# Hermitian Operator has Real Eigenvalues

## 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 $\lambda \in \R$.

$\blacksquare$