Free neutrons are unstable. It decays to proton, electron and an antineutrino via beta decay. Can we not do a quantum field theory calculation to predict the precise the decay width? Its inverse should then tell us the neutron lifetime. However, sometimes I hear that the lifetime is 8 minutes, sometimes I hear it is 10 or 12 minutes. What's the matter here?

What is the precisely calculated number, if calculable? If the lifetime really has such a large uncertainty, what is the reason for that?

  • 1
    $\begingroup$ Neutron decay is probabilistic just like radioactive decay. We can give a precise half life, but the lives of individual neutrons will vary. $\endgroup$ Feb 17, 2021 at 12:12
  • $\begingroup$ So $\Gamma^{-1}(n\to p e^-\bar{\nu}_e)$ gives the half life where $\Gamma(n\to p e^-\bar{\nu}_e)$ is the beta decay width? $\endgroup$ Feb 17, 2021 at 12:15
  • $\begingroup$ If you could measure the total energy released that would be true but we cannot measure the energy of the antineutrino, or at least not to any useful accuracy. So it is experimentally impossible to determine the line width. $\endgroup$ Feb 17, 2021 at 12:17
  • $\begingroup$ Heard where? Wikipedia says $881.5 sec.$ Do you mean theoretically calculate or experimentally measure? $\endgroup$
    – Qmechanic
    Feb 17, 2021 at 12:30
  • $\begingroup$ @Qmechanic I do not have any source to cite. I have heard different people quoting different times of neutron lifetime. I never did a calculation of neutron decay with though I calculated pion and muon decay widths in quantum field theory. Do you have any idea where the number 881.5s comes from? Does it come from putting experimental values to a theoretical decay with formula? $\endgroup$ Feb 17, 2021 at 12:34

1 Answer 1


The 2018 Particle Data Group gives a value of $880.2\pm 1.0$ s for free neutron lifetime, as an average of the seven best measurements.

Neutron lifetime plot from PDG

As can be seen, the measurements have non-overlapping confidence intervals. As discussed in (Wietfeldt 2014) the different experimental methods do not agree on the value.

Wietfeldt gives the formula of the lifetime as $$\tau_n = \left(\frac{2\pi^4\hbar^7}{m_e^5c^4f_R}\right)\frac{1}{G_V^2+3G_A^2}.$$ $f_R$ is a phase space factor for the final state and radiative corrections, $G_V$ and $G_A$ are the nucleon vector and axial vector coupling constants. $G_V=G_FV_{ud}$ where $G_F$ is a universal weak coupling constant and $V_{ud}$ is the first element of the CKM matrix. Good measurements of $\tau_n$ would help determine the values of these constants better; current estimates of $V_{ud}$ are from nuclear or pion decays.

One can do theoretical calculations using weak force diagrams, but I get the feeling that the numerical value still depend on a lot of empirically measured constants.

  • $\begingroup$ Thanks for the answer. I have two questions. 1. What do they do in these experiments? The most straightforward thing would be to start with a bunch of free neutrons at rest and wait till the neutron number decays to half of the initial value. That's it. That will give the half-life. But I don't know if such a simple set up possible to arrange where one can start with a large number of neutrons at rest. Is this what they do in these experiments? 2. The formula that you wrote for $\tau_n$ is a theoretically obtained formula and not an empirical formula to fit the experimental curves. Am I right? $\endgroup$ Feb 17, 2021 at 13:40
  • 1
    $\begingroup$ @mithusengupta123 - (1) Wietfeldt describes the methods used. The most common consists of sending a cold beam of neutrons through a Penning trap and counting the number of accumulated protons. In the bottle method ultracold neutrons are kept in bottles and then counted. (2) Yes, the formula is derived from theory. $\endgroup$ Feb 17, 2021 at 15:03
  • $\begingroup$ Many thanks for clarifications. Just one more. $\tau_n$ is referred to as the neutron lifetime. Is it the half-life as John Rennie suggested? $\endgroup$ Feb 17, 2021 at 15:53
  • 2
    $\begingroup$ @mithusengupta123 no, that figure of 880.2 seconds is the mean lifetime. To get the half life multiply this by $\ln(2)$ i.e. 610.0 seconds. $\endgroup$ Feb 17, 2021 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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