In this article, we outlined the necessary steps in calculating the radial wavefunctions for the Hydrogen atom. Thus, the radial wavefunctions particularly , and are easily obtained  without bothering to normalize it.

We use the formula below to find the wavefunction,

                               ( 1)

where

                                        (2)

while is determined by the recursion formula given by,

                         (3)

and

                                                           (4)

(i) Now, finding

Using equation 1, we need to solve first the coefficient from (eqn. 3), with and .

Knowing the value of , (eqn. 2) can now be easily determined,

                        (5)

Substituting the value of the calculated coefficients to (eqn. 5), we then have

                                        (6)

Thus,

             (7)

Plugging in (eqn. 4) to (eqn. 7), we finally have

Following the same process in (i), the rest of the wavefunctions are just straightforward.

(ii) For

Determining first the coefficients, with and ,

Then,

.

Thus,

(iii) We have the coefficients for with and ,

Using again (eqn. 1), we have

We finally generated the radial wavefunctions (, , ) for the hydrogen atom which is the main aim of this paper.

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.

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.

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.

” The Heisenberg-Bohr concepts leave us all breathless, and have made a deep impression on all theoretically oriented people.”           -Albert Einstein, 1926

The period 1905 to 1925 was a great time for the world’s leading physicists in the race to understand the quantum nature of matter. To explain so many curious and undigestible phenomena about radiation, atoms, molecules and solid materials, some groups worked together and sometimes compete with another. The climax happened around 1925 when the structure of quantum mechanics was finally laid down.

Louis de Broglie

This started with Louis de Broglie’s conjecture in 1924 that particle-like objects such as electrons should display wave properties. Indeed if light which is initially thought to be a wave can behave as a particle or quantum, why not those objects which we normally conceive of as particles display wave-like properties? Why not indeed?

Erwin Schrodinger

Shortly after de Broglie introduced his concept of matter waves, Erwin Schrodinger proposed an answer to the question of what happens to the matter waves when a force acts on it. He came up with a wave equation now known as Schrodinger’s Equation that lies at the heart of quantum mechanics.

Given a particle and the force that acts on it, Schrodinger’s equation  gives the possible waves associated with this particle at a given position and time. And this is designated by the hardest working symbol in modern physics: the wave function .

Max Born

That Schrodinger would be mistaken in the physical interepretation of the wave function is only one of  the many curious twists in this very interesting and engaging conversation.  What took Max Born to interpret the absolute square of as probability density for finding electrons and not  matter density as  Schrodinger intimated?

Werner Heisenberg

Pascual Jordan

Months before Schrodinger was to write down his famous equation, Max Born, with his young students Werner Heisenberg and Pascual Jordan, already created  an entirely different approach from that of Schrodinger.  The matrix formulation of quantum mechanics developed by Born’s group in Gottingen, Germany described matter  and radiation as discrete particles.

The two formulations of quantum mechanics were thought to be different but they were quickly proved to be equivalent by Schrodinger himself.  Soon thereafter, Paul Dirac incorporated the special  theory of  relativity with quantum mechanics and the ‘quantum field theory’ was born.

Paul Dirac

The physical world that was opening up in  1900 was revealed and seen in a radical light. Max Planck’s bold hypothesis that light was emitted in  bundles or  quanta of energy where each quantum’s energy is determined by the frequency was completely  without  precedence. Planck tried for a number of years to fit his quantum hypothesis into the fabric of  classical  physics but failed.

In 1905, Albert Einstein published a paper proposing that light was not only emitted in integral units or bundles of energy but it was also absorbed in such bundles – bundles that came to be known as photons. Again the energy of absorption was equal to the mysterious, h, Planck’s constant multiplied by the light’s frequency, f. Indeed this is the same equation of Max Planck where the quantum of action, h, was introduced for the first time in 1095: E= hf. This is the single equation that changed the world of physics.

Albert Einstein

Nothing could have been more contrary to the prevailing ideas at that time concerning the transfer of energy of a wave. Since light is in the form of waves, it has to got transfer its energy and be absorbed through its intensity and not by units of its frequency. Einstein in this 1905 paper also provided an explanation for a well-known phenomenon known as photoelectric effect which is associated with the absorption and emission of electromagnetic radiation by matter. It was this work that earned Einstein in 1921  the Nobel Prize in physics  not his special theory of relativity  which was also published that same year.

The Rutherford-type model of the atom proposed by Niels Bohr in 1913 was an extraordinary success in accounting for the spectrum, stability and other aspects of the hydrogen atom. Its success hinged on the Einstein-Planck quantum relation. However Bohr’s theory failed when applied to helium and other atoms. Plus the fact that the theory contained inconsistencies that could not been resolved.

Lucretius

Democritus

Fast forward two thousand years later. Lord Kelvin and J.J. Thomson gave the world a simple picture of the atom that can tested by experiment. These gentlemen proposed that the atom’s structure consists of electrons embedded like raisins in a dough or cake of equally positive charge.

William Thomson (Lord Kelvin)

J. J. Thomson

It didn’t take long for Ernest Rutherford in 1911 to modify the cake model of the atom due to his discovery of the atomic nucleus. From his alpha scattering experiments, it was revealed that the atom has a positive charge concentrated minutely at its center with the electrons spread out over a large region beyond it. Much like the solar system with the sun at the center and the planets revolving around the sun. It this model of the atom that persists in the minds of many people today.

Ernest Rutherford

Yet this planetary model of the atom has a very serious flaw. According to the classical laws of physics the electrons going around the atom will emit radiation thereby losing its energy until it will spiral off and collapse to the center of the atom in a short time. Therefore this model cannot hold matter. It is not as stable as the planetary system we are in.

James Clerk Maxwell

Niels Bohr

Isaac Newton

Bohr’s radical proposal had its roots in the work of Max Planck who in 1900 introduced the earliest concept of the quantum of action to explain the age-old problem of radiation produced by a heated body or black body radiation.

Max Planck

The quantum of action hypothesis of Planck marked the beginning of modern physics. We can say that the world was never the same after Planck.