Past exam of the particle and nuclear physics course of the University of Cambridge / 2016 / Paper 1 / 4

An S wave has $L=0$ and positive parity. The deuteron's $J=1$ therefore requires a spin-one-half triplet state, $S=1$. The two-nucleon isospin description makes this the antisymmetric $T=0$ channel. Its existence proves sufficient attraction to bind in the ${}^3{S}_1$-dominated channel.

A spin one-half pair could also form $S=0$. The absence of a bound ${}^1{S}_0$ deuteron indicates that the singlet interaction does not bind, not that it is everywhere repulsive. Together with triplet binding, this shows that a spin-independent central attraction is insufficient. No radial excited bound level also constrains the strength and range: the attraction supports only a shallow low-energy bound state.

For identical neutrons or protons, two-particle isospin has $T=1$ and is symmetric. At even $L$, spatial exchange is symmetric, so the Pauli exclusion principle requires an antisymmetric spin state, $S=0$. A low-energy $nn$ or $pp$ pair cannot occupy the deuteron's $S=1,T=0$ channel. With approximate charge independence of nuclear forces, their singlet strong interaction corresponds to the unbound neutron-proton singlet; proton-proton electrostatic repulsion further disfavors binding. There is no bound dineutron or diproton in these physical low-energy channels. The inference uses charge independence and singlet-channel behavior; the deuteron's quantum numbers alone do not constrain every imaginable potential.

The nuclear binding energy is the separated-nucleon rest energy minus the bound-state rest energy. Consistent nuclear masses avoid atomic-electron bookkeeping:It is positive because the bound deuteron has lower total energy.

In a pure $L=0,S=1$ stretched state both spins point along the total angular momentum, with no orbital contribution. The nuclear magnetic moment would beThe measured $0.857{\mu }_N$ differs. More decisively, a pure S wave spatial state is isotropic and has zero electric quadrupole moment. Its nonzero value requires a nonspherical component of the quantum charge distribution.

Positive parity allows even $L$; coupling $S=1$ to $J=1$ permits $L=0$ and $L=2$. The state contains a deuteron D-state admixture:with radial wavefunctions implicit. A nuclear tensor force couples the orbital components. The quadrupole moment involves their interference as well as a D-state contribution, making it sensitive to a small admixture. This is a quantum angular correlation, not two particles permanently fixed in space.

In the nonrelativistic one-body magnetic-moment approximation,The D-state coupling has $⟨S_z⟩=-1/2$ and $⟨L_z⟩=3/2$ at $J_z=1$; the proton supplies approximately half the relative orbital magnetic moment. ThusThe robust conclusion is a predominantly S wave state with a small tensor-induced D wave component. The 4% estimate is model dependent: exchange currents and other corrections also affect the magnetic moment, and a wavefunction component probability is not itself a uniquely measured observable.^[ref]

Use the relative coordinate and reduced massTake $V=-V_0$ inside, $V=0$ outside, and $E=-B$. For $V_0>B>0$, define inverse lengthsThe radial equation becomes $u^{\prime \prime }+k^{2}u=0$ inside and $u^{\prime \prime }-{\kappa }^{2}u=0$ outside. Finiteness of the full wavefunction at the origin imposes $u(0)=0$; normalizability excludes the growing exterior exponential. HenceContinuity of $u$ was built in. Continuity of its derivative at the finite step givesFor the nodeless ground state, $\pi /2<kb<\pi$ selects the first branch. Higher roots would correspond to different wells with more bound states. Normalization is

For the requested weak-binding estimate, $B/V_0\to 0$ makes $\kappa /k\to 0$. On the first branch $kb\to \pi /2$, and $k^2\simeq 2\mu c^2{V}_0/(\hslash c{)}^2$. The leading threshold estimate isThis is the zero-binding threshold of a spherical square well, so nonzero positive binding needs a depth above it. Corrections start at order $\sqrt{B}$, not just at order $B$: setting $kb=\pi /2+ϵ$ gives $ϵ\simeq \kappa b/(\pi /2)$ and

The actual numbers permit an accuracy check: $\kappa =0.231608{\mathrm{fm}}^{-1}$ and $\kappa b=0.486377$ are not extremely small. Retaining the exterior decay, solveThis gives $x=kb=1.830503$, and the full square-well result isNeglecting $B$ only in the interior kinetic relation while retaining the measured exterior decay instead estimates $V_0\simeq 31.50\mathrm{MeV}$. These approximations should be identified: 23.2 MeV is the leading $B\to 0$ threshold, whereas matching the supplied nonzero binding requires a boundary phase shift and a deeper well.

In neutron-proton radiative capture, the initial pair at rest has total energy $M{c}^2$, where $M=m_p+m_n$. The emitted photon carries momentum, requiring equal and opposite nuclear recoil. Some binding energy becomes deuteron kinetic energy.

Exact two-body kinematics in natural units givesSince $m_d=M-B$,With the stated rounded masses,The photon energy is $2.222683\mathrm{MeV}$. At leading order the difference equals the nonrelativistic recoil estimate $p_{\gamma }^2/(2m_d)\simeq B^2/(2m_d)$; exact kinematics fixes the small distinction between using $m_d$ and $M$ in that approximation.