Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 36B (Applications of Quantum Mechanics)

Let ${\psi }_1$ be the normalized ground state forUsing it as a trial state for $H_2=T+V_2$, the variational principle givesThe pointwise assumption $V_2\le V_1$ makes the last expectation nonpositive. Hence

Use the normalized scaled trial stateIfthen scaling derivatives and changing variables giveIf a finite ground state existed, $\lambda =1$ would minimize this expression, and henceIt would follow thatfor $n>2$. But a bound state of an attractive potential tending to zero at infinity must have negative energy. This is a contradiction.

Indeed, the scaling gives the stronger collapse statementwhen $n>2$. Thus the ideal singular Hamiltonian is not bounded below and has no finite ground bound state.

In the basis $(|{\psi }_1⟩,|{\psi }_2⟩)$,Writing $\Delta =E_2-E_1>0$, its exact eigenvalues are

For the normalized trial state,The minimum of $A\cos 2\beta +h\sin 2\beta$ is $-\sqrt{A^2+h^2}$ with $A=\Delta /2$. ThereforeThe variational estimate is exact because the trial family ranges over all real normalized states in the two-dimensional Hilbert space, and the real symmetric Hamiltonian has a real ground eigenvector.