Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 33A (Principles of Quantum Mechanics)

In position representation,which differentiates only with respect to $x$ and $y$. Multiplication by $z$ therefore commutes with it:Taking the matrix element of this commutator between angular-momentum eigenstates givesThus the matrix element can be nonzero only whenThis is the magnetic part of the electric-dipole selection rules for hydrogen.

WriteTaking the stated double-commutator identity between the two eigenstates gives, if the matrix element is nonzero,For nonnegative integers $\ell ,{\ell }^{\prime }$, this equation has precisely the possibilitiesIndeed, ${\ell }^{\prime }=\ell$ forces $\ell =0$, while for unequal values the difference
$({\ell }^{\prime }-\ell )({\ell }^{\prime }+\ell +1)$ satisfies the equation only when its first factor has magnitude one.

Under parity, $z\mapsto -z$, while a hydrogen state of orbital angular momentum $\ell$ has parity $(-1{)}^{\ell }$. Therefore a nonzero matrix element requiresso ${\ell }^{\prime }$ and $\ell$ must have opposite parity. This excludes every case ${\ell }^{\prime }=\ell$, including ${\ell }^{\prime }=\ell =0$. Combining the results gives

The hydrogen ground-state charge cloud is spherically symmetric and has even parity, so it has no permanent electric dipole moment. A uniform electric field couples through a perturbation proportional to $\mathbf{E}\cdot \mathbf{r}$, which is odd under parity. Its expectation in the ground state therefore vanishes, giving no first-order energy correction.

The field can nevertheless polarize the atom: it admixes odd-parity excited states and creates an induced dipole. Such mixing contributes from second order onward, producing the quadratic Stark effect of the hydrogen ground state.

Choose the $z$-axis along $\mathbf{E}$, so the perturbation isup to an irrelevant sign convention for the electron charge. By second-order nondegenerate perturbation theory,Parts (a) and (b) restrict the contributing bound states toFor hydrogen,and henceThereforeEvery denominator is negative, so the discrete-state contribution lowers the ground-state energy.