In an earlier post about the properties of Hermitian operators, it was noted that quantum operators of physical significance are Hermitian by type. Here we discuss more fully about Hermitian matrices.

A n x n matrix is Hermitian if it is equal to its corresponding adjoint matrix. Now, for each Hermitian matrix, it may be diagonalized by a unitary transformation to its basis. That is by using a unitary matrix composed of eigenvectors of the Hermitian matrix.

But what can be done for two Hermitian matrices?

The good thing is that they may be simultaneously diagonalized. This can be done by finding the eigenvectors common to both. And then by verifying that under a unitary transformation to this basis, both matrices are diagonalized.

Let us consider the following Hermitian matrices.

EIGENVALUES AND EIGENVECTORS

i) For Look first for the eigenvalue by solving it from the determinant,

So that by using the basket rule in solving matrices,

This will give us the values,

a. for

For simplicity, first choose

Next, choose

b. for

So choose ,

ii) For Again look first for the eigenvalue by solving it from the determinant,

Then by using again the basket rule for matrices,

This will give us the values,

a. for

The resulting equation will then be,

This results to,

Now choose so that,

b. for

Then choose the values to be

c. for

The resulting equation will then be,

This will give us,

Since the second term will cancel out to zero, so that this will only then become,

Then choose so that,

So here are the common eigenvectors of and

UNITARY TRANSFORMATION

Now for the Unitary transformation matrix,

Verify if

Finally, using this unitary transformation, find out if and are diagonalized.

Thus we have described the properties of Hermitian operators in terms of its eigenvalues and eigenvectors. We have also shown that two Hermitian matrices can both be diagonalized through a unitary transformation.

About the Author:

Maryjane D. Madulara is presently pursuing a masters degree in physics at MSU-Iligan Institute of Technology (MSU-IIT) in Iligan City, Philippines. Computational physics research is her subject of interest. “Something new for the scientific community” is her motivation to continue, dream big, and do more. She hopes to finish a doctoral degree abroad.

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.