Why doesn't the nucleus have “nucleus-probability cloud”?What prevents the nucleus from wandering into the electron cloud?Finding the maximum of the radial probability density in the hydrogen atomWhat precisely is the wave function a probability density of?Form of Schrödinger equation for the probability densityWhat is the probability for an electron of an atom on Earth to lie outside the galaxy?Can Quantum Mechanical Potential have a Probability DistributionHow to evaluate the probability when a particle is detected?
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Why doesn't the nucleus have “nucleus-probability cloud”?
What prevents the nucleus from wandering into the electron cloud?Finding the maximum of the radial probability density in the hydrogen atomWhat precisely is the wave function a probability density of?Form of Schrödinger equation for the probability densityWhat is the probability for an electron of an atom on Earth to lie outside the galaxy?Can Quantum Mechanical Potential have a Probability DistributionHow to evaluate the probability when a particle is detected?
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While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus?
quantum-mechanics wavefunction nuclear-physics probability subatomic
edited Oct 13 at 8:32
Qmechanic♦
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asked Oct 13 at 8:11
Tim CrosbyTim Crosby
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add a comment
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While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus?
quantum-mechanics wavefunction nuclear-physics probability subatomic
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
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How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
2
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10
add a comment
|
$begingroup$
While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus?
quantum-mechanics wavefunction nuclear-physics probability subatomic
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
1792 silver badges7 bronze badges
$endgroup$
While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus?
quantum-mechanics wavefunction nuclear-physics probability subatomic
quantum-mechanics wavefunction nuclear-physics probability subatomic
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
1792 silver badges7 bronze badges
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
1792 silver badges7 bronze badges
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
edited Oct 13 at 8:32
Qmechanic♦
116k14 gold badges229 silver badges1378 bronze badges
116k14 gold badges229 silver badges1378 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
1792 silver badges7 bronze badges
asked Oct 13 at 8:11
Tim CrosbyTim Crosby
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asked Oct 13 at 8:11
Tim CrosbyTim Crosby
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3
$begingroup$
How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
2
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10
add a comment
|
3
$begingroup$
How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
2
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10
3
3
$begingroup$
How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
$begingroup$
How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
2
2
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10
add a comment
|
5 Answers
5
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oldest
votes
$begingroup$
Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
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10
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Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
add a comment
|
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Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
edited Oct 13 at 9:02
Steeven
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answered Oct 13 at 8:34
my2ctsmy2cts
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The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
$$Psi_p(textbfx|=(textconst.)Psi_e(-alphatextbf x),$$
where $Psi_p$ is the wavefunction of the proton, $Psi_e$ is the wavefunction of the electron, and $alpha$ is the ratio of the masses. Because $alpha$ is large, one can often approximate the proton as being fixed at one point.
answered Oct 13 at 19:42
Ben CrowellBen Crowell
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This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
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– tparker
Oct 13 at 20:29
1
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@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
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– Ben Crowell
Oct 14 at 0:27
2
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While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
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– Michael Seifert
Oct 14 at 14:04
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
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– tparker
Oct 14 at 23:57
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
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– tparker
Oct 15 at 0:00
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show 3 more comments
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The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
answered Oct 13 at 8:26
Marco OcramMarco Ocram
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Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
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– Jetpack
Oct 14 at 0:26
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
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– Luaan
Oct 14 at 7:49
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Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
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5 Answers
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5 Answers
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Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
$endgroup$
10
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
add a comment
|
$begingroup$
Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
$endgroup$
10
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
add a comment
|
$begingroup$
Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
$endgroup$
Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
edited Oct 14 at 11:34
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
answered Oct 13 at 18:59
fraxinusfraxinus
5011 silver badge5 bronze badges
5011 silver badge5 bronze badges
10
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
add a comment
|
10
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
10
10
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
Molecules, likewise, have a probability cloud. It's just even smaller. Taken to the extreme, you could calculate the probability cloud for a whole person (and it'd be so small as to not exist; the plank length being what it is). Have an article about the superposition of a 5000-atom molecule (astronomical in size compared to doing it with photons!).
$endgroup$
– Draco18s
Oct 14 at 15:16
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
You can even calculate the deBrolgie wavelength of a bus traveling between two skyscrapers if you wish. IIRC, it’s shorter than the Planck distance (but it’s been 6+ years since I did that calculation)
$endgroup$
– cjm
2 days ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
$begingroup$
Person, bus, elephant, virus, whatever - it is not the Plank constant that makes the calculation pointless. It is the size of the object itself. Most macro object as we know them have uncertainity way above the atom size because of factors different from pure quantum-fu. For example, do we count adsorbed air/water molecules over an 1-cm metal object?
$endgroup$
– fraxinus
9 hours ago
add a comment
|
$begingroup$
Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
edited Oct 13 at 9:02
Steeven
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answered Oct 13 at 8:34
my2ctsmy2cts
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$endgroup$
add a comment
|
$begingroup$
Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
edited Oct 13 at 9:02
Steeven
30.7k8 gold badges74 silver badges125 bronze badges
answered Oct 13 at 8:34
my2ctsmy2cts
8,4192 gold badges8 silver badges24 bronze badges
$endgroup$
add a comment
|
$begingroup$
Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
edited Oct 13 at 9:02
Steeven
30.7k8 gold badges74 silver badges125 bronze badges
answered Oct 13 at 8:34
my2ctsmy2cts
8,4192 gold badges8 silver badges24 bronze badges
$endgroup$
Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
edited Oct 13 at 9:02
Steeven
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answered Oct 13 at 8:34
my2ctsmy2cts
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edited Oct 13 at 9:02
Steeven
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edited Oct 13 at 9:02
Steeven
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edited Oct 13 at 9:02
Steeven
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30.7k8 gold badges74 silver badges125 bronze badges
answered Oct 13 at 8:34
my2ctsmy2cts
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answered Oct 13 at 8:34
my2ctsmy2cts
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answered Oct 13 at 8:34
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$begingroup$
The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
$$Psi_p(textbfx|=(textconst.)Psi_e(-alphatextbf x),$$
where $Psi_p$ is the wavefunction of the proton, $Psi_e$ is the wavefunction of the electron, and $alpha$ is the ratio of the masses. Because $alpha$ is large, one can often approximate the proton as being fixed at one point.
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
$endgroup$
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
1
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
2
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
|
show 3 more comments
$begingroup$
The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
$$Psi_p(textbfx|=(textconst.)Psi_e(-alphatextbf x),$$
where $Psi_p$ is the wavefunction of the proton, $Psi_e$ is the wavefunction of the electron, and $alpha$ is the ratio of the masses. Because $alpha$ is large, one can often approximate the proton as being fixed at one point.
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
$endgroup$
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
1
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
2
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
|
show 3 more comments
$begingroup$
The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
$$Psi_p(textbfx|=(textconst.)Psi_e(-alphatextbf x),$$
where $Psi_p$ is the wavefunction of the proton, $Psi_e$ is the wavefunction of the electron, and $alpha$ is the ratio of the masses. Because $alpha$ is large, one can often approximate the proton as being fixed at one point.
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
$endgroup$
The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
$$Psi_p(textbfx|=(textconst.)Psi_e(-alphatextbf x),$$
where $Psi_p$ is the wavefunction of the proton, $Psi_e$ is the wavefunction of the electron, and $alpha$ is the ratio of the masses. Because $alpha$ is large, one can often approximate the proton as being fixed at one point.
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
edited Oct 18 at 10:21
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
answered Oct 13 at 19:42
Ben CrowellBen Crowell
63.3k7 gold badges183 silver badges354 bronze badges
63.3k7 gold badges183 silver badges354 bronze badges
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
1
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
2
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
|
show 3 more comments
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
1
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
2
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
$begingroup$
This is a slight oversimplification. In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum; it just means that the different total-momentum sectors time-evolve independently (and so in particular, whatever probability distribution over momentum sectors the initial state has is preserve over time).
$endgroup$
– tparker
Oct 13 at 20:29
1
1
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
$begingroup$
@tparker: In quantum mechanics, conservation of momentum (for example) doesn't necessarily mean that any system is always in an eigenstate of total momentum This isn't very clear. There is nothing in what I wrote that assumes the system to be in an eigenstate of total momentum. What I've written here is not a simplification or oversimplifcation. It's simply some basic facts in freshman physics. Frankly, your comment comes across to me as nonsense -- but if there is actually some substance to it, I would be glad to see you explain it in a less obfuscatory way.
$endgroup$
– Ben Crowell
Oct 14 at 0:27
2
2
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
$begingroup$
While conservation of momentum requires that the center of mass of the atom remains fixed (or more accurately, moving at a constant velocity) at the classical level, the center of mass will have its own quantum-mechanical wavefunction and will not necessarily be localized at zero. I'm pretty sure that the only way to get $|Psi_p(vecx)| = |Psi_e(-alphavec x)|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff.
$endgroup$
– Michael Seifert
Oct 14 at 14:04
1
1
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
$begingroup$
@BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_CM + H_rel$, where $H_CM = fracP^2_CM2M$ and $H_rel = fracp^22mu + V(r)$, which as you say act on different degrees of freedom. So we can always find a complete basis of eigenstates that are tensor products of a CM and relative-position wavefunction. But a general state will be a superposition of such product states; i.e. the CM position and relative position degrees of freedom will be entangled together. In this case, the wavefunction will not take the form that you state.
$endgroup$
– tparker
Oct 14 at 23:57
1
1
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
$begingroup$
Also, I would disagree that "There is nothing in what [you] wrote that assumes the system to be in an eigenstate of total momentum." I think that your statement "Conservation of momentum requires that the center of mass of the electron and proton remain fixed" assumes exactly that; that the total momentum is a zero-momentum eigenstate.
$endgroup$
– tparker
Oct 15 at 0:00
|
show 3 more comments
$begingroup$
The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
$endgroup$
2
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
add a comment
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$begingroup$
The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
$endgroup$
2
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
add a comment
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$begingroup$
The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
$endgroup$
The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
answered Oct 13 at 8:26
Marco OcramMarco Ocram
2,8507 silver badges16 bronze badges
2,8507 silver badges16 bronze badges
2
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
add a comment
|
2
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
2
2
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
$begingroup$
Is that because the uncertainty in the position of the protons is small compared to the distance to the electrons + the uncertainty of the electron position?
$endgroup$
– Jetpack
Oct 14 at 0:26
1
1
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
$begingroup$
@Jetpack Not the distance, really - electrons can even be inside the nucleus. But the uncertainty, yes. The uncertainty in the position of the nucleus is small enough that we can ignore it for most purposes in chemistry; keep in mind that it's very related to the mass of the particle, and electrons mass a thousandth of a proton. Anyway, all those numbers we attach to chemical elements and isotopes were determined experimentally, not from first principles, so even though you need to do the full thing to get the exact right result, you don't bother, since you already have the value in a table.
$endgroup$
– Luaan
Oct 14 at 7:49
add a comment
|
$begingroup$
Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
$endgroup$
add a comment
|
$begingroup$
Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
$endgroup$
add a comment
|
$begingroup$
Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
$endgroup$
Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
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answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
answered Oct 15 at 12:06
timeisnotrealtimeisnotreal
111 bronze badge
111 bronze badge
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3
$begingroup$
How do you know you don't? If the displacement is relative to the electron and the nucleus.
$endgroup$
– marshal craft
Oct 14 at 0:57
2
$begingroup$
Just wait 'til you run into the avant-gardeists who propose there is no mass anywhere, just localized high probability density!
$endgroup$
– Carl Witthoft
Oct 15 at 14:10