Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 33C / Solution

For a normalized trial state $|\psi ⟩={\sum }_n{c}_n|n⟩$ in the form domain of a bounded-below Hamiltonian, the quantum variational principle gives $⟨H⟩={\sum }_n|c_n{|}^2{E}_n\ge E_0$. Thus each trial expectation is an upper bound for the ground-state energy. The same spectral argument covers a continuum: for a decaying potential whose essential spectrum begins at $0$, a negative trial expectation guarantees a negative-energy bound state.

For the stated exponential trial function, normalization follows from $4{\alpha }^3{\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=1$. The kinetic expectation, using $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)$. The potential expectation is $-4A{\alpha }^3{\int }_0^{\infty }r{e}^{-(\mu +2\alpha )r}dr$. ThereforeThe exponential variational bound for the Yukawa potential at $\alpha =\mu /2$ gives $E=(\mu /8)({\hslash }^2\mu /m-A)<0$ if $\mu <Am/{\hslash }^2$. Thus a bound state necessarily exists in the stated range.

For $\mu \ge Am/{\hslash }^2$, this single-exponential family cannot prove binding: the maximum of $4A\alpha /(\mu +2\alpha {)}^2$ is $A/(2\mu )$, so every $E(\alpha )\ge 0$. This does not prove absence of bound states.

The equality case can in fact be strengthened. At $\mu =Am/{\hslash }^2$, use normalized exponentials ${\psi }_{\mu /2},{\psi }_{\mu }$ and write $Q={\hslash }^2{\mu }^2/m$. Their matrix elements areThe Rayleigh numerator for ${\psi }_{\mu /2}+ϵ{\psi }_{\mu }$ is negative for sufficiently small $ϵ>0$, and its norm is nonzero. Hence binding also holds at equality, and by continuity in a small interval beyond it. For arbitrary larger $\mu$, these trials alone do not settle existence; a nonnegative variational upper bound cannot establish a nonnegative ground-state energy.