Hermitian Operator has Orthogonal Eigenvectors

Theorem
The eigenvectors of a Hermitian operator are orthogonal.

Proof
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the complex numbers $\C$, with a simple spectrum:


 * $\hat H \left\vert{x_i}\right\rangle = \lambda_i \left\vert{x_i}\right\rangle$


 * $\lambda_i \ne \lambda_j$


 * $\forall i, j \in \N: i \ne j$

Now we compute the following:

and:

From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product:
 * $\braket {x_i} {x_j} = \braket {x_j} {x_i}^*$

this becomes:


 * $\left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^* = \lambda_j \braket {x_j} {x_i}$

It can be shown that the following relation holds since $\hat H = \hat H^\dagger$:


 * $\left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*$

This now gives us the equations:


 * $(1): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_i \braket {x_j} {x_i}$


 * $(2): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_j \braket {x_j} {x_i}$

Subtracting $(2)$ from $(1)$ gives:


 * $\left({\lambda_i - \lambda_j}\right) \braket {x_j} {x_i} = 0$

Note that $\left({\lambda_i - \lambda_j}\right) \ne 0$ since we were given $\lambda_i \ne \lambda_j$.

Therefore:


 * $\braket {x_j} {x_i} = 0$

Two vectors have inner product $0$ if and only if they are orthogonal.

Therefore the eigenvectors of $\hat H$ are orthogonal.