We examine a simple system in quantum mechanics. A particle is in a one dimensional infinite square well potential where the potential at a given length say L is zero and infinite elsewhere.

The solution to Schrodinger Equation for such a simple system consists of first knowing the initial wave function of the particle. That is, we first solve for wave function at time, t=0 which is given in details by:

This particular initial state is sketched below. We need to determine the initial wave function by finding the normalization constant A.

To determine A, we substitute the given wavefunction to the normalization condition and carry out the calculations as

Solution to the Schrodinger Equation,

The wave function for an infinite square well is then given as

where

From the wavefunction above, we must calculate the constant Cn,

At time ,the wave function reduces to

which we can write as

where

Then, cn can be calculated by applying inner product, that is,

And using the normalized initial wave functions

Recall that the integral can be solved using integral by parts,

let

then

This is easy to evaluate and obtain

but

Thus,

Now we can answer the question as to the probability that a measurement of the energy will yield the value E1?

The energy levels of an infinite square well is given as

For the ground state, that is n=1 the energy is

This is the probability of getting the ground state energy is more than 98 %.

Expectation Values of the Hamiltionian Operator

The Hamiltonian of the quantum system is given by

where the potential energy function V(x) is equal to,

We first solve for the expectation value of the total energy.

The cross terms will vanish since the energy eigenstates are orthogonal to each other.

ABOUT THE AUTHOR:

JUN BONITA is finishing his M.S. Physics degree in the Mindanao State University-Iligan Institute of Technology (MSU-IIT), Iligan City, Philippines.

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

.

For or we have

.

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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]]> http://www.quantumsciencephilippines.com/579/eigenvectors-and-eigenvalues-of-a-perturbed-quantum-system/feed/ 20 http://www.quantumsciencephilippines.com/216/simultaneous-diagonalization-hermitian-matrices/ http://www.quantumsciencephilippines.com/216/simultaneous-diagonalization-hermitian-matrices/#comments Sat, 09 May 2009 05:38:40 +0000 mjayyy_85 http://www.quantumsciencephilippines.com/?p=216 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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