In quantum mechanics, operators are required to be Hermitian. A reason for this is that the eigenvalues of a quantum mechanical operator are interpreted as physically measurable quantities such as positions, momenta, energies, etc. and therefore they are required to be real. As is well-known, Hermitian operators have all real eigenvalues. Hermitian operators also have another nice property. The eigenvectors of a Hermitian operator are mutually orthogonal. Here, we prove this only for the case of matrix operators. Let $A$ be a Hermitian operator and let $|\lambda_i\rangle$ be the eigenvectors of $A$ with distinct eigenvalues $\lambda_i$. Then we have $A|\lambda_i\rangle=\lambda_i|\lambda_i\rangle$. So, $\langle\lambda_j|A|\lambda_i\rangle=\lambda_i\langle\lambda_j|\lambda_i\rangle$. On the other hand,

\begin{align*} \langle\lambda_j|A&=(A^\dagger|\lambda_j)^\dagger\\ &=(A|\lambda_j\rangle)^\dagger\ (A^\dagger=A)\\ &=(\lambda_j|\lambda_i\rangle)^\dagger\\ &=\bar\lambda_j\langle\lambda_j|\\ &=\lambda_j\langle\lambda_j|\ (\mbox{the $\lambda_i$ are real}) \end{align*}

From this we also obtain $\langle\lambda_j|A|\lambda_i\rangle=\lambda_j\langle\lambda_j|\lambda_i\rangle$. This means that $\lambda_i\langle\lambda_j|\lambda_i\rangle=\lambda_j\langle\lambda_j|\lambda_i\rangle$, i.e. we have

$$(\lambda_i-\lambda_j)\langle\lambda_j|\lambda_i\rangle=0$$

If $i\ne j$ then $\lambda_i\ne\lambda_j$ and so $\langle\lambda_j|\lambda_i\rangle=0$.

*Example*. Let us consider the matrix $A=\begin{pmatrix}

3 & 1+i\\

-1+i & -3

\end{pmatrix}$. The adjoint (i.e. the conjugate transpose of this matrix) of $A$ is $A^\dagger=\begin{pmatrix}

3 & -1-i\\

1-i & -3

\end{pmatrix}$. Since $A\ne A^\dagger$, $A$ is not Hermitian. Although $A$ is not Hermitian, it has real eigenvalues $\pm\sqrt{7}$ and the eigenvectors corresponding to the eigenvectors $\sqrt{7}$ and $=-\sqrt{7}$ are, respectively,

$$v_1=\begin{pmatrix}

\frac{1+i}{3-\sqrt{7}}\\

1

\end{pmatrix},\ v_2=\begin{pmatrix}

\frac{1+i}{3+\sqrt{7}}\\

1

\end{pmatrix}$$

$\langle v_1|v_2\rangle=2$, so they are not orthogonal. Interestingly, $v_1$ and $v_2$ are orthogonal with respect to the inner product

\begin{equation}

\label{eq:jproduct}

\langle v_1|J|v_2\rangle

\end{equation}

where

$$J=\begin{pmatrix}

1 & 0\\

0 & -1

\end{pmatrix}$$

The matrices of the form $$H=\begin{pmatrix}

a & b\\

-\bar b & d

\end{pmatrix},$$ where $a$ and $d$ are real and $(a-d)^2-4|b|^2>0$ (this condition is required to ensure that such a matrix has two distinct real eigenvalues), are self-adjoint with respect to the inner product \eqref{eq:jproduct} i.e. $H$ satisfies \begin{equation}\label{eq:jself-adjoint}\langle Hv_1|J|v_2\rangle=\langle v_1|J|Hv_2\rangle\end{equation} which is equivalent to $$H^\dagger J=JH$$

*Exercise*. Prove that the eigenvectors of a matrix which satisfies \eqref{eq:jself-adjoint} are mutually orthogonal withe respect to the inner product \eqref{eq:jproduct}.