# Demonstrate orthogonality of eigenstates without knowing Hermiticity

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 $$\int H^* \Psi^*_n \Psi_m dx = \int \Psi_n^* H\Psi_m dx$$?

• Just substitute your hamiltonian and integrate by parts for the kinetic energy part Oct 10, 2021 at 6:03
• More generally, left eigenstates are orthogonal to the right eigenstates. It is only for hermitian Hamiltonian that the two are related via simple hermitian conjugation. Oct 10, 2021 at 6:19
• @MaxWell can you please expand on the integration by parts? If I substitute in $H$ to $H(\Psi_n^* \Psi_m)$ and compare to $\Psi_n^* H \Psi_m$, I wind up with an extra $\frac{\hbar^2}{2m} [\Psi_m\frac{d}{dx^2}\Psi_n^* + 2(\frac{d}{dx}\Psi_m)(\frac{d}{dx}\Psi_n^*)]$. I've been trying to integrate that and show it disappears with integration by parts but I seem to be going in circles Oct 10, 2021 at 19:54

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.

• Why does it go to zero? Oct 12, 2021 at 0:08
• @DanielC because the wave function needs to be square-integrable and thus it most tend to zero as coordinates go to infinity. Oct 12, 2021 at 1:17
• If it is square-integrable, it needn't go to $0$ at infinity, so taking it to go to zero at infinity is nothing but a convenient assumption in the context in which operators are ignored. Oct 12, 2021 at 2:13
• In this case we're looking at energy eigenstates of a harmonic oscillator, so unless I'm gravely mistaken they will indeed go to zero at infinity due to confinement by the potential. @DanielC are you talking about a general case? My intuition is that maxwell's right that any normalizable function should go to zero at infinity Oct 12, 2021 at 3:27
• @DanielC I see that I am mistaken and in general a square-integrable function does not necessarily goes to zero. I found some questions and answers about it but couldn't follow. Do you know somewhere to read about it? Oct 12, 2021 at 4:37