Wave particle duality and gravity
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Is a particle's center of gravity at the center of its wave function or is it where we would measure the particle to be? When we measure a particle does its center of gravity shift to where the particle is measured?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
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add a comment |
$begingroup$
Is a particle's center of gravity at the center of its wave function or is it where we would measure the particle to be? When we measure a particle does its center of gravity shift to where the particle is measured?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
$endgroup$
add a comment |
$begingroup$
Is a particle's center of gravity at the center of its wave function or is it where we would measure the particle to be? When we measure a particle does its center of gravity shift to where the particle is measured?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
$endgroup$
Is a particle's center of gravity at the center of its wave function or is it where we would measure the particle to be? When we measure a particle does its center of gravity shift to where the particle is measured?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
edited Dec 25 '18 at 10:06
Qmechanic♦
107k121991239
107k121991239
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
261
add a comment |
add a comment |
1 Answer
1
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oldest
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The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction $Ψ$ is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the $Ψ$ .
In any case at present the standard model of particle physics has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, $Ψ*Ψ$, when enough events/measurements with the same boundary conditions are accumulated.
Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a wavepacket wave function ,
but as the center of gravity/mass is where the electron is, the wave function squared will give the probability of finding it away from the maximum probability, which is the center for this wavepacket.
Complex systems, as the proton and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The models for hadrons are complicated .
In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, the wave function itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled.
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
$endgroup$
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
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I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction $Ψ$ is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the $Ψ$ .
In any case at present the standard model of particle physics has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, $Ψ*Ψ$, when enough events/measurements with the same boundary conditions are accumulated.
Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a wavepacket wave function ,
but as the center of gravity/mass is where the electron is, the wave function squared will give the probability of finding it away from the maximum probability, which is the center for this wavepacket.
Complex systems, as the proton and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The models for hadrons are complicated .
In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, the wave function itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled.
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
$endgroup$
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
add a comment |
$begingroup$
The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction $Ψ$ is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the $Ψ$ .
In any case at present the standard model of particle physics has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, $Ψ*Ψ$, when enough events/measurements with the same boundary conditions are accumulated.
Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a wavepacket wave function ,
but as the center of gravity/mass is where the electron is, the wave function squared will give the probability of finding it away from the maximum probability, which is the center for this wavepacket.
Complex systems, as the proton and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The models for hadrons are complicated .
In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, the wave function itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled.
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
$endgroup$
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
add a comment |
$begingroup$
The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction $Ψ$ is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the $Ψ$ .
In any case at present the standard model of particle physics has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, $Ψ*Ψ$, when enough events/measurements with the same boundary conditions are accumulated.
Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a wavepacket wave function ,
but as the center of gravity/mass is where the electron is, the wave function squared will give the probability of finding it away from the maximum probability, which is the center for this wavepacket.
Complex systems, as the proton and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The models for hadrons are complicated .
In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, the wave function itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled.
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
$endgroup$
The wave function only predicts the probability distribution of finding a particle or a system of particles at an (x,y,z). The center of a wavefunction $Ψ$ is not a well defined concept, as it is different for different boundary conditions. It obeys the symmetries imposed by the interactions that define the $Ψ$ .
In any case at present the standard model of particle physics has elementary particles defined as point particles,i.e. no dimensions, and the center of mass/gravity of course is at that point, and is the point whose measurement will verify the probability distribution, $Ψ*Ψ$, when enough events/measurements with the same boundary conditions are accumulated.
Such elementary paticles are described by plane wave wavefunctions which are spread all over the place. A free electron has to be represented by a wavepacket wave function ,
but as the center of gravity/mass is where the electron is, the wave function squared will give the probability of finding it away from the maximum probability, which is the center for this wavepacket.
Complex systems, as the proton and the neutron, will have an extension in space, and the instantaneous center of mass would be part of the model calculating the quantity of interest to measure. The models for hadrons are complicated .
In general, in quantum mechanical problems, it is the symmetries of the interaction that are embedded in the wavefunction, the wave function itself may be all over a large space, depending on the problem under study, and its center has no usable meaning with respect to the center of mass of the particle-system modeled.
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
edited Dec 25 '18 at 9:18
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
answered Dec 25 '18 at 8:53
anna vanna v
161k8153453
161k8153453
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
add a comment |
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
Surely though one could extend the concept of center of mass to a superposition of states instead of position vector. It would be probabilistic, but I don’t immediate see an issue with just replacing the position vectors with the states in Hilbert space and then projecting onto position space.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:06
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
$endgroup$
– anna v
Dec 25 '18 at 9:22
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
$begingroup$
I see, for some reason I didn’t see the figure at the first pass for some reason.
$endgroup$
– InertialObserver
Dec 25 '18 at 9:24
1
1
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
$begingroup$
@InertialObserver I put it in after your comment !!
$endgroup$
– anna v
Dec 25 '18 at 9:50
add a comment |
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