Orthogonality of two wave functions

Question

I have given two normalized wave functions $|\psi\rangle$ and $|\eta\rangle$ (including the eigenfunctions of the operator $A$) and a hermitian operator $A$ with non-degenerate eigenfunctions and eigenvalues. I have to determine whether the two wave functions are orthogonal or not without any major calculations, but how do I do that? Could I argue that because the eigenfunctions are orthogonal, the wave functions containing them are too? I know that the integral of the two functions need to be $0$ to be orthogonal.

Answer 1

Given $|\eta\rangle = a|\phi_{1}\rangle + b|\phi_{2}\rangle$ and $|\psi\rangle = c|\phi_{1}\rangle + d|\phi_{2}\rangle$ we work out the inner product:

$$\langle \eta | \psi \rangle = (a^{\star}\langle\phi_{1}| + b^{\star}\langle\phi_{2}|)(c|\phi_{1}\rangle + d|\phi_{2}\rangle)$$

You will use the fact that the eigenvalues of a self adjoint operator are orthogonal and can be chosen to be orthonormal: expand the inner product (using linearity) and apply $\langle \phi_{i} | \phi_{j} \rangle = \delta_{ij}$ so that cross terms will vanish and "non-crossed" terms will combine to cancel.

Does that help?

If your comment is correct and the states $|\psi\rangle $ and $|\eta\rangle$ are constructed from disjoint members of the set of eigenfunctions then orthogonality follows trivially.

Answer 2

You're on the right track. Inserting an identity $\int dx|x\rangle\langle x|$ in the middle of $\langle\psi|\eta\rangle=0$ gives$$0=\int dx\langle\psi|x\rangle\langle x|\eta\rangle=\int dx\psi^\ast(x)\eta(x).$$This integral is the usual inner product on wavefunctions.