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

  • $\begingroup$ Just substitute your hamiltonian and integrate by parts for the kinetic energy part $\endgroup$
    – MaxWell
    Oct 10, 2021 at 6:03
  • $\begingroup$ 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. $\endgroup$
    – Roger V.
    Oct 10, 2021 at 6:19
  • $\begingroup$ @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 $\endgroup$
    – ttshaw1
    Oct 10, 2021 at 19:54

1 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

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


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

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.

  • $\begingroup$ Why does it go to zero? $\endgroup$
    – DanielC
    Oct 12, 2021 at 0:08
  • $\begingroup$ @DanielC because the wave function needs to be square-integrable and thus it most tend to zero as coordinates go to infinity. $\endgroup$
    – MaxWell
    Oct 12, 2021 at 1:17
  • $\begingroup$ 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. $\endgroup$
    – DanielC
    Oct 12, 2021 at 2:13
  • $\begingroup$ 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 $\endgroup$
    – ttshaw1
    Oct 12, 2021 at 3:27
  • $\begingroup$ @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? $\endgroup$
    – MaxWell
    Oct 12, 2021 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.