Neutron mass: 1.008664 u

Proton mass: 1.007276 u

Why the discrepancy?

On a related note, how does one go about measuring the mass of a neutron or proton, anyway?
['https://physics.stackexchange.com/questions/85', 'https://physics.stackexchange.com', 'https://physics.stackexchange.com/users/104/']
A recent paper [1] has computed the mass difference ab initio from the theory, using a technique know as lattice, where one writes the equations on a discretised spacetime (this is conceptually similar to what engineers do with finite elements method to study how a beam of metal flexes for example). They took into account 

* the contributions of electromagnetism and of the strong interaction,
* the difference of mass between up and down quarks,
* up, down, strange, and charm in the quark sea [\*],

Their result is

$$m\_n-m\_p = 1.51(16)(23) \text{MeV},$$

to be compared with the experimental value,

$$m\_n - m\_p = 1.29,$$

with an error orders of magnitude smaller. This is pretty impressive: an accuracy of 390 keV, and according to the paper a proof that $m\_n > m\_p$ at the level of 5 standard deviations. Again, ab initio, just from the Standard Model.

The authors of that paper also give the respective contributions of electromagnetism and of the strong interaction. This is very interesting as the electrostatic energy, just based on the quark charges, would naively be thought to be negative for the neutron but zero for the proton, on average [\*\*]. Their result is that

* $(m\_n-m\_p)\_\text{QCD}=2.52(17)(24)$ MeV,
* $(m\_n-m\_p)\_\text{QED}=-1.00(07)(14)$ MeV,

where QCD stands for Quantum ChromoDynamics, our best theory of the strong interaction, and QED for Quantum ElectroDynamics, our best theory of electromagnetism. So indeed, the different of electrostatic energy is in the intuitive direction I have just highlighted above. This is compensated by the contribution of the strong interaction which goes in the other direction. I don't have an intuitive reason for that to share.

[\*] Protons and neutrons are made of 3 so-called valence quarks, but also of a "soup" of other quarks-antiquark pairs which is called the sea.

[\*\*] The sum of the product of the charges are $\frac{2}{3}\frac{2}{3} + \frac{2}{3}\frac{-1}{3}+ \frac{2}{3}\frac{-1}{3}=0$ for the proton but $\frac{2}{3}\frac{-1}{3} + \frac{2}{3}\frac{-1}{3}+ \frac{-1}{3}\frac{-1}{3}=-\frac{1}{3}$ for the neutron.

[1] Sz. Borsanyi, S. Durr, Z. Fodor, C. Hoelbling, S. D. Katz, S. Krieg, L. Lellouch, T. Lippert, A. Portelli, K. K. Szabo, and B. C. Toth. Ab initio calculation of the neutron-proton mass difference. Science, 347(6229):1452–1455, 2015. [arxiv.org/abs/1406.4088](https://arxiv.org/abs/1406.4088)
We can write approximately assuming strong energy $E\_s$ contribution inside proton and neutron is almost the same:

$m\_n c^2 = m\_d c^2 +m\_d c^2 +m\_u c^2 +E\_s $

$m\_p c^2 = m\_u c^2 +m\_u c^2 +m\_d c^2 +E\_s +E\_c $
in terms of quark u and d masses and $E\_c$ - electrostatic energy around and inside the proton, which can be calculated classically $E\_c = \frac{3}{5} \frac{k e^2} {R} = 1 MeV $, and R=0.87 fm is charge radius from scattering experiments.
Thus we have
 $m\_n c^2 -m\_p c^2 $ = $ m\_d c^2 - m\_u c^2 - E\_c $ = 4.8 MeV - 2.3 MeV - 1 MeV= 1.5 MeV

On the other hand we have $m\_n c^2 -m\_p c^2 $ = 1.3 MeV i.e very close (and quark masses are not know so precisely)