Variational method

The Rayleigh quotient of any normalized trial state is at least the ground-state energy.

For a regular one-dimensional confining potential, the bound-state eigenfunctions can be ordered by their number of nodes. The unique nodeless normalizable eigenfunction is the ground state.

Forthe nodeless wavefunction $e^{-x^4/4}$ is an exact eigenstate with energy ${\hslash }^2/m$, and hence is the ground state.

For the trial state $e^{-\alpha x^2/2}$, the dimensionless Rayleigh quotient isIts unique minimizer isand at the minimum $\epsilon ({\alpha }_{\ast })=2+2{\alpha }_{\ast }/3-1/{\alpha }_{\ast }$.

If $V_1\ge V_2$ pointwise for Hamiltonians with the same kinetic term, using the ground state of $H_1$ as a trial state for $H_2$ gives $E_1\ge E_2$.

For a normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{d/2}\psi (\lambda x)$ and a homogeneous potential of degree $-n$, the energy is ${\lambda }^{2}T+{\lambda }^{n}V$. Stationarity at an eigenstate gives $2T+nV=0$.

For an attractive potential $-\alpha /|x{|}^n$ with $n>2$, dilations make the negative potential energy grow faster than kinetic energy, sending the Rayleigh quotient to minus infinity.

If a normalized trial family covers the entire two-dimensional Hilbert space, minimizing its Rayleigh quotient gives the exact lower eigenvalue rather than merely an upper bound.

The eigenvalues of $\left(\begin{array}{cc}{E}_1& h\\ h& E_2\end{array}\right)$ are their mean plus or minus $\sqrt{(E_2-E_1{)}^2/4+h^2}$, so nonzero coupling increases their separation.