Wave particle duality and gravity
$begingroup$
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♦
103k121851175
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♦
103k121851175
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♦
103k121851175
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♦
103k121851175
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
edited Dec 25 '18 at 10:06
Qmechanic♦
103k121851175
asked Dec 25 '18 at 7:03
Ben ArnaoBen Arnao
261
edited Dec 25 '18 at 10:06
Qmechanic♦
103k121851175
edited Dec 25 '18 at 10:06
Qmechanic♦
103k121851175
edited Dec 25 '18 at 10:06
Qmechanic♦
103k121851175
103k121851175
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
active
oldest
votes
$begingroup$
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
157k8149446
$endgroup$
$begingroup$
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 |
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
});
}
});
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
$begingroup$
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
157k8149446
$endgroup$
$begingroup$
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
157k8149446
$endgroup$
$begingroup$
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
157k8149446
$endgroup$
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
157k8149446
edited Dec 25 '18 at 9:18
answered Dec 25 '18 at 8:53
anna vanna v
157k8149446
answered Dec 25 '18 at 8:53
anna vanna v
157k8149446
answered Dec 25 '18 at 8:53
anna vanna v
157k8149446
157k8149446
$begingroup$
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 |
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.
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
