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?
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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 craftCommented Oct 14, 2019 at 0:57
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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 WitthoftCommented Oct 15, 2019 at 14:10
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6 Answers
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Yes, the nucleus is composed of subatomic particles that have a 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.)
Uncertainty principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and its constituents are way smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
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11$\begingroup$ 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!). $\endgroup$ Commented Oct 14, 2019 at 15:16
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$\begingroup$ 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) $\endgroup$– cjmCommented Oct 25, 2019 at 18:48
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1$\begingroup$ 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? $\endgroup$– fraxinusCommented Oct 27, 2019 at 13:15
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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.
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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
$$\Psi_p(\textbf{x}|=(\text{const.})\Psi_e(-\alpha\textbf{ x}),$$
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.
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$\begingroup$ 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). $\endgroup$– tparkerCommented Oct 13, 2019 at 20:29
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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. $\endgroup$– user4552Commented Oct 14, 2019 at 0:27
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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 $|\Psi_p(\vec{x})| = |\Psi_e(-\alpha\vec{ x})|$ is to have the center of mass be definitively at zero, which is inconsistent with the uncertainty principle and all that good stuff. $\endgroup$ Commented Oct 14, 2019 at 14:04
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1$\begingroup$ @BenCrowell In the usual two-body relative coordinates, the Hamiltonian factorizes into $H = H_{CM} + H_{rel}$, where $H_{CM} = \frac{P^2_{CM}}{2M}$ and $H_{rel} = \frac{p^2}{2\mu} + V(r)$, 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. $\endgroup$– tparkerCommented Oct 14, 2019 at 23:57
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2$\begingroup$ 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. $\endgroup$– tparkerCommented Oct 15, 2019 at 0:00
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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, 2019 at 8:26
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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? $\endgroup$– JetpackCommented Oct 14, 2019 at 0:26
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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. $\endgroup$– LuaanCommented Oct 14, 2019 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.
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$\begingroup$ One additional question. If the nucleus of the atom were to be a composite boson, such as Helium-4’s nucleus, would the probability cloud be higher considering the nucleus now behaves more like on single entity than several separated ones? $\endgroup$ Commented Apr 8, 2021 at 7:54
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As already pointed out in the other answers, nuclei do have a probability cloud. It is only a lot smaller than compared to electrons because the mass of the nucleus (or even a single proton or neutron) is a lot higher.
The uncertainty principle is given by
$$\Delta x \Delta p \ge \frac{h}{4\pi}$$
where $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum. If a particle has a large probability cloud, this is the same as saying it has a large uncertainty in position.
For the equation to be true, it either $\Delta x$ or $\Delta p$ increases or decreases, the other must decrease or increase.
Since $p=mv$, we can see that if the mass is higher, momentum is aswell. Thus, a higher mass means higher uncertainty in momentum and likewise, smaller uncertainty in position.
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$\begingroup$ That‘s a bit oversimplified since the velocity of the electrons is far greater too. $\endgroup$– SilasCommented Jan 20, 2023 at 7:56