Hermitian Operators have Real Eigenvalues and Orthogonal Eigenvectors

Hermitian Operators have Real Eigenvalues

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 $\lambda \in \R$.

$\blacksquare$

Hermitian Operators have Orthogonal Eigenvectors

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 field of complex numbers, $\C$, with a simple spectrum, i.e. for:

$\hat{H} \mid x_i \rangle = \lambda_i \mid x_i \rangle$

$\lambda_i \not= \lambda_j$

$\forall i,j \in \N,$ $i \not= j$.

Now we compute the following:

$\langle x_j \mid \hat{H} \mid x_i \rangle = \langle x_j \mid (\hat{H} \mid x_i \rangle) = \langle x_j \mid \lambda_i \mid x_i \rangle = \lambda_i \langle x_j \mid x_i \rangle$

and

$\langle x_i \mid \hat{H} \mid x_j \rangle^* = (\langle x_i \mid (\hat{H} \mid x_j \rangle))^* = \langle x_i \mid \lambda_j \mid x_j \rangle^* = (\lambda_j \langle x_i \mid x_j \rangle)^*$

From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product, $\langle x_i \mid x_j \rangle = \langle x_j \mid x_i \rangle^*$ this becomes:

$\langle x_i \mid \hat{H} \mid x_j \rangle^* = \lambda_j \langle x_j \mid x_i \rangle$

It can be shown that the following relation holds since $\hat{H} = \hat{H}^\dagger$:

$\langle x_j \mid \hat{H} \mid x_i \rangle = \langle x_i \mid \hat{H} \mid x_j \rangle^*$

This now gives us the equations:

$(1)$ $\langle x_j \mid \hat{H} \mid x_i \rangle = \lambda_i \langle x_j \mid x_i \rangle$

$(2)$ $\langle x_j \mid \hat{H} \mid x_i \rangle = \lambda_j \langle x_j \mid x_i \rangle$

Subtracting $(2)$ from $(1)$ gives

$(\lambda_i - \lambda_j) \langle x_j \mid x_i \rangle = 0$

Note that $(\lambda_i - \lambda_j) \not= 0$ since we were given $\lambda_i \not= \lambda_j$.

Therefore

$\langle x_j \mid x_i \rangle = 0$

Two vectors have inner product $0$ if and only if they are orthogonal, therefore the eigenvectors of $\hat{H}$ are orthogonal.

$\blacksquare$