Are there individual protons and neutrons in a nucleus? [duplicate]

Question

The popular science material always talks about the number of protons and neutrons in a nucleus, but I've always wondered if that's a real thing nuclear physicists believe or if it is just a convenient model. In other words, is there some reason to believe that an atomic nucleus isn't just a single particle with mass and charge equal to a certain number of protons and neutrons instead of a clump of individual protons and neutrons? Are there experiments to distinguish one situation from the other? It's hard to imagine what such an experiment would be like. Anything that produces a fragment of a nucleus could be interpreted in either of two ways: a certain number of protons and neutrons broke off of the clump, or the single particle transmutes into smaller particles.

Answer 1

It's a fair question. Let's start with the related question:

Are there individual electrons in an atom?

Here the answer is generally "no." The answer is more emphatically "no" for nucleons, but I want to talk about electrons first because they are generally less mysterious to people.

By definition, all of the electrons in an atom are indistinguishable, which can arguably be rephrased to say that any electron in the electron cloud is the same electron as any other one. When you come up with some multi-electron wavefunction that describes an atom with $n$ electrons, you technically need to add $n^2\text{-ish}$ cross terms to make sure that your wavefunction is antisymmetric under exchange of any two electrons. There's a way to do this, which I think is named for Slater, by taking the determinant of a matrix of labeled-electron states.

But it's not just a matter of entanglement. When we distill quantum mechanics for a non-mathematical audience, we say things like "the electron absorbs a photon and jumps to a higher energy level." That's clearly an oversimplification. A free electron famously can't absorb or emit photons, because the electron doesn't have any internal excited states that could be used to conserve energy. It's more correct to say that the external photon is absorbed by the electromagnetic field of the electron and the nucleus, whose relative motions are changed by this absorption so that the atomic electromagnetic field stores the energy. Whether we refer explicitly to the atomic electromagnetic field, versus more simply to "the potential energy term," is just a matter of semantics. In a multi-electron atom, the configuration of the electromagnetic field depends on the entire electron cloud, rather than on any particular electron.

In a multi-electron atom, there are a few types of excited states where the electron cloud can be described by one- or two-electron excitations relative to the ground state. One family is the Rydberg states, where all but one electron acts as a "core" with charge $+1$ and the remaining electron has a hydrogen-like excitation spectrum. That "hydrogen-like spectrum" is fairly well-behaved for enormous principal quantum numbers, where the last electron is just barely bound to the core ion. But for low-lying excitations, even if you can identify one series as Rydberg-like, the energies are strongly modified by multi-electron interactions with the "core." You get the least modification in the Group I elements, where the electronic "core" should have the same quantum numbers as the ground state of a noble gas.

The other family of single-particle excitations are the extremely energetic excitations associated with the emission of Auger electrons. These excitations are well-modeled as single-particle vacancies in the innermost electron orbitals. We say that some high-energy process, usually a nuclear decay, ejects one electron from an inner shell. There is a cascade of high-energy photons emitted as electrons from outer shells "fall" into the inner shells to fill the vacancies and emit the x-rays. The jargon, whose origin I don't know, is that the shells are enumerated K, L, M, et cetera. If a vacancy in the K shell is filled by "an electron" which leaves a vacancy in the L or M shell, that's called a K$\alpha$ or K$\beta$ transition. These x-rays give you the energy differences among the innermost shells.

However, it's probably more correct to say that the cascade of Auger x-rays is due to a single-hole excitation, as the vacancy rises up through the electron cloud like a bubble. I suspect that this interpretation becomes more important as the vacancy gets to the higher shells, where there are angular momentum states available for the vacancy to anti-occupy. If you want to refer to a hole state as a single-particle excitation, the single particle is the hole, not any of the electrons around it. We talk about a bubble floating up through the water, not about all the water flowing down around the bubble. Hole motion is a collective phenomenon in electron motion.

The "meat and potatoes" of atomic spectroscopy lies in the middle: medium-energy excitations which become very difficult to describe in a "shell model" of these electrons moving from those orbitals to these other orbitals. Electronic excited states of atoms are named based on the atomic quantum numbers, rather than based on the shell designations. This choice is partly because the energy eigenstates are the observable states. But it's also because a particular energy eigenstate may be a complicated superposition of different shell-model states.

For example, if you look at the NIST Atomic Spectra Database for hydrogen, you find spectral information that you can predict based on an introductory quantum-mechanics course. But if you look at iron-I (already a particular configuration of the valence electrons), you find that the ground state has configuration $3d^6 4s^2$ outside of its neon-like core. (That's six $({}^6)$ electrons in the outermost (third) $d$ orbital, plus two electrons in the outermost (fourth) $s$ orbital, where the orbitals are named in the "spdf" sequence.) This ground state has term symbol $a\,{}^5D_4$. I'm not sure what the $a$ means; the rest of the symbol means that the entire atom has a spin quintuplet with quantum number obeying $2S+1=5$, total orbital angular momentum $L=2$ also identified by a letter in the "spdf" sequence, and total angular momentum $\vec J = \vec S + \vec L = 4$ (although all of $J\in\{0,\cdots,4\}$ are allowed). As you look at higher energies, you find iron-I states where the shell-model description is a complicated mix of terms. For instance, there's a state at $\rm 3.4\,eV$ whose shell-model wavefunction is apparently

$$ \newcommand{\ket}[1]{\left|#1\right>} \ket{ a\,^1P_1\ (\mathrm{3.4\,eV})} = \sqrt{0.62}\ket{3d^7({^2P})4s} + \sqrt{0.23}\ket{3d^7({^2D2})4s\ [{}^3D]} + \cdots, $$ where the $3d^7$ electrons are in a superposition of different configurations with different internal term symbols. There are at least three nontrivial contributions: note that 15% of the wavefunction is undescribed in this table. The different term symbols are allowed because, due to so-called "spin-orbit" interactions, it's only approximately correct to treat the spin and orbital quantum numbers $S$ and $L$ as being conserved separately. The presence of $D2$, as opposed to just $D$, suggests that there are at least two different shell-model ways for the seven $d$-shell electrons to end up with total orbital angular momentum $L=2$. You can even find states where the $4s$ orbital is in a superposition of filled and half-filled, or filled and empty, or half-filled and empty.

I would be surprised if there isn't some lanthanide or actinide where the ground state is a nontrivial superposition of electrons in the $d$ and $f$ orbitals.

Nuclear structure has all of this complication and more. The spin-orbit interaction is lots stronger among nucleons versus among electrons, so a state with definite $J$ is almost always a mix of states with different $L,S$. Even in the deuteron, the smallest bound nucleus, you can't assign an identity to either nucleon. The deuteron is a singlet in "isotopic spin," a quantum number in which the nucleon is an "isodoublet" whose projections are the proton and the neutron. The two-particle isosinglet state is $$ \ket{(T=0)} = \frac{ \rm \ket{np} - \ket{pn} }{\sqrt2} $$ so that you can't even say which particle is the proton and which is the neutron. You can draw a Feynman diagram for $\pi^\pm$ exchange and see that the neutron and proton swap identities. To the extent that meson exchange is a good model for the nucleon-nucleon interaction, pion exchange is the most important piece of it.

Sometime as a grad student I ran across an old paper which had "solved" the helium-nucleus ground state as a mixture of at least a dozen different terms. I don't remember any details; I was looking for something else and that paper started on the page where the paper I had printed out ended.

Anything that produces a fragment of a nucleus could be interpreted in either of two ways: a certain number of protons and neutrons broke off of the clump, or the single particle divided into separate pieces.

This is called a "cluster nucleus" model. The most famous example is Gamow's tunneling model of alpha decay, where the alpha particle already exists in the nucleus and leaks out of the Coulomb barrier. There are also cluster models for entire low-mass nuclei, such as $^6$Li as alpha plus deuteron. Carbon has a high-energy "triple alpha" state which is responsible for essentially all of the helium fusion in the entire history of the universe.

The popular science material always talks about the number of protons and neutrons in a nucleus, but I've always wondered if that's a real thing nuclear physicists believe or if it is just a convenient model.

The number of nucleons in a nucleus is absolutely a constant; it's called the "baryon number." There is no evidence for any baryon-number-changing interaction since the earliest parts of the Big Bang. The total charge --- that is, the proton number --- is likewise absolutely known for any given nucleus. But you can't say which nucleon is which proton: a fundamental consequence of indistinguishability.

Answer 2

So the detailed answers already given cover most of it, but I'll summarize: all protons are indistinguishable fermions, so if you have two unbound protons (labeled $b$ for beam and $t$ for target) elastically scattering:

$$ p_b(p_t, p_b)p_t $$

means you detect the scattered beam proton, and know by kinematics that the undetected target is still a proton.

This is indistinguishable from detecting the recoil proton:

$$ p_b(p_t, p_t)p_b $$

so in the experiment: both reactions occur, and you detected proton is in a superposition of beam and target proton.

The same is true for neutrons--that's the nature of indistinguishable particles.

So in a nucleus: all the protons (neutrons) are in an entangled state where no individual proton (neutron) has an identity.

As mentioned in other answers, there is also (slightly broken) isospin symmetry, where $p$ and $n$ are considered identical particles with different $I_3$ (or $T_3$, and sometimes $\tau_3$) projections.

In the notation of the above reactions, if you scatter a neutron beam off a liquid hydrogen target, and detect the recoil proton:

$$ n(p, n)p $$

that final state is a mixture of the $n, p$ exchanging a neutral pion to become $(n, p)$ and exchanging a charged pion to become $(p, n)$--so you can't know if the proton was from the target, or is a transmuted neutron from the beam.

In a nucleus, all neutrons and protons are in this entangled state where their isospin projection is mixed. The reason it's called isospin is because the math is identical to adding quantum spins: to be in a total (iso)spin eigenstate, the individual particles are (usually) NOT in $S_3$ ($I_3$) eigenstates.

Now the question as to whether the nucleus becomes a "blob"...the answer is "no"--as that would be a quark gluon plasma, and it is definitely not the ground state. Nevertheless, the observation that the quark structure functions of individual nucleons are modified was made in 1983, and remains an unsolved problem in nuclear physics. It is called "The EMC Effect":

https://en.wikipedia.org/wiki/EMC_effect

of which, much has been written.

Answer 3

Nuclear fission discovered in 1938 by Otto Hahn, Fritz Strassmann, Lise Meitner and Otto Robert Frisch showed that Uranium nucleus can be divided into two separate parts, through the input of slow neutron : $$ {}_{92}U + n \to {}_{56}Ba + {}_{36}Kr + 3n + 200~\text{MeV}\tag 1$$

So Uranium nucleus has divided into Barium nucleus and Krypton nucleus, conserving total nucleus charge ($92e^+=56e^++36e^+$).

This proves experimentally that Uranium nuclei (and nuclei in general) has charge composite structure since you can't divide monolithic/unitary thing by definition.

From here you can easily arrive at the proton existence in the nuclei fact if you keep breaking apart smaller and smaller nucleus ($Ba,Kr$, and so on further) until all what you've got is a single $e^+$ charge.

EDIT.

@david has gave an interesting contra-argument that this reaction could be interpreted as a single particle transmutation into two or more constituent (also single) particles like it happens in a neutron transmutation into proton in beta decay: $n \to p~+ e^-~+~\overline \nu_e \tag 2.$ (Albeit nor proton, neither neutron are single particles, but rather composed from three quarks).

Anyway, the main point why we can't argue that (1) reaction is a simple transmutation like in (2), is that Uranium nuclei upon breakup, releases $200~MeV$ energy in the form of residual $Ba,Kr$ parts kinetic energy. If ${}_{92}U$ nuclei would be a monolithic/indivisible single charge, then we couldn't explain where this additional energy (which is a mass deficit $M_{U}-m_{Ba}-m_{Kr}-m_{3n}\approx 200 MeV$) comes from.

If you plunge the all nucleon binding energies in the Uranium atom, then you will get these exact 200 megaelectronvolts which are released as per $\Delta E= \Delta mc^2$ mass deficit in the post mentioned reaction.

Surely, it's not a bullet-proof conclusion, since nobody forbids to build a new theory which would explain that emitted energy in other ways, but considering other advancements in atom and nuclear Physics it's as close as we can get into it using standard model.

Answer 4

Some very detailed and convincing arguments have been given from both sides, please don't judge this answer too harshly in comparison as I am not a particle physicist and this is maybe more of a question than an answer:

Does it matter?

This is a riff off the "shut up and calculate" argument: I would argue that it does not matter. We know that the core can be split by inducing an imbalance or by just waiting for long enough for that imbalance to occur randomly, so we can as well assume either opinion - if the math gives valid results, we do not care about "actual reality" - and that reality at this scale may even be multiple things at the same time.

But how is it "really"?

This is a theory / interpretation: since protons and neutrons demonstrably form from smaller parts and only their assembly gives them their individual properties or even kinds, we could argue that also the nucleons have to remain intact in this assembly, as we can perform calculations based on these emergent properties. I cannot prove this, but I assume if it was just a QGP, that would have very different properties. This is not to say the core is just a pile of particles, just as my watch is not just a pile of atoms. Also, the higher level order can give new properties to the assembly, even if the individual parts do really exist.

I am looking forward to some serious counter arguments :)

Answer 5

Given that QFT predicts particle numbers are not conserved, the nucleus is much more of a complicated affair than popular books suggest it is on close examination.

Nevertheless, we should also admit that there is a strong degree of truth in the popular assertion. This is because physics operates in a variety of scales with emergent features at various scales. There is no fundamental scale as a variety of scales is fundamental.