If eigenstate for a Hermitian operator are orthonormal why are the energy eigenstates not so?

My Quantum Mechanics notes says:

We found the orthonormality relation holds for any Hermitian operator eigenstates: Upon reading this I would assume that this holds for Energy eigenstates too. Then i did a question and looked at the solution and saw the following: u0(ground state) and u1(first exited state) are energy eigenstates of the quantum Harmonic oscillator and so my question is if they are multiplied by one another in the final integral how come it does not go to zero. Any Help would be appreciated.

• The $\psi$'s are eigenstates of $\hat H$ not of $x$; the action of $x$ does not produce a multiple of the original state so you cannot invoke orthogonality of eigenstates of $\hat H$. – ZeroTheHero Dec 18 '17 at 20:24
• ψ is a particle wave function. could you give a bit more detail as I am still not quite sure what you mean. thanks – David Abraham Dec 18 '17 at 20:42
• Presumably $\psi_n(x,t)$ satisfies $\hat H\psi_n(x,t)=E_n\psi(x,t)$ so that $\int \psi_n(x,t)^*\psi_m(x,t) dx=\delta_{mn}$. This holds because of orthogonality of eigenstates of $\hat H$ but the $\psi_n(x,t)$ are not eigenstates of $\hat x$, i.e. $\hat x\psi_{n}(x,t)\ne \lambda \psi_n(x,t)$ for $\lambda$ a real constant so you cannot invoke orthogonality. – ZeroTheHero Dec 18 '17 at 23:08

Because your final integral (the third one) is not the inner product of two states. Basically I mean: \begin{align} \langle u_0|u_1\rangle \equiv \int dx\cdot u_0(x)u_1(x) = 0 \end{align} this is the orthogonality of two states. While: \begin{align} \langle u_0|x|u_1\rangle \equiv \int dx\cdot u_0(x)u_1(x)\cdot x \neq 0 \end{align} since this integral is not the inner product of them at all, but with an extra $x$ in the kernel.