Hermitian Operators have Orthogonal Eigenvectors

From ProofWiki
Jump to: navigation, search

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 field of complex numbers, $\C$, with a simple spectrum, i.e. for:

$\hat{H} \mid x_i \rangle = \lambda_i \mid x_i \rangle$

$\lambda_i \not= \lambda_j$

$\forall i,j \in \N,$ $i \not= j$.

Now we compute the following:

$\langle x_j \mid \hat{H} \mid x_i \rangle = \langle x_j \mid (\hat{H} \mid x_i \rangle) = \langle x_j \mid \lambda_i \mid x_i \rangle = \lambda_i \langle x_j \mid x_i \rangle$

and

$\langle x_i \mid \hat{H} \mid x_j \rangle^* = (\langle x_i \mid (\hat{H} \mid x_j \rangle))^* = \langle x_i \mid \lambda_j \mid x_j \rangle^* = (\lambda_j \langle x_i \mid x_j \rangle)^*$

From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product, $\langle x_i \mid x_j \rangle = \langle x_j \mid x_i \rangle^*$ this becomes:

$\langle x_i \mid \hat{H} \mid x_j \rangle^* = \lambda_j \langle x_j \mid x_i \rangle$

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

$\langle x_j \mid \hat{H} \mid x_i \rangle = \langle x_i \mid \hat{H} \mid x_j \rangle^*$

This now gives us the equations:

$(1)$ $\langle x_j \mid \hat{H} \mid x_i \rangle = \lambda_i \langle x_j \mid x_i \rangle$

$(2)$ $\langle x_j \mid \hat{H} \mid x_i \rangle = \lambda_j \langle x_j \mid x_i \rangle$

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

$(\lambda_i - \lambda_j) \langle x_j \mid x_i \rangle = 0$

Note that $(\lambda_i - \lambda_j) \not= 0$ since we were given $\lambda_i \not= \lambda_j$.

Therefore

$\langle x_j \mid x_i \rangle = 0$

Two vectors have inner product $0$ if and only if they are orthogonal, therefore the eigenvectors of $\hat{H}$ are orthogonal.

$\blacksquare$