6
$\begingroup$

Some people naturally assume that atomic nuclei are made of protons and neutrons. That is, they are basicly clumps of protons and neutrons that each maintain its separate existence, like pieces of gravel maintain their existence if you mold them together in a ball with mud for a binding force.

How come neutrons in a nucleus don't decay?

This is a natural assumption. A hydrogen nucleus can have one proton as its nucleus. Nuclei can absorb neutrons to become other isotopes. It's natural to assume that nuclei are clumps of protons and neutrons.

Sometimes if an atomic nucleus gets broken by application of large amounts of energy, typically applied with a fast-moving subatomic particle, they might release a neutron or a proton. So for example, smash an alpha particle into a beryllium nucleus and a neutron comes out. Doesn't that imply that the neutron was in there all along, waiting to get out?

But that reasoning implies that electrons, positrons, muons etc are also inside the nucleus all the time, waiting to get out.

There's an idea that protons and neutrons inside a nucleus swiftly transfer charges. This is analogous to a theory from organic chemistry, where sometimes single and double bonds switch back and forth, increasing stability. We could have quarks getting exchanged rapidly between protons and neutrons, increasing stability. I can see that as increasing stability for the nucleus, but I just don't see it as making the protons and neutrons more stable. If ten Hollywood couples get repeated divorces and marry each other's exes, you wouldn't say that the original marriages are stable.

In the extreme, the quarks might just wander around in a nuclear soup, and the protons and neutrons have no more identity than a bunch of used computers disassembled with the parts on shelves for resale. Maybe you could collect enough parts to take a working computer out of the store with you, but it probably won't be one of the old computers.

So -- my question -- is there experimental evidence that strongly implies protons and neutrons maintain their separate identities inside atomic nuclei? Or is there data which can be interpreted that way but which can also be easily interpreted another way?

$\endgroup$
3
  • $\begingroup$ quantum nuclear models phys.libretexts.org/Bookshelves/Nuclear_and_Particle_Physics/… $\endgroup$
    – anna v
    Aug 2, 2022 at 18:13
  • $\begingroup$ Briefly I'd mention that rather than treating nuclei as "quark soups" at least one model treats them as collections of paired-nucleons ( the interacting boson model ). $\endgroup$ Aug 2, 2022 at 21:48
  • $\begingroup$ @StephenG-HelpUkraine It makes sense to me that there could be lots of different models that fit the indirect evidence available. Or maybe there's solid evidence to throw out some of the models, that I haven't noticed? $\endgroup$
    – J Thomas
    Aug 2, 2022 at 22:32

5 Answers 5

11
$\begingroup$

When you ask "do protons and neutrons retain their identity in nuclei" and fail to mention isospin, there is clearly a misunderstanding.

So, a simpler question: Do electrons in a neutral helium atom maintain their spin alignment? Where the electrons are added to the $S$ shell one by one, first spin down, and then spin up to complete the shell without violating Pauli's exclusion principle.

Well that is wrong. The electrons fill the shell with a totally antisymmetric wave function, and all the antisymmetry is in the spins:

$$ \chi = \frac{|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle}{\sqrt 2} =|S=0, S_z=0\rangle$$

Each electron is in an indefinite state of $s_z$, but the total state is an eigenvalue of $\hat S^2$ and $\hat S_z$.

Isospin is called isospin ($\vec I$) because it replaces spin ($\vec S$) up/down with the exact mathematical formalism of spin, except with proton/neutron being eigenstates of $\hat I_3$.

So what's a deuteron? It's mostly $S$-wave (with some $D$-wave)...so it's even in spatial coordinates. The spin is one, so it's even in spin. That mean the antisymmetry falls on isospin, where it's clearly $I_3=0$. To be antisymmetric, it must be $I=0$, too:

$$|I=0, I_3=0\rangle = \frac{|p\rangle|n\rangle-|n\rangle|p\rangle}{\sqrt 2}$$

Since the 1st (2nd) bra refers to particle one (two), it is clear that individual nucleons are not in definite states of $I_3$ in a deuteron. In other words, they're 1/2 proton and 1/2 neutron, just as two electron spins aren't definite in the $S_z=0$ states.

There's no need to reference quarks to resolve this, as:

$$ p \rightarrow \pi^+ + n \rightarrow p $$ $$ n \rightarrow \pi^- + p \rightarrow n $$

is perfectly fine (where the intermediate state is a virtual pion binding stuff).

Since a helium atom was mentioned, a good exercise is to workout the spatial, spin, and isospin wave functions: it is very spherical in real space and iso-space...which is why it is so stable.

$\endgroup$
2
  • $\begingroup$ I hoped for some sort of experimental evidence. You have given an entirely theoretical answer, which says why we should not consider protons and neutrons to have their own existence inside a nucleus. There's nothing wrong with that -- if your theory is correct then anything derived from your theory will fit reality also. But I get nothing from this to tell me about the evidence, only the theory. $\endgroup$
    – J Thomas
    Aug 8, 2022 at 18:21
  • 2
    $\begingroup$ @JThomas The fact that the deuteron exists while $^2{\rm He}$ and the di-neutron don't is experimental data. That isospin works at all means neutrons and protons don't retain identity in nuclei, they're 2 states of a single particle: the nucleon. Of course the symmetry is not exact. For more on experiments, see "The EMC Effect" $\endgroup$
    – JEB
    Aug 9, 2022 at 11:53
3
$\begingroup$

Here I will begin waving my hands:

When a neutron is bound in a helium nucleus, its overall energy level is reduced to the point where it's not energetically possible for it to decay the way it would if it were a "free" neutron. This means that 1) helium nuclei are "forever" unless blasted with gigantic energies, and 2) successive elements in most of the periodic table are made by adding helium nuclei as subunits to the next lighter element. the long-term persistence of elements furnishes evidence that protons and neutrons maintain their identities within the nucleus.

Note also that a proton is lighter than a neutron which means that while unbound neutrons can decay to protons and electrons (plus antineutrinos), the reverse isn't possible because there is no lower energy hadron into which the proton can decay. This means that the nucleons cannot randomly switch identities while hanging around inside the nucleus.

In addition, while it is possible to strike a quark inside a proton hard enough to mess with the quark and possibly alter the identity of the proton, at those energies both the proton and the nucleus itself fly apart so the point is moot.

Now, we know it isn't geometrically possible to stack & fit helium nuclei together in a way which makes everyone perfectly happy in a heavier nucleus, which means that the binding energy for a nucleus will have noticeable peaks and valleys in it as you move up the mass scale to bigger and bigger nuclei, and it becomes possible to build up nuclei called isotopes which have the same number of protons in them but include an extra neutron or two jammed into them. Even at the uncomplicated end of the periodic table you can load a hydrogen with two extra neutrons and get tritium, but it is unhappy and will rid itself of one of those extra guys if given the right chance.

There are exceptions to these general rules because the topic is more complicated than I have made it out to be here, and I invite the higher-level handwavers here to weigh in.

$\endgroup$
4
  • $\begingroup$ @niels_nielsen - would you say that the strong force makes a He_4 nucleus more like a 12 quark soup than like 2 protons and 2 neutron? Wave away ... $\endgroup$
    – Paul Young
    Aug 2, 2022 at 22:27
  • $\begingroup$ @niels nielsen You have presented a theoretical argument that seems at least reasonable to me. A bound neutron can't decay the way a free one would, because it just doesn't have the energy required to do that. I don't find it completely convincing. A crystal of table salt requires tremendous energy to melt and then evaporate, but it can separate into ions in water. Maybe there could be something in a nucleus that's analogous to water, that lets neutrons lose their identity without emitting lots of energy outside the nucleus. But thank you! You described it clearly. $\endgroup$
    – J Thomas
    Aug 2, 2022 at 22:31
  • 1
    $\begingroup$ See Fermi's Golden Rule. If the density of final states is zero because of energy, no amount of decay amplitude can overcome that. $\endgroup$
    – JEB
    Aug 2, 2022 at 23:25
  • 2
    $\begingroup$ @paul young, I pass, and defer to the experts!! $\endgroup$ Aug 3, 2022 at 0:40
2
$\begingroup$

"is there experimental evidence that strongly implies protons and neutrons maintain their separate identities inside atomic nuclei? Or is there data which can be interpreted that way but which can also be easily interpreted another way?"

First, I would like to stress that nuclei as well as neutrons and protons (although all composites) are quantum entities that cannot be thought of as classical "bags of marbles." I thus interpret your question to be "do protons and neutrons maintain their quantum identities inside nuclei? The best evidence that I know of for a "yes" answer comes from careful analysis of the isotones Tl-205 and Pb-206 that extract the charge density difference obtained from experimental elastic electron scattering cross sections of these nuclei. According to the nuclear relativistic mean field model (see L. D, Miller and A. E. S. Green, Phys Rev C, p. 241-252, 1972, my PhD dissertation BTW) the extra proton in Pb-206 should reside in a 3s orbital. Quoting from the abstract of the following paper: Frois, B., et al, Nuclear Physics, Section A, Volume 396, p. 409-418, March 1983. "Elastic electron scattering cross sections have been measured for 205Tl and 206Pb at an incident energy of 502 MeV, covering a momentum transfer range from 1.4 to 3.2 fm -1. The charge density difference of the two isotones is accurately determined. We show that the shape of the 3s proton orbit can be extracted from this difference without ambiguity. " The shape of the 3s orbital is quite distinctive and its nodal structure guarantees that this proton samples the entire nuclear environment (not just the periphery) despite its being the least bound proton. I can't think of any explanation other than this is the proton maintaining its quantum identity inside a heavy nucleus.

$\endgroup$
1
  • $\begingroup$ i have to admit that I didn't understand much of your answer. But you show clearly that there is experimental data which is compatible with the assumption that there is a proton in a nuclear 3s orbit in Pb-206. You don't say whether some other interpretation could be made, but that's a tall order and one nobody should expect you to fulfill. You have answered exactly the question I intended to ask. Thank you! $\endgroup$
    – J Thomas
    Aug 8, 2022 at 17:40
1
$\begingroup$

In a nucleus like, say, 12C that is stable against beta decay, the stability is very easily explained simply because the nucleus can't decay without violating conservation of mass-energy.

When we model a nucleus as a collection of neutrons and protons, we're making an approximation. They're really composite systems made of quarks, and when such a composite system interacts with another composite system, their structures get disturbed. To see the same effect in a much less exotic context, think of molecular physics. Chemical bonds distort the orbitals and energy levels, so that a hydrogen atom living in an H2 molecule doesn't have the same detailed structure as a free hydrogen atom.

Nuclei normally exist at zero temperature, and at that temperature, the quarks and gluons exist in a phase that's analogous to a superfluid. If you then add a bunch of excitation energy (for example in relativistic collisions of heavy ions), you get a phase transition to a quark-gluon plasma. In the superfluid state, the quarks have very strong three-body correlations that make the clusters behave approximately like neutrons and protons. In the quark-gluon plasma, this is no longer the case.

"So for example, smash an alpha particle into a beryllium nucleus and a neutron comes out. Doesn't that imply that the neutron was in there all along, waiting to get out?"

This is similar to the informal way people have historically described alpha decay. You have a uranium nucleus sitting around, an alpha particle is pre-formed inside, and then after some number of assaults on the potential barrier, it tunnels out. This is OK as a heuristic, but it doesn't actually imply that the uranium nucleus literally consists of clusters of pre-formed alpha particles. In fact, such a structure would violate the Pauli exclusion principle.

niels nielsen says: "successive elements in most of the periodic table are made by adding helium nuclei as subunits to the next lighter element."

This is wrong.

JEB's answer about isospin seems to me to miss the point. Isospin has very little to do with this. This is all about correlations of the sub-units in a system made of composite particles. We get analogous stuff going on in molecular, condensed matter, and plasma physics, where there is clearly nothing like isospin going on. That's why nuclear physicists use terminology that explicitly borrows from these fields: superfluid, quark-gluon plasma.

$\endgroup$
4
  • $\begingroup$ Thank you! I did not understand isospin well enough to want to comment on it, but my impression is that he's arguing that it requires that a deuterium nucleus must be somehow balanced in a way that it can't be if it has a proton and a neutron. Therefore they do not retain their identities as protons and neutrons, and if it isn't true for deuterium there's no reason to assume it's true for anything else. It sounds like a reasonable argument, but I know so little about it that he could spout utter bullshit and I wouldn't know the difference. $\endgroup$
    – J Thomas
    Aug 3, 2022 at 14:31
  • $\begingroup$ You explain that nuclei are normally in a low-energy state, and they can't decay because they don't have sufficient free energy to decay. That makes sense. I tried to ask whether it makes most sense to think of them as protons and neutrons in a low-energy state or something else, and you appear to say that today there are better models. The low-energy state is not exactly protons and neutrons, but something else, which can be called quarks. Thank you. $\endgroup$
    – J Thomas
    Aug 8, 2022 at 17:56
  • $\begingroup$ @JThomas The quark-gluon plasma is an extremely high energy state. The state inside a normal nucleus is very different from that. JEB is certainly not saying that protons & neutrons in a nucleus break down into quarks! $\endgroup$
    – PM 2Ring
    Aug 9, 2022 at 12:16
  • $\begingroup$ I am a dilettante and my understanding of all this is superficial. So I don't know ahead of time whether protons and neutrons break down into some lower-energy-state quark version, or whether protons and neutrons are the lowest-energy-state inside a nucleus. That's closely related to what I was asking in the first place. $\endgroup$
    – J Thomas
    Aug 9, 2022 at 16:05
0
$\begingroup$

I want to discuss the answers without adding long complicated comments that would get sent elsewhere.

First, Lewis Miller provides a definitive answer, with experimental evidence backing it up. In a particular circumstance, something can be measured in a nucleus which appears to be a proton.

I am a little cautious about this because I am not enough of an expert to follow the details well. Maybe physics has gotten to the point that you have to be very much expert in each particular discipline to follow the reasoning.

When they decided there was a nucleus, anybody could follow it. You throw some stuff at atoms and most of it passes through, but very rarely something bounces back. There must be something tiny that it bounces off of, the tiny something is in some specific place, and usually the thrown object misses it.

If they had said that the alpha particles correspond to a quantum wavefunction, and occasionally when the wavefunction collapses it reveals a reflective wall of unknown size, location, and duration, that would have been a more precise statement of what the experiment revealed. But it would have been harder to follow.

Anyway, they establish something like a place for a proton in lead, Pb206. Lead has 81 other protons, and maybe some of them don't have a stable form.

JEB reaches a different conclusion. Deuterium has a kind of symmetry that you can't get from a proton and a neutron. Therefore its nucleus is not a proton and a neutron.

Again the details are hard to follow, but the argument has a pleasing clarity.

But I'm still cautious. As I understand it, there used to be an argument in organic chemistry about single and double bonds. Clearly both of those existed. But what happens in benzene rings, and carboxyl groups? They could be represented as pairs of single and double bonds, but were they really? No, not stably. There was the argument that the single and double bonds switched places very fast, and the argument that really it was just something inbetween. Eventually they decided it didn't really matter and they stopped arguing about it. Could deuterium have a proton and a neutron that switch back and forth so fast that on the isospin timescale they are symmetrical? I can ask that question but I don't have the expertise to answer it.

Some people have been saying that protons and neutrons in nuclei swap quarks back and forth real fast. Apparently they definitely do that sometimes. Do they do it when their energy levels are too low to watch them do it? I don't know. Perhaps it could be argued that it would take too much energy, but then maybe there's a way for them to tunnel through anyway. I bet nobody knows enough about it to prove there's absolutely no way for quarks to tunnel through despite the energy barriers.

My original concept was protons and neutrons that can sit around in a nucleus for a billion years without changing their form. Like a bag of marbles.

Swapping back and forth so fast that they look symmetric, is not that. The original question is starting to make less sense.

Putting it together, my tentative conclusion is that sometimes it makes sense to think of some protons as having an actual proton identity inside a nucleus. And sometimes it doesn't. It varies with atoms, and it might vary with the particular role for a proton in that particular nucleus.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.