In this video it is stated that:
It can easily be verified that the wavefunction of a non-degenerate quantum mechanical system will be real.
However the presenter does not explain why this statement is true. How can we prove this? Does the professor assume a real Hamiltonian, i.e. one that includes only kinetic energy and Coulomb interaction terms?
We prove this by a reductio ad absurdum. We start by assuming that the wavefunction of a non-degenerate ground state is complex, then show this means the wavefunction must be degenerate.
Suppose we have a complex ground state. Then we can write it as a sum of real and imagniary parts:
$$ \psi = \psi_r + i\psi_i \tag{1} $$
The ground state obeys Schrodinger's equation:
$$ H\psi - E\psi = 0 $$
and if we use equation (1) to substitute for $\psi$ we get:
$$ H\psi_r + iH\psi_i - E\psi_r - Ei\psi_i = 0 $$
For a complex number to be zero both its real and imaginary parts must be zero, so we get the two equations:
$$\begin{align} H\psi_r - E\psi_r &= 0 \\ H\psi_i - E\psi_i &= 0 \end{align}$$
But this means that $\psi_r$ and $\psi_i$ are also eigenstates with the same energy $E$ as $\psi$. That means $\psi$ is degenerate, and that contradicts our initial assumption.
