I imagine that there is a pretty simple answer to my question, but I have just never gotten it straight. If a proton is comprised of two up quarks and a down, and neutrons are comprised of two down and an up, how can a neutron be a proton and electron?


3 Answers 3


A neutron is not "a proton and an electron". A neutron is not composed of a proton and an electron inside of the neutron.

In quantum mechanics, particles can appear and disappear or change into other particles. With the neutron, one of the down quarks can decay change into an up quark by emitting a W boson, turning into a proton. The W boson quickly decays into an electron and an electron antineutrino. The new up quark didn't exist until the down quark turned into it. The W boson is what is called a virtual particle. It doesn't exist in the classical sense, it's just kind of there in the ambiguous region of spacetime where the decay occurs. The electron and antineutrino didn't exist until the decay.

Here is a Feynman diagram of the process, from here:

enter image description here

  • 1
    $\begingroup$ The W boson exists, but there will never be a W boson that isn't a virtual particle. There are up and down quarks aren't virtual particles. $\endgroup$ Jul 26, 2017 at 15:18
  • 12
    $\begingroup$ @user21820 the W boson in this diagram is more massive than the neutron, proton, electron, and neutrino combined -- and it is not just "a little more massive" either; it's ~80 times more massive than the neutron in the diagram. So there is not enough energy to create a literal W boson. What is really happening is an instance of a phenomenon called quantum tunneling, which says "even though quantum mechanics obeys conservation of energy in its final results, the probability of those final results can be influenced by processes which violate conservation of energy in intermediate stages." $\endgroup$
    – CR Drost
    Jul 26, 2017 at 15:25
  • 2
    $\begingroup$ @user21820 a virtual particle is any particle that doesn't exist at the beginning or end of an interaction. They're nothing more than a mathematical construct. They are what is called off shell, which means that their energy, mass, and momentum don't obey the usual $m^2=E^2-p^2$ relationship that on shell real particles obey. The idea of "borrowing energy from the universe" is a cat oversimplification that greatly distorts what's really going on. $\endgroup$ Jul 26, 2017 at 15:45
  • 2
    $\begingroup$ @user21820 "borrowing energy" isn't real, so things that are only explained through "borrowing energy" are also not real. $\endgroup$
    – hobbs
    Jul 26, 2017 at 20:06
  • 3
    $\begingroup$ @user21820 why should we provide a counterexample to a bad analogy? "Borrowing energy" is not how virtual particles behave. End of story. If you want the details, takes a semester or two of graduate level QFT. And despite what pop sci magazines may call it, Hawking radiation is not virtual particles. $\endgroup$ Jul 27, 2017 at 2:50

A neutron isn't a proton and an electron.

The reaction involved in beta decay is $$n \to p + e^- + \bar{\nu}_e$$ where $\bar{\nu}_e$ is an electron anti-neutrino.

But even that doesn't mean a neutron is a proton plus an electron plus a anti-neutrino. It means that a neutron's quantum numbers are the same as a state consisting of a proton an electron and an electron anti-neutrino with the proper angular momentum relationship. And that plus the fact that a neutron's mass energy exceeds that of the products means that the decay is both allowed and mandatory.

  • 5
    $\begingroup$ Choice of "mandatory" required by law or mandate; compulsory is intriguing. Compulsory decay. Maybe "inevitable" is less anthropomorphic? $\endgroup$
    – uhoh
    Jul 26, 2017 at 0:48
  • 6
    $\begingroup$ I think the word "mandatory" is a little misleading. There is a characteristic decay time for that process (around 15 minutes for a free neutron, much longer if the neutron is stabilized by the strong force from neighboring hadrons in an atomic nucleus). Over time scales much longer than this decay time, the probability of the decay occurring comes arbitrarily close to 100%, but it never becomes exactly 100% - there's always some tiny probability that the neutron will fail to decay over any arbitrarily long time period. $\endgroup$
    – tparker
    Jul 26, 2017 at 1:51
  • 4
    $\begingroup$ "Mandatory" was a reference to Gell Man's totalitarian principle, though as uhoh noticed I misremembered the quote. $\endgroup$ Jul 26, 2017 at 2:03
  • 1
    $\begingroup$ @dmckee I suggested "compulsory" as humor, it's dumb luck that it turns out to be Gell-Mann's precise term! Thank you for the history of QM min-lesson — "totalitarian" sounds even more frightening considering the era :) $\endgroup$
    – uhoh
    Jul 26, 2017 at 4:17
  • $\begingroup$ Thank you both for answering my question and eliminating my misconceptions. Super helpful $\endgroup$ Jul 26, 2017 at 13:07

The other answers are right, but I'd like to offer the opposite statement, too: A neutron is a proton plus an electron.

One wonderful thing about nuclear and particle physics is that you can do "arithmetics" with particles. In some sense, you can say a proton and an electron add up to a neutron, $p + e^{-}= n$. This is what you have in beta decay (if you are missing something, bear with me). But you can also "do the same on both sides" like in math (aka equivalence transformations). Lets add a positron:

$$\begin{aligned} n &= p + e^{-} \qquad &\big|_{+e^{+}} \\ \Rightarrow \quad n + e^{+} &= p + e^{-} + e^{+} = p & \end{aligned}$$

Note that the positron and the electron cancel out. The resulting formula $p = n + e^{+}$ can also happen in nature, and is called beta-plus decay. You can do all kinds of transformations, for example subtracting particles (which is the same as adding antiparticles). You can also turn an electron into a muon by removing an electron neutrino and adding a muon neutrino. I like to think of it as removing the electron-ness, and adding muon-ness:

$$e - \nu_{e} + \nu_{\mu} = \mu$$ or more conventionally: $$e + \bar\nu_{e} + \nu_{\mu} = \mu$$

This calculus works on the scale of nuclei, nucleons, and even quarks. This is the Feynman diagram from Johnathan Gross' answer:

$$\begin{aligned} p &= u+u+d \\ &= u+d+(u+W^{-}) \\ &= u+d+u+(\bar\nu_e+e^-) \\ &= n + \bar\nu_e+e^- \end{aligned}$$

Now we see that the formula in the beginning is incomplete, we were missing the neutrino. The reason it seemed to work without is that we were only considering the electric charge, but the neutrino is electrically uncharged.

The reason these cute calculations work is essentially a property called crossing symmetry, and the fact that quantum numbers are preserved. I think of this like a beginners version of Feynman diagrams (and in fact I think this is usually taught first. I figured this out in school, in the context of nuclear decay, and this was a major "wow" moment that increased my interest in particle physics.)

Of course, there are some downsides to this simplistic view. The most important is that there is no consideration of masses, and mass defects. Only heavier particles can decay into lighter ones, plus energy. But apart from that, this "calculating with particles" can be very useful, for example if you forgot whether to put a neutrino or an anti-neutrino.


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