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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.

For a normalized trial state $\psi$, expand it in an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$. The Rayleigh quotient is thenThis is the quantum variational principle: minimizing the expectation of the Hamiltonian operator over any trial family gives an upper bound on the ground-state energy.

Let $\psi$ be the normalized ground-state wavefunction and form the normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{1/2}\psi (\lambda x)$. Its kinetic and potential expectations areThe true ground state makes the energy stationary at $\lambda =1$, soThus the power-law quantum virial theorem gives

If $n\le -3$ and $A>0$, the two sides have opposite signs, because $T>0$ and $V>0$. If $A<0$, the scaled energy ${\lambda }^{2}T+{\lambda }^{-n}V$ tends to $-\infty$ as $\lambda \to \infty$, since $-n>2$, so the Hamiltonian is not bounded below. If $A=0$, the free particle has no normalizable ground state. Hence there is no normalizable ground state for $n\le -3$.

For $n=1$ take the normalized Gaussian functionThe Gaussian integral givesand thereforeIts unique minimum satisfiesAt this point $⟨V⟩=2⟨T⟩$, consistently with the virial theorem, and the resulting variational estimate is