Two eigenvectors with different eigenvalues are orthogonal.
Summary
Any two eigenfunctions corresponding to distinct eigenvalues of an Hermitian operator $A$ are orthogonal to each other.
$$ \begin{cases} A\psi_{n}=a_{n}\psi_{n} \\ A\psi_{m}=a_{m}\psi_{m} \end{cases} $$
If $a_{n} \ne a_{m}$,
$$ \braket{\psi_{n} | \psi_{m}} = 0 $$
Proof
Since eigenvalues of Hermitian operators are always real, the following holds:
$$ \braket{A\psi_{n}|\psi_{m} } ={a_{n}}^{\ast}\braket{\psi_{n}|\psi_{m} } =a_{n}\braket{\psi_{n}|\psi_{m}} $$
Moreover, according to the definition of Hermitian operator,
$$ \braket{A\psi_{n}|\psi_{m}}=\braket{\psi_{n}|A^{\dagger}\psi_{m}}=\braket{\psi_{n}|A\psi_{m}} = a_{m} \braket{\psi_{n} | \psi_{m}} $$
Therefore, subtracting the above two equations, we get:
$$ 0 = \braket{A\psi_{n}|\psi_{m}}- \braket{A\psi_{n}|\psi_{m}} = a_{m}\braket{\psi_{n}|\psi_{m}} - a_{n}\braket{\psi_{n}|\psi_{m}} $$ $$ \implies (a_{m}-a_{n})\braket{\psi_{n}|\psi_{m}}=0 $$
Given that $a_{m} \ne a_{n}$, we have $\braket{\psi_{n}|\psi_{m}}=0$.
■