Demonstrate orthogonality of eigenstates without knowing Hermiticity

Question

I'm at a point in a course where we haven't talked about Hermitian operators or matrix mechanics. I have a Hamiltonian $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{m\omega^2x^2}{2}$ (a quantum harmonic oscillator) and need to show the orthogonality of the eigenstates.

What I've been trying to do is write down the Schrodinger equation, multiply by another state, then integrate: $$H\Psi_m = E_m \Psi_m \to \int \Psi_n^* H\Psi_m dx = E_m \int \Psi_n^* \Psi_m$$ $$H^* \Psi^*_n = E_n^* \Psi^*_n = E_n \Psi_n^* \to \int H^* \Psi^*_n \Psi_m dx = E_n \int \Psi_n^* \Psi_m$$ then I can subtract the right equation of the first line from the second. I know that since the Hamiltonian is Hermitian the LHS will be 0 - but I haven't yet been able to prove that the Hamiltonian can move and act on $\Psi_n^*$ for $n\neq m$

How can I show that $\int H^* \Psi^*_n \Psi_m dx = \int \Psi_n^* H\Psi_m dx$?

Answer 1

Substituting $H$, the $\frac{1}{2} m \omega^2 x^2$ terms are trivial. We want to prove

$$ \int \left( \frac{d^2}{dx^2} \psi_n^\ast \right)\psi_m dx = \int \psi_n^\ast \left( \frac{d^2}{dx^2} \psi_m \right) dx $$

A fancy way

The previous is equivalent to

$$ I = \int \left[ \left( \frac{d^2}{dx^2} \psi_n^\ast \right)\psi_m - \psi_n^\ast \left( \frac{d^2}{dx^2} \psi_m \right) \right] dx = 0 $$

using

$$ \frac{d}{dx}\left[\left(\frac{d}{dx}\psi_n^\ast\right) \psi_m - \psi_n^\ast\left(\frac{d}{dx}\psi_m\right)\right] = \left( \frac{d^2}{dx^2} \psi_n^\ast \right)\psi_m - \psi_n^\ast \left( \frac{d^2}{dx^2} \psi_m \right) $$

then follows

$$ I = \left. \left[\left(\frac{d}{dx}\psi_n^\ast\right) \psi_m - \psi_n^\ast\left(\frac{d}{dx}\psi_m\right)\right]\right|_{-\infty}^{\infty} = 0 $$

thus proven.

Integration by parts

$$ \int \left( \frac{d^2}{dx^2} \psi_n^\ast \right)\psi_m dx = \left. \psi_m\frac{d}{dx}\psi_n^\ast \right|_{-\infty}^{\infty} - \int \left (\frac{d}{dx}\psi_n^\ast \right) \left( \frac{d}{dx}\psi_m \right) dx = \\ \left. \left[\left(\frac{d}{dx}\psi_n^\ast\right) \psi_m - \psi_n^\ast\left(\frac{d}{dx}\psi_m\right)\right]\right|_{-\infty}^{\infty} + \int \psi_n^\ast \left( \frac{d^2}{dx^2} \psi_m \right) dx $$

the term between square brackets goes to zero, leaving us with what we wanted to prove.