13.1: Variational Principle
( \newcommand{\kernel}{\mathrm{null}\,}\)
Suppose that we wish to solve the time-independent Schrödinger equation Hψ=Eψ,
E0≤⟨ψ|H|ψ⟩.
Let us prove the variational principle. Suppose that the ψn and the En are the true eigenstates and eigenvalues of H: that is,
Furthermore, let
⟨ψ|H|ψ⟩=⟨∑ncnψn|H|∑mcmψm⟩=∑n,mc∗ncm⟨ψn|H|ψm⟩=∑nc∗ncmEm⟨ψn|ψm⟩=∑nEn|cn|2,
where use has been made of Equations ??? and ???. So, we can write
⟨ψ|H|ψ⟩=|c0|2E0+∑n>0|cn|2En.
However, Equation ??? can be rearranged to give
|c0|2=1−∑n>0|cn|2.
Combining the previous two equations, we obtain
⟨ψ|H|ψ⟩=E0+∑n>0|cn|2(En−E0).
The second term on the right-hand side of the previous expression is positive definite, because En−E0>0 for all n>0 (Equation ???). Hence, we obtain the desired result
⟨ψ|H|ψ⟩≥E0.
Suppose that we have found a good approximation, ˜ψ0, to the ground-state wavefunction. If ψ is a normalized trial wavefunction that is orthogonal to ˜ψ0 (i.e., ⟨ψ|˜ψ0⟩=0) then, by repeating the previous analysis, we can easily demonstrate that
⟨ψ|H|ψ⟩≥E1.
Thus, by varying ψ until the expectation value of H is minimized, we can obtain approximations to the wavefunction and the energy of the first excited state. Obviously, we can continue this process until we have approximations to all of the stationary eigenstates. Note, however, that the errors are clearly cumulative in this method, so that any approximations to highly excited states are unlikely to be very accurate. For this reason, the variational method is generally only used to calculate the ground-state and first few excited states of complicated quantum systems.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)