# The Eigenvectors of a Hermitian Operator Are Mutually Orthogonal

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:pt-product}
\langle v_1|J|v_2\rangle
\end{equation}
where
$$J=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$
The inner product \eqref{eq:pt-product} is called PT-inner product. The matrices of the form $\begin{pmatrix} a & b\\ -\bar b & -a \end{pmatrix}$, where $a$ is real and $a^2>|b|^2$ (this condition is required to ensure that the eigenvalues are real), are said to be PT-symmetric.

Exercise. Prove that the eigenvectors of a PT-symmetric matrix are mutually orthogonal with respect to the PT-inner product \eqref{eq:pt-product}.