Wave particle duality and gravity
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?
gravity wave-particle-duality
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Ben Arnao
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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?
gravity wave-particle-duality
asked 2 hours ago
Ben Arnao
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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?
gravity wave-particle-duality
asked 2 hours ago
Ben Arnao
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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?
gravity wave-particle-duality
gravity wave-particle-duality
asked 2 hours ago
Ben Arnao
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ben Arnao
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ben Arnao
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ben Arnao
111
asked 2 hours ago
Ben Arnao
111
111
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ben Arnao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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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 53 mins ago
anna v
156k8148445
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
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 53 mins ago
anna v
156k8148445
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
add a comment |
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 53 mins ago
anna v
156k8148445
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
add a comment |
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 53 mins ago
anna v
156k8148445
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 53 mins ago
anna v
156k8148445
edited 27 mins ago
answered 53 mins ago
anna v
156k8148445
answered 53 mins ago
anna v
156k8148445
answered 53 mins ago
anna v
156k8148445
156k8148445
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
add a comment |
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
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.
– InertialObserver
39 mins ago
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.
– InertialObserver
39 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
24 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
22 mins ago
add a comment |
Ben Arnao is a new contributor. Be nice, and check out our Code of Conduct.
Ben Arnao is a new contributor. Be nice, and check out our Code of Conduct.
Ben Arnao is a new contributor. Be nice, and check out our Code of Conduct.
Ben Arnao is a new contributor. Be nice, and check out our Code of Conduct.
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