Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 15D (Quantum Mechanics)

Use $[AB,C]=A[B,C]+[A,C]B$ and the canonical commutation relation. With $L_i={\epsilon }_{iab}{x}_a{p}_b$,Using these two transformation laws on $L_j={\epsilon }_{jab}{x}_a{p}_b$ givesThese are the orbital angular momentum commutation relations.

NowThe factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:

The same commutators show that $p^2=p_j{p}_j$ and $r^2=x_j{x}_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric ${\epsilon }_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and forwe have $[H,L_i]=0$, henceIn particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By simultaneous diagonalization, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the rotational invariance of a central-potential Hamiltonian.

For a bound state, $E<0$ and therefore $\gamma >0$. As $r\to \infty$, the terms proportional to $1/r$ are subleading and the Radial Schrodinger equation has dominant balanceIts exponential behaviours are $e^{\gamma r}$ and $e^{-\gamma r}$. The growing solution is not normalizable, so

Put $R=f{e}^{-\gamma r}$. ThenAfter substitution and cancellation of the ${\gamma }^{2}f$ terms, the equation becomesFor the power series $f={\sum }_{n=0}^{\infty }{a}_n{r}^n$, the coefficient of $r^{n-1}$ isThus

If the series does not terminate, then $a_n/a_{n-1}\sim 2\gamma /n$, so $f$ acquires the large-$r$ behaviour $e^{2\gamma r}$ and $R$ grows like $e^{\gamma r}$. Normalizability therefore requires termination, which occurs whenfor some positive integer $N$. HenceThe ground state has $N=1$, so the hydrogen ground-state energy isThis is the series-termination quantization of the Coulomb radial equation.

Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is thenThereforeEquivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=({\gamma }^3/\pi {)}^{1/2}$.

The expected radius isFor the ground state, $\gamma =m{q}^2/{\hslash }^2$. The Bohr radius is $a_0={\hslash }^2/(m{q}^2)=1/\gamma$, soThe mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the radial normalization of the hydrogen ground state.