Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 1 / b

Uniform probability per volume does not give a uniform radial distribution. A shell of radius $r$ and thickness $dr$ occupies volume $4\pi r^{2}dr$, so the radial probability density function isIt integrates to one. The second radial moment isThe radial uncertainty scale in a uniform sphere supplies the position-uncertainty scale as instructed by the toy model. It is not literally the standard deviation of the scalar radius: $⟨r⟩=3R/4$ and $\sqrt{⟨r^2⟩-⟨r{⟩}^2}=R\sqrt{3/80}$. Nor is a uniform spatial probability density sufficient to determine a unique momentum space wavefunction. This explicit distinction keeps the order-of-magnitude assumption separate from a fully specified quantum bound-state calculation. Replacing the radial density by a uniform density in $r$ would give a different second moment because it omits the spherical volume element.

Adopt the lower-bound scale of the Heisenberg uncertainty principle, $\Delta r\Delta p\sim \hslash /2$, with the prescribed radial scale from a. Using $\hslash c=197.327$ MeV fm,For a one-sided exponential distribution of a positive momentum magnitude, $f(p)=p_0^{-1}{e}^{-p/p_0}$, both its mean and its standard deviation are $p_0$. Under the additional identification made in this estimate, set the characteristic momentum to $\Delta p$. This identification concerns a positive magnitude; the mean of a signed momentum component in an isotropic stationary state would instead vanish.

Use the relativistic energy-momentum relation, not a nonrelativistic kinetic-energy approximation, because the momentum-energy scale is much larger than the stipulated quark rest energy:This is a characteristic energy estimated at the characteristic momentum. It is not an exact evaluation of $⟨\sqrt{p^2{c}^2+m_q^2{c}^4}⟩$ over a fully specified three-dimensional wavefunction. An order-of-magnitude convention $\Delta p\sim \hslash /\Delta r$ would give an answer about twice as large; the uncertainty principle by itself specifies a bound, not a unique confinement energy.

Adding the three characteristic quark energies from b givesThe estimate recovers a sub-GeV scale but underestimates the stated neutron mass by a factor of a few. This is consistent with its deliberately crude uncertainty and geometry assumptions; it is not a precision mass prediction.

A better model needs the actual spatial and momentum correlations, a confining interaction, and the energy in the quark and gluon fields. In Quantum chromodynamics, hadron masses are not simply the sum of three freely moving particles with tiny masses. The supplied $5$ MeV rest-energy scale resembles a light current-quark scale; a phenomenological constituent-quark mass includes substantial dressing and is a different parameter. Color, spin correlations, relativistic dynamics and sea components all matter when defining a more realistic bound-state Hamiltonian operator or performing a lattice QCD calculation. The prescribed uniform-sphere model supplies none of those extra dynamical inputs, so its discrepancy should not be repaired by an unannounced change of the given radius or mass.