Im learning about the variation method for solving quantum mechanics problems. The principle is that eigenfunctions minimize the expectation value of the Hamiltonian on that state (provided $\Psi$ is normalized):
$$E_1\leq\left<\Psi|H|\Psi\right>,$$
with equality meaning $\Psi=\Psi_1$. Next eigenstates are found sequentially by noting that $E_n$ will still be a local minimum and $\Psi_n$ will be orthogonal to all previously found eigenstates.
The professor did the derivation for $E_1$ in class. His argument was: $\Psi_n$ will form a basis for any trial solution $\Psi$. Thus using fouriers trick we can find an expansion for $\Psi$ in the basis $\Psi_n$ with coefficients $b_n$. Then the energy expectation value for the trial is
$$|E|=\sum|b_n|^2E_n.$$
Clearly, the smallest $E_n$ is $E_1$, so the smallest $|E|$ is $E_1$, and the coefficients are $|b_1|=1$ and $|b_n|=0$ for all other $n$.
But what happens if I take a trial wave function that cannot be expressed in this basis? For instance, say I have a square well from $0$ to $L$ and I have a text function which is normalized but nonzero at some location with $x>L$. Obviously this is a stupid trial function and it is not a possible solution. I will not be able to find $|E|$ to check if the functional is minimized. But what part of the math above will break down? Where does the variational principle say, for instance, that in addition to being normalizable, the test function must be zero in quantum mechanically disallowed regions?
Best Answer
The professor's derivation shows that any trial solution will have energy larger than the ground state energy corresponding to the lowest energy eigenstate. But the fact that the ground state energy is lower than any trial function's can even be treated as a definition. If there exists a lowest energy and a lowest energy wavefunction, of course all other wavefunctions will provide an upper bound on that lowest energy.
The goal is to try out many trial solutions with the aim of getting closer to the ground state energy. Each trial solution provides an upper bound.
Trying out any test wavefunction which is nonzero when $V = \infty$ will add unwelcome energy. By the variational method, you'd simply prefer a lower energy trial function, which you can always achieve by moving that portion of the trial wavefunction to a place where $V$ is finite.
The professor's derivation broke down because you considered an infinite energy wavefunction. The variational principle prevents non-zero $V=\infty$ test functions because (as you pointed out) such functions would be obviously unproductive in forming upper bounds to the ground energy state.