MSPhysics1

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We are going to prove the following vector identity using Levi-Civita symbol:

Solution:

Let    ,     ,   ,   .

Then,

By definition:

We have to let m=n so that,

Levi-Civita symbol can be expressed in terms of Kronecker delta given by:

Thus,

Let ,

and where and .

1. Prove that .

Proof:

Now, . Since i=j for the divergence of normal unit vector n,

but (i=j). Moreover, for three dimensions, , so

Therefore, .

2. Prove that .

Proof:

but (index i is repeated).

Therefore, .

in Proving the Vector Formula: 

By: Quennie J. Paylaga

Prove:

using Kronecker Delta Function and Levi-Civita Symbol.

To prove this, we let

We can write the expression for in summation form as:

where

where i, j, l are dummy summation variables. Each of which can be any letter (a,b,c) or number (1,2,3).

In the same way, we can write as:

Thus, we have prove that

Here is another method of verifying a vector formula using the Levi-Civita symbol. Levi-Civita symbol is a tensor of rank three and is defined by +1 if the indices are in even permutation of , -1 if the indices are in odd permutation, and 0 if any two indices are the same.

Prove that .

Proof:

Let .

Using the Levi-Civita symbol, we rewrite the cross products,

.

Since each term is only scalar, they can be rearranged such that,

where is summed over.

Note that is only nonzero if are all different from . There are two possibilities of its implications, either,

and or

and .

Consider the case of  and , which gives  for any value of . If and , . Therefore,

.

We can now write the proof so that the th component is,

.

Therefore,

.

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Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.

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/ 22 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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]]> http://www.quantumsciencephilippines.com/216/simultaneous-diagonalization-hermitian-matrices/feed/ 12 http://www.quantumsciencephilippines.com/94/schwarz-inequality/ http://www.quantumsciencephilippines.com/94/schwarz-inequality/#comments Thu, 23 Apr 2009 09:04:10 +0000 debbieclaire http://www.quantumsciencephilippines.com/?p=94

Schwarz Inequality, also known as Cauchy–Schwarz inequality, Cauchy inequality, or the Cauchy–Schwarz–Bunyakovsky inequality, is useful in many Mathematical fields such as Linear Algebra. This Inequality was formulated by Augustin Cauchy (1821), Viktor Yakovlevich Bunyakovsky (1859) and Hermann Amandus Schwarz (1888).

The uncertainty principle of quantum mechanics, which relates the incompatibility of two operators, rests on this important theorem of Schwarz.

This is a theorem that arise from the inner product of two vectors which sates that the square magnitude of the inner product of two vectors is less than or equal to the product of the square magnitude of any vector, i. e.,

where and are any vectors which obey the four axioms of inner product. The four axioms are:

where α and β are scalar constants.

Exercise (1):

By going through the derivation of Schwarz Inequality, show that the inequality becomes an equality if

where μ is an arbitrary constant.

Solution:

Starting with the Schwarz Inequality

with the general equation

From the axiom;

we let the axiom equal to zero and substitute the value V so then we have,

Doing algebra and simple transformation we arrive to the equation

and from the general equation we have, we derived this

with the condition

About the Author

Debbie Claire R. Sanchez is currently a student of MSU-IIT pursuing her graduate study and hopefully will be graduating soon. She is very much interested in the field of Materials Science more specifically on Polymers. She plans to pursue her Ph. D in the United States and dreams on working in a well known company.

The perturbation theory is best applied in the determination of the approximate correction to the energy levels and eigenstates after a certain perturbation is introduced to a real quantum system. To understand this deeply, let us look at this example.

which can be written further as

.

We recall that it was shown in the properties of quantum oscillators that

and

and substituting these to our equation , we then get

.

We also have the relation that

.

Since m = n+1 (not equal to n), then we now have

so,

Finally,

.

Thus, the first-order correction is indeed equal to 0.

For the second-order correction, it is found using the fundamental equation of the second order perturbation theory which is

where

.

Following the same procedure as in getting the first-order correction in simplifying the numerator of the equation, that is, using the raising and lowering operators, we get

and simplifying, we now have

.

With the delta function, it is important to note that

,

and the above equation becomes

.

Substituting this to our fundamental equation, it becomes

and for a harmonic oscillator,

and

.

Then, our second-order equation becomes

.

Simplifying the numerator, we now have

.

It is important to note that

So, now we have the equation,

.

Finally,

.

This is the second-order correction to the energy levels.

b) The Schrödinger equation (SE) can be solved exactly in this case by a change of variables. Find the exact energies and show that they are consistent with the perturbation theory approximation.

Solutions:

The Schrödinger equation for this potential is:

By change of variables, we let

.

Considering first the potential part of the SE and changing the variables, we have

.

Thus, substituting this to our SE, it becomes,

and rearranging terms, we get

which is the SE for simple harmonic oscillator in the variable x’.
We know that,

and finally

.

In the above equation, the second term is the second order correction to the energy level and since we found that the first order correction is zero, thus this solution is consistent with the perturbation theory approximation.

About the author:

Ann finished her BS Physics degree at MSU main campus in Marawi City and is pursuing now a graduate degree at MSU-IIT, Iligan City. She is into performing experiments in Material Science and hopes to become one of the experimental physicists of the country someday.