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According to the Wikipedia article on atomic nucleus, captioned on an impression of helium atom, it states that

This depiction shows the particles as separate, whereas in an actual helium atom, the protons are superimposed in space and most likely found at the very center of the nucleus, and the same is true of the two neutrons. Thus, all four particles are most likely found in exactly the same space, at the central point.

How is this possible? Does this not violate Pauli's exclusion principle?

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    $\begingroup$ Very relevant: physics.stackexchange.com/q/36469 and links therein. $\endgroup$ – dmckee♦ Mar 23 '17 at 19:14
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    $\begingroup$ Why would this violate the Pauli principle? Even provided they are to be found in the same space (whatever this means) they still need not have the same quantum state (not to mention that different types of particles by definitions have different states). $\endgroup$ – gented Mar 24 '17 at 9:24
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    $\begingroup$ Would this statement hold for nuclei heavier than helium? I would think that once you had more than two of each type of nucleon, you'd have to have at least one nucleon in a state with $\ell \geq 1$, and such states have the wavefunction going to zero at the origin (or at least some of them do). Or does my intuition from atomic orbitals not carry over to nuclear orbitals? $\endgroup$ – Michael Seifert Mar 24 '17 at 16:26
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    $\begingroup$ @MichaelSeifert: Your intuition from atomic orbitals can be applied to nuclear orbitals, but with some reservations. See Nuclear shell model. Even the simplest compound nucleus, the deuteron, is tricky. See Isospin singlet state of the deuteron. $\endgroup$ – PM 2Ring Mar 25 '17 at 10:42
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    $\begingroup$ If I had to take a guess the problem here is picturing the nucleus as a set of billiard balls of two different colors put together. That picture is not precise in much the same sense the atomic planetary model is not precise: wavefunctions in general don't need to satisfy such a picture with arbitrary certainty (classical physics) you'd get if it was possible to tell exactly the position of a set of nucleons/electrons. Position is an OBSERVABLE in quantum mechanics. Position/momentum is not what characterizes the particle state in quantum mechanics. I think this is the source of the confusion. $\endgroup$ – Vendetta Mar 28 '17 at 19:18
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This does not violate the exclusion principle because the exclusion principle merely states that there cannot be more than one fermion in the same quantum mechanical state. In the case of two protons and two neutrons, the different particle species don't exclude each other to begin with (because a neutron state is different from a proton state).

Furthermore, that they have the same expectation value for position doesn't mean that they are in the same state. States can coincide with their expectation values for some observables but not for others. In this specific case, the states likely differ by their spin (one proton/neutron has "spin up" and the other "spin down").

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  • $\begingroup$ How does this relate to the variety of shapes of nuclei? $\endgroup$ – Zwolf Mar 23 '17 at 19:10
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    $\begingroup$ @Zwolf What kind of "shape" do you mean? (If you have an additional question that's not merely a clarification, please ask it separately as a new question instead of a comment here) $\endgroup$ – ACuriousMind♦ Mar 23 '17 at 19:12
  • $\begingroup$ If the neutrons and protons occupy the same positions, shouldn't the shapes of nuclei having equal numbers of protons and neutrons be the same (not sure if this should be a separate question)? $\endgroup$ – Zwolf Mar 23 '17 at 19:14
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    $\begingroup$ Having the same expectation is not the same thing as being in the same space. As a macroscopic example, consider a chair shared by two employees on different shifts. One of them spends half their time in the chair, the other spends half their time in the char. If you want to find one of them, their expected position is "in the chair." Quantum mechanics makes these sorts of situations even easier, but even in the macroscopic world you can get away with it. $\endgroup$ – Cort Ammon Mar 24 '17 at 0:05
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    $\begingroup$ @htmlcoderexe A specific one works too. If you want to know where Bob is, 50% of the time he's in the chair, and the remainder of the time his position is diffused elsewhere. (Technically not quite an expectation, but it would be trivial to turn it into one if you really wanted to) $\endgroup$ – Cort Ammon Mar 24 '17 at 23:54
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Pauli's exclusion principle states that two fermions can't occupy the exact same quantum state simultaneously. Two fermions can have spatial wavefunctions that overlap with nonzero values at common locations. That is fine - the point is that the entire spatial wavefunctions (along with spin states) can't be the same for both particles.

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Thus, all four particles are most likely found in exactly the same space, at the central point.

This doesn't seem to say that the particles overlap entirely. This sates that wave functions of all the particles are centered around a common central point. Pauli's principle doesn't forbid that.

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  • $\begingroup$ That is not the point, though. Even if they were exactly in the same spatial location (whatever this means) still the general states are different. $\endgroup$ – gented Mar 25 '17 at 10:26
  • $\begingroup$ @GennaroTedesco: But they are never in the same exact spatial location, regardless of which interpretation of quantum mechanics you take! So it is deserved to be pointed out that identical wavefunctions do not imply overlap in the layman sense. $\endgroup$ – user21820 Mar 25 '17 at 12:34
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    $\begingroup$ Again, that's not the point of the question. The point is that even if there were indeed a position overlap (although not possible) this still wouldn't imply that the quantum states are the same. My remark is that the position (whatever it means) doesn't specify the whole quantum state. $\endgroup$ – gented Mar 25 '17 at 16:44

protected by Qmechanic♦ Mar 24 '17 at 9:41

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