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Further Use of Quantum Ladder Operators

jessica — Thu, 17 Sep 2009 09:08:09 +0000

by JESSICA IRISH LOPEZ

The use of ladder operators in simple quantum oscillators was discussed by Simon Jude Burgos in an earlier post. We can further look at other quantum oscillators properties using again the ladder operator concept.

For example, we can derive the nth eigenstate from the ground state by applying the creation or raising operator n times, i.e.

To prove the above statement, we consider an eigenvector where . We operate the raising operator, , to , -times. By applying the operator this way, we will always get . To clear things out, we check the behavior of the eigenvector when we operate to it, considering several values of ,

By doing the above steps as alpha goes to n we will get,

Expressing yields,

Further Use of Ladder Operators

Using the ladder operator concept allows us to calculate, in a few algebraic steps, the expectation values of more complicated quantities such as:

From the definition of ladder operators as given earlier, it allows us to express the position operator in terms of the raising operator and the lowering operator and this is given by

As the problem requires us the cube of the position operator so we can then simply write,

Substituting the latter equation to the given problem:

The left hand side can also be written as

where

The next step we need to do is to operate each term to the state |n=2>. We also remember the result of the ladder operators when operated to an eigenvector yields,

With these we can easily obtain the following results:

It will just be a simple step to get the inner product of each of these terms with the state |n=3>. However, before applying the results obtained above, we can directly check if the terms will satisfy the orthogonality condition

It is easy to see that for the orthogonality condition to be satisfied the only terms that should remain are those with eigenvector |3>. Therefore,

So that finally we obtain,

which is the desired answer.

About the Author:

JESSICA IRISH LOPEZ is a graduate student in Physics at the Mindanao State University-Iligan Institute of Technology (MSU-IIT) in Mindanao, Philippines. She will be finishing his masters degree soon and hope to go on to Ph.D. physics research 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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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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