# What experiment(s) have or can refute the existence of an electron-particle “system” over the separate existence of a neutron within itself?

This question actually came about from a discussion of another question posed here

The neutron is known to be comprised of an electron and a proton, and there are observations that the neutron can be created by these particles or alternately decay into these particles. Furthermore the stability of the neutron is vastly improved if it remains bound within the nucleus. So presumably the nuclear force helps to maintain the neutron's stability. Once outside the nucleus the neutron has a much harder time keeping itself together, and once again becomes a proton and electron. And just to keep matters, and this post more simple and shorter, I'm neglecting the additional fact that the electron antineutrino is also produced in the decay.

Now getting to my question, I'm wondering why we have to consider the neutron as a separate particle within itself rather than a proton-electron "system" of particles? Is there an experiment that refutes this way of looking at things? (note I'm asking for an experiment or set of experiments, not just theories)?

• "The neutron is known to be comprised of an electron and a proton ..." dmckee didn't say that (in the comment you linked to). – 299792458 Apr 28 '15 at 14:37
• Thanks for the down-votes guys. I'm just asking a legitimate question from my own point of view. I'm not a particle physicist. Just trying to learn. – docscience Apr 28 '15 at 14:57
• I believe experiments have been done that show protons and neutrons are composed of fractionally-charged quarks, so presumably that would refute the neutron being composed of proton + electron? – Time4Tea Apr 28 '15 at 15:07
• "The neutron is known to be comprised of an electron and a proton" In fact, no one is saying that but docscience because this is not the understanding of the scientific community. – dmckee --- ex-moderator kitten Apr 28 '15 at 15:38
• As an experimental high energy physicist I can get as many particles out of a neutron as the energy of my government funded accelerator allows, so I know that docscience is completely wrong about there being one electron and one proton in there. There is all kinds of crap in there, and it all comes out if I just hit it hard enough. :-) – CuriousOne Apr 28 '15 at 16:12

An early bit of evidence for the neutron as an uncharged constituent of the nucleus, with approximately the mass of the proton, actually comes from the exclusion principle, and the low-temperature heat capacities and excitation spectra of atomic gases. The argument is a little bit subtle, so you'll have to bear with me.

First, we have the exclusion principle for electrons: the probability of finding two different electrons in exactly the same quantum-mechanical state is zero. This is accomplished in the mathematics of quantum mechanics by demanding that electron wavefunctions be "antisymmetric under exchange." That is, if I have my "first" electron in some state $\left|a\right>$, and my "second" electron in some state $\left|b\right>$, then my total wavefunction must be (up to an irrelevant constant) $$\left|\text{total}\right> = \left|a\right>_1 \left|b\right>_2 - \left|b\right>_1 \left|a\right>_2$$ This construction has two effects. First, if someone else comes along and labels the "first" and "second" electrons differently --- or, if they swap places --- then the wavefunction $\left|\text{total}\right>$ changes sign. This is what's meant by "anti"-symmetry. Second, if the two states $\left|a\right>$ and $\left|b\right>$ are the same, then the wavefunction $\left|\text{total}\right>$ is identically zero. This is the "exclusion" part: there is simply no permitted description of two electrons in the same state.

Now, what other particles obey the exclusion principle? The hydrogen molecule, $\rm H_2$, is a nice system, because it is made only of electrons and protons. The molecule is a rotor whose energy is determined by the angular momentum quantum number $L$. The total nuclear spin, $S$, doesn't have much effect on the energy because the interaction between the two proton spins is quite feeble; this feebleness also makes the protons' total spin quite stable against changes in temperature of the gaseous phase of the material. There are four ways for the proton spins to combine (again ignoring a normalization constant): \begin{align} \text{antisymmetric: } \left|S=0\right> &= \quad \left| \downarrow\uparrow \right> - \left| \uparrow\downarrow \right> \\ \text{symmetric: } \left|S=1\right> &= \left\{ \begin{array}{c} \left| \uparrow\uparrow \right> \\ \left| \downarrow\uparrow \right> + \left| \uparrow\downarrow \right> \\ \left| \downarrow\downarrow \right> \\ \end{array} \right. \end{align} Since there are three times as many ways to be symmetric, we expect that hydrogen gas at high temperature will eventually have about 75% of its molecules in the symmetric $S=1$ state.

Under exchange symmetry, states with even $L$ are symmetric, while states with odd $L$ are antisymmetric. If the proton obeyed the exclusion principle, our argument about overall antisymmetry would suggest that hydrogen molecules with $S=0$ should only have even $L$, and molecules with $S=1$ should only have odd $L$, so that every hydrogen molecule is antisymmetric under exchange. And this is in fact what we see. In the excitation spectra of warm hydrogen gas, the transitions between states with odd $L$ are about three times more common than transitions between states with even $L$, and transitions which change $L$ from odd to even or vice-versa are missing. However, when you liquify hydrogen all the molecules will try to enter the $L=0, S=0$ "parahydrogen" ground state (a transition which releases nearly as much heat per molecule as the heat of vaporization), and recently-liquified warm hydrogen gas undergoes only the even-$L$ transitions. This distinction between parahydrogen and "orthohydrogen" (the $S=1$, odd-$L$ component) was the first evidence that protons, like electrons, obey the exclusion principle. (The spin-statistics theorem came later.)

So let's consider a model where

• electrons obey the exclusion principle
• protons obey the exclusion principle
• the neutron is an electron-proton bound state, which is distinct from the neutral hydrogen atom for reasons unspecified.

Right away we see that the neutron must have a different behavior under exchange symmetry than the electron or proton. If exchanging two electrons makes a wavefunction change sign, and exchanging two protons makes a wavefunction change sign, then exchanging two neutrons must make the wavefunction change sign twice --- that is, no change at all! Such a neutron therefore doesn't obey the Pauli exclusion principle. We can use this information to predict the behavior of the other lightweight diatomic gases:

• molecules of oxygen (with an even number of protons) should be symmetric under exchange, and the even-$L$ spectrum should dominate over the odd-$L$ spectrum. The model gets this one right: in fact, in $\rm O_2$ molecules the odd-$L$ spectrum is completely missing, consistent with the oxygen nucleus having total spin zero.

• molecules of fluorine (with an odd number of protons) should be antisymmetric under exchange, and like hydrogen should have more odd-$L$ transitions than even-$L$. The model gets this one right, too: the fluorine nucleus has spin $\hbar/2$, like the lone proton.

• molecules of nitrogen (with an odd number of protons) should be antisymmetric under exchange, and like hydrogen should have more odd-$L$ transitions than even-$L$. Here we are in trouble: reality is the other way around, and the statistics are consistent with the nitrogen nucleus having a spin of $\hbar$.

The contemporary explanation, of course, is that the electron-proton neutron model is wrong: the neutron has spin $\hbar/2$, obeys the exclusion principle, and the exchange symmetry of a nucleus depends on whether the number of nucleons is odd or even, rather than the number of protons alone. Of course neutrons had to be produced before they could be said to have been "discovered," but the need for a neutral nucleon obeying the exclusion principle was already established.

At this point you might like to backpedal and wonder whether the neutron is a metastable bound state of the proton, electron, and antineutrino. But that opens you up to a whole flurry of other questions. Why not a proton and a negative pion? Why are any nuclei stable against beta decay? Can such a model predict (as the standard model does) rates of beta decay and cross sections for neutrino capture? Why is the neutron's size comparable to the proton's size, given the uncertainty in the momentum for its lightweight bound constituents? Why can neutron-nucleus interactions near beta-decay energies trigger strong and electromagnetic interactions, but can't trigger weak interactions? The list goes on and on.

The neutron is in no way "composed" of a proton and an electron. It can decay to a proton, electron, and an antineutrino. But that doesn't mean that these three particles literally co-exist inside the neutron at the beginning. Instead, the decay involves some real transmutation of elementary particles. The only thing that one can say because of the decay is that the neutron contains the same value of conserved charges as (and at least as large energy as) the union of the "proton, electron, and antineutrino".

Such transmutations are occurring all the time, according to quantum field theory. Particles may even be created from pure energy (out of nothing) and annihilated into pure energy. For example, when two protons at the LHC collide, their total energy is 13 TeV (well, it will be in weeks) and 10 new protons, 15 new antiprotons, some neutrons, antineutrons, positrons, and lots of neutrinos may be created, not to mention pions and muons. That's mundane and allowed because the conservation laws – for electric charge, energy, and momentum – are easily obeyed. Energy – the kinetic energy of the colliding protons – may take the form of the mass of new particles, according to Einstein's $E=mc^2$. Because Nature allows "individual particles" to be created as well as annihilated, it allows their "type" to change, too (this change may be viewed as a combination of annihilation and creation), and the proton decay is an example of such a change.

The object composed of a proton and an electron is known as the hydrogen atom. One can make many experiments that prove that the neutron isn't a hydrogen atom. For example, the hydrogen atom may be ionized by 13.6 eV radiation but the neutron won't care about such low-energy radiation at all.

The neutron is as small as the proton itself. The hydrogen atom – which contains the extra electron, aside from the proton – is about 10,000 times larger. This has the implication that the neutron behaves like an uncharged nucleus and exhibits no atomic-physics phenomena, just nuclear physics, while bound states involving electrons always have atomic-physics (chemical...) properties.

It's the presence of the electron in the atom that makes the object rather large. The atoms are large thanks to the uncertainty principle. For a light particle such as an electron to obey the uncertainty principle, a limited kinetic energy – a highly limited momentum – leads to a huge uncertainty in the position, and this uncertainty $\Delta x$ determines the size of the atom. There are no electrons inside a neutron so there's no need to "squeeze" such electrons into a small space.

According to QCD, Quantum Chromodynamics, a proton is composed of 2 up-quarks and 1 down-quark while a neutron is composed of 2 down-quarks and 1 up-quark (plus lots of gluons and particle-antiparticle pairs etc., in both cases). From this QCD viewpoint, the proton and the neutron are exactly equally "composite" (not too elementary). None of them contains an electron.

In fact, even without QCD, your logic would allow to claim that the proton is more composite than the neutron. Just like a neutron may decay to a proton, electron, antineutrino, a proton bombarded by gamma rays (which are "nearly nothing") may decay into a neutron, a positron, and a neutrino. But the proton isn't "made" of a neutron and a positron, either. The situation is analogous to the opposite one (except that we needed the extra photon because the proton was lighter than the neutron, and stable – or almost stable).

• And the specific experiment(s) you cited to refute my position? – docscience Apr 28 '15 at 14:52
• Nice complete, serious answer to a novice. – garyp Apr 28 '15 at 15:33
• Dear @docscience, experiments can only refute a "position" if it is sufficiently well-defined to make at least some predictions that are not guaranteed a priori. With any understanding of the "composition" we have seen, and with some knowledge what the composition means according to quantum mechanics, your theory makes the prediction that the neutron will behave like the hydrogen atom or be indistinguishable, and easy experiments are surely enough to refute this prediction. You may refuse the assumptions "what QM implies about composite states" but then you have to write a whole new theory. – Luboš Motl Apr 28 '15 at 15:34
• @doc You seem to be asking for a small number of experiments which nail down the reliability of both quantum mechanics and quantum field theories. There is not a short list of experiments that do this, there are thousands (literally) of experiments that contribute to the high degree of certainty we put in our understanding of the world of the very small. If you wanted a small number to hint that your idea has direct problems then Time4Tea's suggestion in the comments is good: measurements of the charges of the constituent partons in Drell-Yan. – dmckee --- ex-moderator kitten Apr 28 '15 at 15:36