An operator with domain is ** Hermitian** if

for all .

Eq35 can also be expressed as because

where we have used the Hermitian property of the operator in the 2^{nd} equality.

Another way of defining a Hermitian operator is by noting that the average value of a physical property must be a real number, i.e. or

###### Question

Show that eq37 is equivalent to eq36.

###### Answer

Substitute where is a constant in of eq37 and simplify to give:

Since can be expressed as any linear combination of the basis functions and , can be any number. If we carry out the following:

- Substitute in eq38
- Subsitute in eq38 and divide the resultant equation by
- Sum the two resultant equations

we have eq36.

Hermitian operators have the following properties:

- Eigenfunctions of a Hermitian operator form a complete set
- Eigenvalues of a Hermitian operator are real
- Eigenfunctions of a Hermitian operator are orthogonal (if they have distinct eigenvalues) or can be chosen to be orthogonal (if they describe a degenerate state)

The first property is a postulate of quantum mechanics. For the 2^{nd} property, we assume two eigenvalue equations and . Multiplying the first equation on the left by , multiplying the complex conjugate of the 2^{nd} equation on the left by , and then subtracting the two resultant equations and rearranging, we have:

Using the 2^{nd} equality of eq36,

To show that eigenvalues of a Hermitian operator are real, let

The integral is zero if and only if were zero over its entire domain. However, an eigenfunction is defined as a non-zero function. Therefore, .

For the 3^{rd} property, we rewrite eq39 as

If , we have and so, .

Examples of Hermitian operators are and .

###### Question

Show that and are Hermitian.

###### Answer

We need to show that .

For , we have

where, for the 3^{rd} equality, we used the fact that is the position of a particle which must be real ().

For , we integrate by parts to give:

Since a well-behaved wavefunction that is square-integrable is defined as , i.e. it vanishes at , the 1^{st} term on the RHS of the above equation vanishes and we have:

Finally, two operators and are defined as a ** Hermitian conjugate pair** if