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.

]]> 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.

]]> http://www.quantumsciencephilippines.com/216/simultaneous-diagonalization-hermitian-matrices/feed/ 11 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.

In this post we investigate the properties of a quantum oscillator by using an algebraic tool in quantum mechanics called ‘ladder operators’. Using the ladder operator it becomes easy to find the following properties for a quantum oscillator in a given energy level:  the average position and momentum and the square of these values as well as the average kinetic energy of a simple harmonic oscillator. In formal notation, we are looking for the following respective quantities: , , , and .

Some discussion about ladder operators

We begin by introducing the so-called ladder operators. There are two types: the raising operator, symbolized by , and the lowering operator, symbolized by .  For reasons that will be evident later, the two are also called creation and annihilation operators respectively.

The ladder operators come from the roots of the Hamiltonian for a simple harmonic oscillator. The Hamiltonian is given by

which can be rewritten as

We then take the roots or factors of the expression inside the brackets. We should note however that we are dealing here with operators which do not commute. Simple algebraic factoring yields two roots:

To be clear, we rewrite the two roots separately below as

where the momentum operator is given by

To be able to find the expectation values of (position operator) , (momentum operator) and (kinetic energy),  we express the position operator and momentum operator in terms of the ladder operators and . We add the two roots in order to get the expression for the position operator in terms of the ladder operators as

and then by subtracting the lowering from the raising operator gives the expression for the momentum operator as

Now we consider the product of the two ladder operators. Since operators do not commute there are different results when we change the order when multiplying both operators:

from which we derive the expression for the Hamiltonian as
.
The term in the braces is just the dimensionless Hamiltonian operator which is more convenient for our purposes:

This Hamiltonian operator can be expressed differently by multiplying the ladder operators in a different order. Then we get

and its dimensionless counterpart is just

The Schroedinger eigenvalue equation for a simple harmonic oscillator will then yield

hence it follows that

Now we can operate these ladder operators to and see how the eigenvalues behave. We write down the action of the lowering operator as
.
Its adjoint is given by

Multiplying the latter 2 equations gives us

since is the eigenfunction is normalized and is given, then

we finally arrive at the result that for the raising operator we have

And also for lowering operator the result is
.

When using ladder operators it is imporatnt to note that orthogonality condition must be satisfied. The orthogonality condition  is given by,

Finding the properties of a quantum oscillator

Using the preceding results, we can now find the desired solutions to the problem initially given at the top of this post; which are
a. In finding , we proceed as follows using the derived expression for the position operator in terms of the ladder operators. We note that  where <n| is any eigenvector. So we write,

b. we can find  in the same manner

c. Finding  involves a similar algebraic procedure

d. We repeat the same algebraic procedure in finding for .

e. Finally we can derive the expectation value for the kinetic energy, <T> in a straightforward way as

.

Relation to Heisenberg’s Uncertainty Principle

The quantum oscillator we have described above obeys the Heisenberg uncertainty principle.

We use the results from a) to d) above in proving these statements.

Using the above results, it is easy to see that

We thus have seen that the quantum harmonic oscillator satisfies the Heisenberg uncertainty principle.

About the Author:

SIMON JUDE BURGOS is a graduate student in Physics at the Mindanao State University-Iligan Institute of Technology (MSU-IIT) in Mindanao, Philippines. He goals to work in research facilities in the field of medical physics. He will be finishing his masters degree soon and hope to go on to Ph.D. physics research in the near future.

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.

In quantum mechanics, I have learned that the wavefunctions, , reside in Hilbert’s space.  What is Hilbert’s space? I guess to answer this question requires exploring the basic  properties of Hilbert’s space.

Hilbert’s space is a linear vector space whose elements, entities or components obey certain rules or axioms.  This means firstly than you can add these elements and the resulting sum is also as a member or entity of  that  space. Secondly, you can multiply the elements with any arbitrary scalar and the product yields something which is also a component of that same space.  Additionally, the operations of addition and multiplication obey definite rules. These rules are called axioms for addition and multiplication.

By means of simple problems discussed below, I illustrate these axioms which are obeyed by a linear vector space and to which the wavefunctions, , of quantum mechanics  belongs.

As a simple example, let us consider the set of all entities of the form where are real numbers. Do these form a linear vector space? First, we have know how these elements are added and how they multiply with scalars. If their addition and multiplication are defined respectively as follows:

;

and

,

we can then verify that the axioms required for a linear vector space are satisfied in this case.

From the addition operation, we can write the null vector of the set as:

.

Also from the multiplication operation, we can then write down the inverse of simply as  .

We can now verify that all four axioms for addition of elements of the set are satisfied.

First Axiom: Commutativity Property

The operation of addition in a linear vector space is commutative; which means that we don’t care about the order in which the elements are added because we always get the same result.  This axiom is written as:

(i)

Our proof is as follows. Let and .

Then,

.

Thus in a linear vector, the addition of vectors is commutative.

Second Axiom: Associative Property

The operation of addition in a linear vector space is associative which means that we don’t care about the order in which two elements are added to the third one because we always get the same result. This axiom is expressed as:

(ii) .

To prove this in the case of the set , we let

Then,

.

Therefore the addition of vectors in a linear vector space is associative.

Third Axiom: Existence of an identity element

The third requirement for a set to be a linear vector space is that the identity element exists. The identity element is defined as

(iii) .

The identity element of the set is therefore none other than the null vector

To show this property, we just apply the definition of addition hence

.

Fourth Axiom: Existence of an inverse

The inverse of a vector should exist in a linear vector space. The inverse is defined by the statement

(iv) .

For the set we can then verify the existence of an inverse as follows:

Examples of non-vector spaces

From the four axioms of addition of linear vector space, we can further make the following observations.

(1) If are required to be positive numbers, we can’t construct a vector space because Axiom (iv) will not be satisfied.

(2) The vectors of the form do not form a linear vector space. To show this, we let

where are all real numbers.

Then by Axiom (i),

.

Thus, does not form a linear vector space. The closure property is clearly violated since

.

About the Author:

CARIEL O. MONTALBAN finished his B.S. in Physics from Mindanao State University-Iligan Institute of Technology (MSU-IIT), Iligan City, Philippines in March 2008 and is now a graduate student of the same university. He hopes to become an active researcher in the field of experimental physics in the future.