Hermitian Operators have Real Eigenvalues

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Theorem

Hermitian operators have real eigenvalues.

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 $\mid x \rangle \in V, \mid x \rangle \not= \mid 0 \rangle$ and eigenvalue $\lambda \in \C$,

$\hat{H}\mid x \rangle= \lambda \mid x \rangle$

We know for a general operator $\hat{A}$ on $V$, the following holds:

$\langle x \mid \hat{A} \mid y \rangle = \langle y \mid \hat{A}^\dagger \mid x \rangle^*$ for any $\mid x \rangle , \mid y \rangle \in V$, where $^*$ denotes the complex conjugate.

Noting $\hat{H}=\hat{H}^\dagger$ gives:

$\langle x \mid \hat{H} \mid y \rangle = \langle y \mid \hat{H} \mid x \rangle^*$

Now we compute:

$\langle x \mid \hat{H} \mid x \rangle = \langle x \mid (\hat{H} \mid x \rangle) = \langle x \mid \lambda \mid x \rangle = \lambda \langle x \mid x \rangle$

Using our previous result,

$\langle x \mid \hat{H} \mid x \rangle = \langle x \mid \hat{H} \mid x \rangle^* = \lambda \langle x \mid x \rangle$

gives

$\lambda \langle x \mid x \rangle = (\lambda \langle x \mid x \rangle)^*$

Recalling the conjugate symmetry property of the inner product, we can see that:

$\langle x \mid x \rangle = \langle x \mid x \rangle^*$

which is true if and only if $\langle x \mid x \rangle \in \R$.

So

$\lambda \langle x \mid x \rangle = \lambda^* \langle x \mid x \rangle$

$\lambda = \lambda^*$

Therefore $\lambda \in \R$.

$\blacksquare$