**by BEBELYN A. ROSALES**

Linear operators in quantum mechanics may be represented by matrices. A type of linear operator of importance is the so called Hermitian operator. An operator is Hermitian if each element is equal to its adjoint. Most quantum operators, for example the Hamiltonian of a system, belong to this type.

Now linear operators are represented by its matrix elements. We can therefore easily look at the properties of a Hermitian operator by looking at its matrix representation. A particular Hermitian matrix we are considering is that of below. We can calculate the determinant and trace of this matrix .

**The determinant and trace of a Hermitian matrix**

A. The determinant and trace of the matrix are shown below as:

where , so that

and,

B. Next we then calculate the eigenvalue of . Their sum and product of its eigenvalues are shown to be consistent with its determinant and trace.

To get its eigenvalues, we solve the eigenvalue equation:

Hence, we can easily see that

These results are therefore consistent with the answers in part A.

**Eigenvalues and eigenvectors of a Hermitian operator**

C. Knowing its eigenvalues, we can solve for the eigenvectors of . Within the degenerate sector, we construct two linearly independent eigenvectors. We do this by making the eigenvectors orthogonal to each other. Then we finally normalize all three eigenvectors so that their magnitudes are unity.

Beginning with the

We solve first the eigenvector for =0;

Solving equations (1) and (2) simultaneously leads to

and get

Now, solving equations (2) and (3) yields

and get

Substituting to equation (1),

and we therefore get .

Since is abitrary, we can choose . With this choice we now have

Therefore the eigenvector corresponding to the eigenvalue 0 is

.

Now, solving the eigenvector for , we have

Also since and are arbitrary,

We can choose

and

and get,

or we can also choose

and ;

and get,

Note that we have two eigenvalues which are equal to 3. To solve the corresponding eigenvector, we need to use the Gram Schmidt procedure which is outlined below.

Let

Normalizing,

The corresponding normalized eigenvectors for , , and are then

**The Unitary Transformation**

D. We now construct the unitary matrix that diagonalizes the matrix .

We can also show explicitly that the similarity transformation reduces to the appropriate diagonal form where its eigenvalues can be read directly from its diagonal elements.

Given the eigenvectors

we can construct the unitary matrix by having these eigenvectors as elements, thus:

the adjoint of this matrix is then given by

.

We can apply a similarity transformation of the form

Hence the matrix is transformed into its diagonal form:

About the Author:

**BEBELYN A. ROSALES** is studying for her masters degree in physics at the Mindanao State University-Iligan Institute of Technology (MSU-IIT) in Iligan City, Philippines. She hopes to continue with her doctoral studies in computational and experimental physics in a university abroad.