Hermitian Operators have Orthogonal Eigenvectors
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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:
| \(\displaystyle \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle\) | \(=\) | \(\displaystyle \left\langle{x_j}\middle \vert{\left({\hat H}\middle \vert{x_i}\right\rangle}\right)\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left\langle{x_j}\middle \vert{\lambda_i}\middle \vert{x_i}\right\rangle\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle\) |
and:
| \(\displaystyle \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*\) | \(=\) | \(\displaystyle \left({\left\langle{x_i}\middle \vert{\left({\hat H}\middle \vert{x_j}\right\rangle}\right)}\right)^*\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left\langle{x_i}\middle \vert{\lambda_j}\middle \vert{x_j}\right\rangle^*\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({\lambda_j \left\langle{x_i}\middle \vert{x_j}\right\rangle}\right)^*\) |
Justification for the above is needed.
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From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product:
- $\left\langle{x_i}\middle \vert{x_j}\right\rangle = \left\langle{x_j}\middle \vert{x_i}\right\rangle^*$
this becomes:
- $\left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^* = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$
It can be shown that the following relation holds since $\hat H = \hat H^\dagger$:
- $\left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*$
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This now gives us the equations:
- $(1): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle$
- $(2): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$
Subtracting $(2)$ from $(1)$ gives:
- $\left({\lambda_i - \lambda_j}\right) \left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$
Note that $\left({\lambda_i - \lambda_j}\right) \ne 0$ since we were given $\lambda_i \ne \lambda_j$.
Therefore:
- $\left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$
Two vectors have inner product $0$ if and only if they are orthogonal.
Therefore the eigenvectors of $\hat H$ are orthogonal.
$\blacksquare$
