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?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited 1 hour ago
Qmechanic♦
101k121831141
asked 4 hours ago
Ben Arnao
161
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?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited 1 hour ago
Qmechanic♦
101k121831141
asked 4 hours ago
Ben Arnao
161
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?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited 1 hour ago
Qmechanic♦
101k121831141
asked 4 hours ago
Ben Arnao
161
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?
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
quantum-mechanics newtonian-gravity wavefunction wave-particle-duality wavefunction-collapse
edited 1 hour ago
Qmechanic♦
101k121831141
asked 4 hours ago
Ben Arnao
161
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.
edited 1 hour ago
Qmechanic♦
101k121831141
asked 4 hours ago
Ben Arnao
161
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.
edited 1 hour ago
Qmechanic♦
101k121831141
edited 1 hour ago
Qmechanic♦
101k121831141
edited 1 hour ago
Qmechanic♦
101k121831141
101k121831141
asked 4 hours ago
Ben Arnao
161
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 4 hours ago
Ben Arnao
161
asked 4 hours ago
Ben Arnao
161
161
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
1
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 2 hours 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Ben Arnao is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450263%2fwave-particle-duality-and-gravity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 2 hours 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour 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 2 hours 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour 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 2 hours 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 2 hours ago
anna v
156k8148445
edited 2 hours ago
answered 2 hours ago
anna v
156k8148445
answered 2 hours ago
anna v
156k8148445
answered 2 hours 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour 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
2 hours 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
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
@InertialObserver supeposition of states would be probability distribution also., see the wave packet above.
– anna v
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
I see, for some reason I didn’t see the figure at the first pass for some reason.
– InertialObserver
2 hours ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour ago
@InertialObserver I put it in after your comment !!
– anna v
1 hour 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.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f450263%2fwave-particle-duality-and-gravity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
