Past exam of the physics A course of the University of Cambridge / 2016 / Paper 1 / Section C / C11

For the nonrelativistic Coulomb Hamiltonian with reduced electron-proton mass $\mu$, separation in Spherical coordinate system gives ${\psi }_{n\ell m}(r,\theta ,\varphi )=R_{n\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The quantum numbers obey $n\ge 1$, $0\le \ell \le n-1$ and $-\ell \le m\le \ell$. These are joint eigenstates of energy, $L^2$ and $L_z$, with the latter two eigenvalues $\ell (\ell +1){\hslash }^2$ and $m\hslash$.

The Coulomb bound-state energies and characteristic length areFor example, ${\psi }_{100}=(\pi a_{\mu }^3{)}^{-1/2}{e}^{-r/a_{\mu }}$ is spherically symmetric and has no radial node. In general the radial part contains an exponential times a Generalized Laguerre polynomial, with $n-\ell -1$ radial nodes. The spherical harmonic specifies angular structure and azimuthal phase; real linear combinations of degenerate $m$ states give familiar directional orbitals.

The leading Coulomb energy depends on $n$ alone, giving $n^2$ spatial states for each principal shell, or $2n^2$ including electron spin. Fine structure, Lamb shift and Hyperfine structure lift parts of this ideal degeneracy. An atomic orbital is a probability amplitude, not a classical planetary path.^[ref]

Quantum tunnelling is supported by transport and decay through barriers that classical particles of the same energy cannot traverse. Three complementary examples are:

Together these test barrier penetration in nuclear decay, vacuum-gap transport and semiconductor transport. Their detailed currents depend on state supply and matching as well as the barrier transmission probability.

For the Quantum harmonic oscillator defineThe Canonical commutation relation gives $[\hat{a},{\hat{a}}^{†}]=1$, andThus the creation operator and annihilation operator raise or lower the energy by one quantum. Positivity of ${\hat{a}}^{†}\hat{a}$ and termination of downward stepping give $E_n=\hslash \omega (n+1/2)$, $n=0,1,\dots$. Normalized actions are $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The annihilated ground state obeysThe nonzero Ground state energy is the zero-point energy, consistent with position-momentum quantum uncertainty. Since $\hat{x}=\sqrt{\hslash /(2m\omega )}(\hat{a}+{\hat{a}}^{†})$, a perturbation linear in position connects only neighbouring number states. The algebra is especially useful for matrix elements and selection rules.^[ref]