Hermitian Operator has Orthogonal Eigenvectors

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The eigenvectors of a Hermitian operator are orthogonal.


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:

\(\ds \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle\) \(=\) \(\ds \left\langle{x_j}\middle \vert{\left({\hat H}\middle \vert{x_i}\right\rangle}\right)\)
\(\ds \) \(=\) \(\ds \left\langle{x_j}\middle \vert{\lambda_i}\middle \vert{x_i}\right\rangle\)
\(\ds \) \(=\) \(\ds \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle\)


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

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:

$\paren {\lambda_i - \lambda_j} \braket {x_j} {x_i} = 0$

Note that $\paren {\lambda_i - \lambda_j} \ne 0$ since we were given $\lambda_i \ne \lambda_j$.


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

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

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