Quantum Science Philippines » Eigenvectors

Taylor Series Expansion of Operators

bien — Fri, 31 Jul 2009 06:31:18 +0000

by BIENVENIDO M. BUTANAS JR.

Some properties and examples of Hermitian and unitary operators were discussed in previous posts given by Mary Jane Madulara and Bebelyn Rosales. Hermitian operators are such that its matrix elements are equal to the elements of its corresponding adjoint matrix. Also that a matrix is classified as a unitary if its adjoint is equal to its inverse, among other properties.

Given the matrix representation of an operator, the procedure in extracting the eigenvalues and corresponding eigenvectors of this operator was shown.

An example of a unitary transformation which allows a Hermitian matrix to become diagonal by constructing a unitary matrix from its eigenvectors is also shown in the procedure involving the simultaneous diagonalization of two Hermitian matrices.

Here we will show another commonly used mathematical expression in the form of Taylor Series. Taylor series are used to define functions and operators in diverse areas of mathematics. In particular, this is true in areas where the classical definitions of functions break down. Using Taylor series, one may define analytical functions of matrices and operators such as matrix exponential or matrix algorithm (See for example http://en.wikipedia.org/wiki/Taylor_expansion#Taylor_series_as_definitions).

Here we show a few calculations involving the Taylor series expansions of two matrices.

To do these calculations, we use the fact that functions of matrices are defined by their Taylor expansions, i.e.:

a. We can then find the expression exp(M) if given that

by substituting into the Taylor series expansion and doing the necessary matrix multiplications, we get

In the last expression above, it is easily seen that the fourth term yields zero. Therefore, the subsequent terms which are multiples of this are also equal to zero. Hence, we arrive at the result,

For the next example, we take the matrix M to be

Again we apply Taylor series expansion to get

re-grouping these terms we obtain

We can now see that each expression in the braces can be represented in compact form as

Next we show that

We first take the first given matrix Mand note that

we get the determinant also and substituting these calculated values we arrive at the following results

For the next sample matrix, we have

so we have

evaluating the right hand side, this becomes

Functions of Hermitian and Unitary Operators

Lastly, we show that if H is hermitian, show that is unitary.

The proof of the above statement, starts with the definition of hermitian operators, that is;
and that of unitary operators which is: . We let then we can show that .

Simplifying the last equation, we have
but and simplifying the left hand side of the equation, we get
so and . Therefore, is unitary.

ABOUT THE AUTHOR:

BIENVENIDO BUTANAS JR. is a graduate student in physics at the MSU-Iligan Institute of Technology. He is fascinated with quantum mechanics in particular and would like to do finish more advanced studies in the near future.

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Eigenvectors and Eigenvalues of a Perturbed Quantum System

henrilen — Wed, 24 Jun 2009 14:17:07 +0000

by HENRILEN A. CUBIO

Finding the eigenvectors and eigenvalues of the state of a quantum system is one of the most important concepts in quantum mechanics. And it is here where many students get confused.

In order to learn this by heart, one has to do several exercises. There are many ways that can be employed when we deal with these concepts. Let us have an example problem of determining the eigenvectors and eigenvalues of a perturbed quantum system.

A perturbed quantum system

We consider a quantum system with just three linearly independent states. The Hamiltonian, in matrix form, is

where V0 is a constant and is some small number manifesting the perturbation such that .

We have learned in quantum mechanics that the perturbed system describes a complicated quantum system but can be expressed in terms of a simpler one. The trick then is to begin with a simpler system for which a solution is known, and add an additional perturbing Hamiltonian that represents a small disturbance to the system. In this problem we are tasked to solve for the eigenvalues and eigenvectors of the perturbed quantum system.

First we need to write down the eigenvalues and eigenvectors of the unperturbed Hamiltonian.

The unperturbed Hamiltonian in this case is just

For the undisturbed system, it is straightforward to solve the eigenvalue equation

We just solve the characteristic equation in order to get the eigenvalues corresponding to the unperturbed Hamiltonian

In matrix form the above equation is written as

From the above matrix we can easily obtain the determinant so that we can get this expression

The solution to this algebraic equation provides us with the different eigenvalues of the simpler, unperturbed Hamiltonian.

Now solving for , we have the solution set as

The eigenvalues now of the simple quantum system are just

For each eigenvalue of a transformation, there is a corresponding eigenvector. The eigenspace of a given transformation for a particular eigenvalue is the set of the eigenvectors associated to this eigenvalue. After we have successfully obtained the eigenvalues, we are now tasked to find the corresponding eigenvectors for each eigenvalue.

For , the corresponding matrix equation gives

Therefore

The remaining two eigenvectors remain arbitrary. The resulting eigenvector for is then

Since the two are arbitrary we have the freedom to choose what their values are and to make things simple we choose 1 and 0 so that the eigenvectors become

Similarly,

The linear combination of these eigenvectors is the eigenvector for

For we have the following matrix,

It is easy to see that

Since it is arbitrary we can let any value for it and the most non-trivial and simplest value would be

Therefore

The eigenvectors corresponding to the different eigenvalues of the unperturbed hamiltonian are then written as follows

For or we have

If there is a basis defined in a vector space, the vectors can be expressed in terms of components. If we have finite dimensional vector spaces for example with dimension n, the transformations can be represented with n x n square matrices.

Next we solve for the exact eigenvalues of H. We expand each of them as power series in up to second order.

Using the characteristic equation again for solving now the Hamiltonian for the perturbed system we have

Solving for the determinant of this matrix we can easily arrived to this equation

We can equate the first factor above to zero giving the expression

This expression yields the first eigenvalue which is

Now, equating the second factor to zero again we have

This would require us to use the quadratic formula to get the desired roots and so by applying we can have this expression

Simplifying the right hand side algebraically results to

The term with the radical sign may be written as

This is because of the power series expansion, up to second order as was asked, given by

Therefore the expression results to

The roots are easily read out separating the + and – signs

We now have the second eigenvalue which is

Solving for the third eigenvalue

This expression results to

Finally, writing down the three desired eigenvalues of the perturbed system

The first one is,

The second eigenvalue results to,

and the third and last eigenvalue is

The eigenvalue problem simply tells us that under the transformation, the eigenvectors experience only changes in magnitude and sign. The result of the eigenvalue shows the amount of stretch or shrink to which a vector is subjected when transformed.

About the author:

Henrilen is a graduate student of physics at MSU-IIT . She hopes to do many researches someday that could truly benefit the people not only in this country but as well as for the whole world.

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Simultaneous Diagonalization of Hermitian Matrices

mjayyy_85 — Sat, 09 May 2009 05:38:40 +0000

by MARYJANE D. MADULARA

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.

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Properties of Hermitian Operators

bebelyn — Fri, 13 Mar 2009 13:12:38 +0000

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.

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