Hermitian Operator has Real Eigenvalues
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Theorem
The eigenvalues of a Hermitian operator are real.
Proof
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $\C$. That is, $\hat H = \hat H^\dagger$.
Then for an eigenvector $\left\vert{x}\right\rangle \in V, \left\vert{x}\right\rangle \ne \left\vert{0}\right\rangle$ and eigenvalue $\lambda \in \C$:
- $\left.{\hat H}\middle\vert{x}\right\rangle = \left.{\lambda}\middle\vert{x}\right\rangle$
We know for a general operator $\hat A$ on $V$, the following holds:
- $\forall \left\vert{x}\right\rangle, \left\vert{y}\right\rangle \in V: \left\langle{x}\middle\vert{\hat A}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat A^\dagger}\middle\vert{x }\right\rangle^*$
where $^*$ denotes the complex conjugate.
Noting $\hat H = \hat H^\dagger$ gives:
- $\left\langle{x}\middle\vert{\hat H}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*$
Now we compute:
| \(\ds \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle\) | \(=\) | \(\ds \left\langle{x}\middle\vert{ \left({\hat H}\middle\vert{x}\right\rangle}\right)\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left\langle{x}\middle\vert{\lambda}\middle\vert{x}\right\rangle\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda \left\langle{x}\middle\vert{x}\right\rangle\) |
Using our previous result:
| \(\ds \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle\) | \(=\) | \(\ds \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*\) |
Equating the two previous equations gives:
- $\lambda \left\langle{x}\middle\vert{x}\right\rangle = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*$
Recalling the conjugate symmetry property of the inner product, we can see that:
- $\left\langle{x}\middle\vert{x}\right\rangle = \left\langle{x}\middle\vert{x}\right\rangle^*$
which is true if and only if $\left\langle{x}\middle\vert{x}\right\rangle \in \R$.
So
- $\lambda \left\langle{x}\middle\vert{x}\right\rangle = \lambda^* \left\langle{x}\middle\vert{x}\right\rangle$
and so:
- $\lambda = \lambda^*$
Therefore $\lambda \in \R$.
$\blacksquare$