Properties of the Eigenvalues and Eigenvectors of a Hermitian Operator | Quantum Science Philippines

# Properties of the Eigenvalues and Eigenvectors of a Hermitian Operator

We can say that an operator A is a Hermitian Operator if it satisfies that: $A^\dagger = A$.

Given the eigenvalues and eigenvectors of the Hermitian Operators, the following properties apply:

1. The eigenvalues of a Hermitian operator are real.
2. Two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal.

In this section,we are to prove these statements.

### Proof:

(1.)  The eigenvalues of a Hermitian operator are real.

Suppose that we have a Hermitian operator $H$, then we have

$H|u\rangle=\lambda |u\rangle$                            [1]

where $|u\rangle$ is the eigen vector and $\lambda$ is the eigenvalue of the operator $H$.

Taking the scalar product of [1] with $\langle u|$ yields,

$\langle u| H |u\rangle = \langle u|\lambda|u\rangle$                 [2]

Here, we have $\langle u|H|u\rangle$ to be real because we assumed our operator $H$ to be a Hermitian operator already. And hence this implies that $\lambda$ is also real.

Now, we note that $H$ is a Hermitian operator and by definition, $H^\dagger=H$.

Then for the conjugate we can have,

$\langle u| H^\dagger |u\rangle = \langle u|\lambda^\star|u\rangle$              [3]

Since the definition establishes that $H^\dagger=H$, then we equate the scalar products of these 2 operators.

$\langle u|H|u\rangle = \langle u| H^\dagger |u\rangle$              [4]

Using equations [2] and [3], we have

$\langle u|\lambda|u\rangle = \langle u|\lambda^\star|u\rangle$                  [5]

$\lambda \langle u|u\rangle = \lambda^\dagger \langle u|u \rangle$                    [6]

We know already that $\langle u|u \rangle =1$. So we are left with,

$\mathbf{\lambda} = \mathbf{\lambda^\star}$

Since we have seen that the eigenvalues of the conjugate of the Hermitian operator to be  real.

Therefore we can say that the eigenvalues of a Hermitian operator are real. $\Box$

(2.) Two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal.

Let $|u \rangle$ and $|v \rangle$ be eigenvectors of a Hermitian operator $H$, corresponding to the eigenvalues $\lambda$ and $\mu$ respectively.

For both eigenvectors to be orthogonal, it must satisfy that: $\langle u|v \rangle = \langle v|u \rangle =0$.

Now, we have

$H |u \rangle = \lambda |u \rangle$                  [7]

$H |v \rangle = \mu |v \rangle$                  [8]

Since $H$ is Hermitian, we can write [8] in the form

$\langle v| H = \langle v| \mu$                  [9]

If we take the scalar product of [7] with $\langle v|$ and [9] with $|u \rangle$, then

$\langle v| H |u \rangle= \langle v | \lambda |u \rangle$          [10]

$\langle v| H |u \rangle= \langle v | \mu |u \rangle$          [11]

Equating equations [10] and [11] yields.

$\langle v | \lambda |u \rangle = \langle v | \lambda |u \rangle$           [12]

$\lambda \langle v | u \rangle = \mu \langle v | u \rangle$             [13]

$(\lambda - \mu) \langle v | u \rangle = 0$           [14]

Since $(\lambda - \mu) \neq 0$, then the only way for equation [14] to be true is to set $\mathbf{\langle v | u \rangle} = \mathbf{0}$ .

Thus, two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal$\Box$

By: Joshua J. Ordeniza, MS Physics Student, MSU-IIT