponding operator. However two non-commuting operators do not have a set of common eigenvectors, then the theoretical impossibility of a simultaneous measurement.
In recent years, people have started to analyze quantitatively such impossibility, asking questions about how good could be, on theoretical basis, a joint measurement of two non-commuting observables. See for instance the paper by Cyril Branciard on PNAS and references therein contained.
Under such new viewpoint, it is possible to recover in a semiquantitative form the original Heisenberg's formulation, although the exact value of the "uncertainty" may be slightly different.
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
$endgroup$
add a comment |
$begingroup$
The content of the Uncertainty Principle (UP) is apparently simple. Nevertheless, after its statement by Werner Heisenberg, it underwent an importand mutation after Robertson's derivation of a general inequality for the product of the variances of the statistical distributions of the velues of two non-commuting operators.
After Robertson's work UP relations are presented as a statement about the distribution of the values of two no-commuting observables, when measured in a given quantum state. As such, there is no reference to experimental disturbs and even the observer effect plays a minor role. Actually, the statistical intepretation of UP implies only that, if an ensemble of quantum systems has been prepared in a given state $left| 0 right>$, independent measurements of observable represented by operators $A$ and $B$ would imply that the distribution of measurements of $A$ and $B$ should be subject to the inequality
$$
sigma^2_A sigma^2_B geq big| dfrac12i langle 0[A,B] 0rangle big|^2.
$$
As such, the UP would say something quite different from the original content of Heisenberg's microscope. Therefore, it is usual to find, even here on PhysicsSE, vigorous denial that UP are directly connected with experimental errors.
However, Heisenbergs pont of view should not be confused with the unavoidable presence of uncertainty in any practical measurement. It was stressing something else, which has a common origin in the non-commutation of some pairs of operators representing observables, but does not coincides with Robertson's statistical result.
This last point has emerged quite clear by a revival of interest, in the last couples of decades, for the physical content of UP with respect to the prblem of (almost) simultaneous mesurements of non-communting observables.
Indeed, the non commuting of two operators, according to the basic postulates of QM implies that it is not possible to measure at the same time the two quantites. The reason is that one of the basic postulates of QM says that the effect of a measurement of a quantity A is to bring the quantum system into one of the eigenstates of the corresponding operator. However two non-commuting operators do not have a set of common eigenvectors, then the theoretical impossibility of a simultaneous measurement.
In recent years, people have started to analyze quantitatively such impossibility, asking questions about how good could be, on theoretical basis, a joint measurement of two non-commuting observables. See for instance the paper by Cyril Branciard on PNAS and references therein contained.
Under such new viewpoint, it is possible to recover in a semiquantitative form the original Heisenberg's formulation, although the exact value of the "uncertainty" may be slightly different.
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
$endgroup$
The content of the Uncertainty Principle (UP) is apparently simple. Nevertheless, after its statement by Werner Heisenberg, it underwent an importand mutation after Robertson's derivation of a general inequality for the product of the variances of the statistical distributions of the velues of two non-commuting operators.
After Robertson's work UP relations are presented as a statement about the distribution of the values of two no-commuting observables, when measured in a given quantum state. As such, there is no reference to experimental disturbs and even the observer effect plays a minor role. Actually, the statistical intepretation of UP implies only that, if an ensemble of quantum systems has been prepared in a given state $left| 0 right>$, independent measurements of observable represented by operators $A$ and $B$ would imply that the distribution of measurements of $A$ and $B$ should be subject to the inequality
$$
sigma^2_A sigma^2_B geq big| dfrac12i langle 0[A,B] 0rangle big|^2.
$$
As such, the UP would say something quite different from the original content of Heisenberg's microscope. Therefore, it is usual to find, even here on PhysicsSE, vigorous denial that UP are directly connected with experimental errors.
However, Heisenbergs pont of view should not be confused with the unavoidable presence of uncertainty in any practical measurement. It was stressing something else, which has a common origin in the non-commutation of some pairs of operators representing observables, but does not coincides with Robertson's statistical result.
This last point has emerged quite clear by a revival of interest, in the last couples of decades, for the physical content of UP with respect to the prblem of (almost) simultaneous mesurements of non-communting observables.
Indeed, the non commuting of two operators, according to the basic postulates of QM implies that it is not possible to measure at the same time the two quantites. The reason is that one of the basic postulates of QM says that the effect of a measurement of a quantity A is to bring the quantum system into one of the eigenstates of the corresponding operator. However two non-commuting operators do not have a set of common eigenvectors, then the theoretical impossibility of a simultaneous measurement.
In recent years, people have started to analyze quantitatively such impossibility, asking questions about how good could be, on theoretical basis, a joint measurement of two non-commuting observables. See for instance the paper by Cyril Branciard on PNAS and references therein contained.
Under such new viewpoint, it is possible to recover in a semiquantitative form the original Heisenberg's formulation, although the exact value of the "uncertainty" may be slightly different.
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
answered Jun 11 at 10:45
GiorgioPGiorgioP
6,1892934
6,1892934
add a comment |
add a comment |
$begingroup$
Such example is misleading (though very common in all quantum mechanics textbooks) as if focused the attention on the experimental details, seemingly addressing the uncertainty to practical measurements practices.
The uncertainty principle is solely due to the fact that two observables do not commute; whenever that is the case you may see that the product of their standard deviations must be at least equal to their non-zero commutator: whenever their non-zero commutator happens to be a constant (like in the case of position and momentum) this implies that the smaller the uncertainty on one, the bigger the uncertainty on the other (in order for the product to be a constant).
The uncertainty in quantum mechanics is theoretical, by definition, and not due to experimental errors.
answered Jun 11 at 9:41
gentedgented
4,959920
$endgroup$
1
$begingroup$
I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
$endgroup$
– Umaxo
Jun 11 at 11:22
3
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
$endgroup$
– gented
Jun 11 at 11:50
add a comment |
$begingroup$
Such example is misleading (though very common in all quantum mechanics textbooks) as if focused the attention on the experimental details, seemingly addressing the uncertainty to practical measurements practices.
The uncertainty principle is solely due to the fact that two observables do not commute; whenever that is the case you may see that the product of their standard deviations must be at least equal to their non-zero commutator: whenever their non-zero commutator happens to be a constant (like in the case of position and momentum) this implies that the smaller the uncertainty on one, the bigger the uncertainty on the other (in order for the product to be a constant).
The uncertainty in quantum mechanics is theoretical, by definition, and not due to experimental errors.
answered Jun 11 at 9:41
gentedgented
4,959920
$endgroup$
1
$begingroup$
I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
$endgroup$
– Umaxo
Jun 11 at 11:22
3
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
$endgroup$
– gented
Jun 11 at 11:50
add a comment |
$begingroup$
Such example is misleading (though very common in all quantum mechanics textbooks) as if focused the attention on the experimental details, seemingly addressing the uncertainty to practical measurements practices.
The uncertainty principle is solely due to the fact that two observables do not commute; whenever that is the case you may see that the product of their standard deviations must be at least equal to their non-zero commutator: whenever their non-zero commutator happens to be a constant (like in the case of position and momentum) this implies that the smaller the uncertainty on one, the bigger the uncertainty on the other (in order for the product to be a constant).
The uncertainty in quantum mechanics is theoretical, by definition, and not due to experimental errors.
answered Jun 11 at 9:41
gentedgented
4,959920
$endgroup$
Such example is misleading (though very common in all quantum mechanics textbooks) as if focused the attention on the experimental details, seemingly addressing the uncertainty to practical measurements practices.
The uncertainty principle is solely due to the fact that two observables do not commute; whenever that is the case you may see that the product of their standard deviations must be at least equal to their non-zero commutator: whenever their non-zero commutator happens to be a constant (like in the case of position and momentum) this implies that the smaller the uncertainty on one, the bigger the uncertainty on the other (in order for the product to be a constant).
The uncertainty in quantum mechanics is theoretical, by definition, and not due to experimental errors.
answered Jun 11 at 9:41
gentedgented
4,959920
answered Jun 11 at 9:41
gentedgented
4,959920
answered Jun 11 at 9:41
gentedgented
4,959920
answered Jun 11 at 9:41
gentedgented
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I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
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– Umaxo
Jun 11 at 11:22
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@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
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– gented
Jun 11 at 11:50
add a comment |
1
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I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
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– Umaxo
Jun 11 at 11:22
3
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
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– gented
Jun 11 at 11:50
1
1
$begingroup$
I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
$endgroup$
– Umaxo
Jun 11 at 11:22
$begingroup$
I always thought the example focuses the attention on interactions themselves, not on the expermental details. The example talks about interaction of light with electron without any reference to the experimental setup and thus the uncertainty relation depends only on the nature of the interaction itself. And since physics is not math, i like this approach better, then just talking about noncommutative operators. The theory is rooted in experimental results, not the other way around.
$endgroup$
– Umaxo
Jun 11 at 11:22
3
3
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
$endgroup$
– gented
Jun 11 at 11:50
$begingroup$
@Umaxo Yes and no. From the example of scattering of light could you infer that the same property holds for any other system and not only for scattering of light? Moreover even if that were the case (and it isn't) you would have proven the uncertainty principle for position and momentum only and not for any pair of non-commuting observables. As you see at the end of the day we do need a certain degree of abstraction to infer general results.
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– gented
Jun 11 at 11:50
add a comment |
2
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Maybe my answer to this similar (aka "duplicate"?) question physics.stackexchange.com/q/229168 might help?
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– John Forkosh
Jun 11 at 9:40
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Observer effect.
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– Cosmas Zachos
Jun 11 at 10:13
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Also related physics.stackexchange.com/q/114133/162193 in particular one answer gives an explanation for why the Heisenberg microscope was introduced.
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– Alchimista
Jun 11 at 12:44