I understand that if my operator $A$ is a Hermitian, then suppose it has the following eigen vectors $$A|v_1\rangle=\lambda|v_1\rangle:(A).(V_1)=\lambda(V_1)\\ A|v_2\rangle=\beta|v_2\rangle:(A).(V_2)=\beta(V_2)\\ \langle v_2|A|v_1\rangle=\lambda\langle v_2|v_1\rangle:(V_2)^{\dagger}.(A).(V_1)=\lambda(V_2)^{\dagger}.(V_1)\\ \text{since}(A).(V_2)=\beta(V_2)\quad\text{hence}\quad (V2)^{\dagger}.(A)^{\dagger}=\beta(V_2)^{\dagger}$$ Now, if $A$ is a hermitian operator, then $A=A^{\dagger}$, hence $$(V_2)^{\dagger}.(A)^{\dagger}=(V_2)^{\dagger}.(A)=\beta(V_2)^{\dagger}\\ \text{Hence}\quad (V_2)^{\dagger}.(A).(V_1)=\beta(V_2)^{\dagger}(V_1)\\ \text{implies} \quad\langle v_2|A|v_1\rangle =\lambda\langle v_2|v_1\rangle = \beta\langle v_2|v_1\rangle:\lambda\neq\beta\\ \text{Hence}\quad \langle v_2|v_1\rangle=0$$ So, the vectors must be orthogonal to each other and their inner product must be zero, but how do I know in a differential equation if my operator is a hermitian or not and whether I will have quantized solutions whose inner products will be zero? For example the schrodinger wave equation leads to quantized orthogonal solutions. Or rather, how do I extend a matrix based Hermitian operator to a differential equation?