Quantum Science Philippines
Quantum Science Philippines

Quantum Mechanics in Three-Dimensions: The Radial Equation

Hananish Joy G. Odarve and Majvell Kay G. Odarve

The wavefunction, or quantum state, is a complete description that can be given into a physical system. The Schrodinger equation can describe how the wavefunction changes as time propagates.

A particle state, for example, can be determined by solving the Schrodinger equation. Since the potential of a system, [eq] V(r) [/eq],  is a function of the distance from the origin, the shperical coordinates [eq] (r, \;\theta, \;\phi) [/eq] can be employed . The time-independent Schrodinger equation in spherical coordinates has the expression

[eq]\frac{-\hbar^2}{2m}[{\frac{1}{r^2}}\frac{\partial}{\partial r}\;(r^2 \frac{\partial \Psi}{\partial r})[/eq] [eq]+\frac{1}{{r^2}\sin \theta}\;\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial \Psi}{\partial \theta}) [/eq][eq]+\frac{1}{{r^2}\sin^2 \theta}(\frac{\partial^2 \Psi}{\partial \phi^2})] + V\Psi = E\Psi[/eq]     (1)

where [eq]\Psi[/eq] is a separable solution to Eqn.(1) given as

[eq]\Psi (r,\theta,\phi) = R(r)Y(\theta,\phi)[/eq].

Performing the separation of variables leads to the determination of the angular and the radial equations given, respectively as

[eq]\frac{1}{Y}\left[\frac{1}{\sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial Y}{\partial \theta})+\frac{1}{\sin^2 \theta}\frac{\partial^2 Y}{\partial \phi^2}\right]= -l(l+1)[/eq]                                  (2)

and

[eq]\frac{1}{R}\frac{d}{dr}(r^2\frac{dR}{dr})-\frac{2mr^2}{\hbar^2}\left(V(r)-E\right)=l(l+1)\;.[/eq]                                   (3)

The actual shape of a given potential, [eq]V(r)[/eq], only affects the radial part of the wavefunction, [eq]R(r)[/eq], which can be determined from Eqn. (3). This can be further be simplified by changing of variables where we let

[eq]U(r)=rR(r),[/eq]

leading to the general expression of the radial equation,

[eq]\frac{-\hbar^2}{2m}\;\frac{d^2U}{dr^2}+\left[V(r)+\frac{\hbar^2}{2m}\;\frac{l(l+1)}{r^2}\right]U = EU.[/eq]                              (4)

From here, we can compute the ground state of a particle with [eq]l=0[/eq]. To demonstrate this, we can study a particle of mass m placed in a finite spherical well having a potential

[eq]V(r) = \left\{
\begin{array}{rl}
-V_o, & if\;\;\;r\leq a\\
0, & if\;\;\;r > a
\end{array} \right[/eq].

We also have to show that no bound states exists if

[eq]V_o\;\;a^2\;\;<\;\;\frac{\pi^2\hbar^2}{8m}.[/eq]

Now, the Schrodinger equation yields bound states when [eq]E < V_o[/eq]. We first investigate the region at [eq]r\leq a[/eq]. Eqn. (4) with [eq]l=0[/eq]  becomes,

[eq]\frac{-\hbar^2}{2m}\;\frac{d^2U}{dr^2} – V_o U = EU[/eq]

[eq]\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}(E+V_o)U[/eq]

where we let [eq]\epsilon = \sqrt{{\frac{2m}{\hbar^2}}(E+V_o)} [/eq]. Now we have,

[eq]\frac{d^2U}{dr^2}+\epsilon^2 U = 0[/eq]

The general solution for this expression is given as

[eq]U_{in}(r)=A\sin(\epsilon r)+ B\cos(\epsilon r).[/eq]

Thus, the radial equation inside the finite spherical well can be expressed as

[eq]R_{in}(r)=\frac{U_{in}(r)}{r}= \frac{A\sin(\epsilon r)}{r}+ \frac{B\cos(\epsilon r)}{r}.[/eq]                                   (5)

We impose the boundary condition that as [eq] r \rightarrow 0 [/eq], the potential has to have a finite value. However, the term [eq]\frac{B\cos(\epsilon r)}{r} [/eq] blows up so we set B=0. Thus,

[eq]R_{in}(r)= \frac{A\sin(\epsilon r)}{r}.[/eq]                                                                          (6)

Next, we investigate the region where [eq]r>a[/eq]. In this region, [eq]V(r)[/eq] is zero so Eqn. (4) becomes

[eq]\frac{-\hbar^2}{2m}\frac{d^2U}{dr^2} = EU[/eq]

[eq]\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}EU.[/eq]

We then let [eq]\beta = \sqrt{-\frac{2mE}{\hbar^2}}[/eq] so we have

[eq]\frac{d^2U}{dr^2}-\beta^2 U = 0.[/eq]

The general solution for this expression is

[eq]U_{out}(r)=Ce^{\beta r}+ De^{-\beta r}.[/eq]                                  (7)

The radial equation outside the spherical well  is then expressed as

[eq]R_{out}(r)=\frac{U_{out}(r)}{r}= \frac{Ce^{\beta r}}{r}+ \frac{De^{-\beta r}}{r}.[/eq]

From here, we impose another boundary condition that as [eq]r\rightarrow\infty[/eq] the potential must be finite. However, the term [eq]e^{\beta r} \rightarrow\infty[/eq], so we set [eq]C=0[/eq]. Thus,

[eq]R_{out}(r)=\frac{De^{-\beta r}}{r}.[/eq]                                                                       (8)

The continuity of [eq]R[/eq] and [eq]\frac{dR}{dr}[/eq] at the interface region, [eq]r=a[/eq], requires that,

i. [eq]R_{in}(r) = R_{out}(r)[/eq] and

ii. [eq]\frac{dR_{in}(r)}{dr}[/eq] = [eq]\frac{dR_{out}(r)}{dr}[/eq]

From condition (i):

[eq]\frac{A\sin(\epsilon a)}{a}=\frac{De^{-\beta a}}{a}[/eq]

[eq]A\sin(\epsilon a) = {De^{-\beta a}}.[/eq]                                                (9)

and from condition (ii):

[eq]\frac{dR_{in}(r)}{dr} = \frac{Aa\epsilon \cos(\epsilon a) – A \sin (\epsilon a)}{a^2}[/eq]

[eq]\frac{dR_{out}(r)}{dr} = \frac{-Da\beta e^{-\beta a} – De^{-\beta a}}{a^2}.[/eq]

equating [eq]\frac{dR_{in}(r)}{dr}[/eq] and [eq]\frac{dR_{out}(r)}{dr}[/eq],

[eq]Aa\epsilon \cos(\epsilon a) – A \sin (\epsilon a)=-Da\beta e^{-\beta a} – De^{-\beta a}[/eq]

[eq]A(a\epsilon \cos(\epsilon a) – \sin (\epsilon a))=-De^{-\beta a}(a\beta +1).[/eq]                                 (10)

We divide Eqn. (10) by (9),

[eq]\frac{a\epsilon \cos(\epsilon a) – \sin (\epsilon a)}{\sin\epsilon a} = -(a\beta+1)[/eq]

[eq]a\epsilon\frac{\cos{\epsilon a}}{\sin{\epsilon a}} – 1= -a\beta – 1[/eq]

[eq]a\epsilon \cot(\epsilon a) = -a\beta[/eq]

we let [eq]k_1 = \epsilon a[/eq] and [eq]k_2 = \beta a[/eq] so,

[eq]k_1 \cot(k_1) = -k_2.[/eq]                                                                                   (11)

From our representation that [eq]\epsilon = \frac{k_1}{a}[/eq]  and [eq]\beta=\frac{k_2}{a}[/eq], it tells us that the value of [eq]k_1[/eq] and [eq]k_2[/eq] should be positive.

[eq]k_1^2 + k_2^2 =\epsilon^2 a^2 + \beta^2 a^2 = a^2 \left[\frac{2m}{\hbar^2} (E+V_o) + (\frac{-2mE}{\hbar^2})\right][/eq][eq] = \frac{a^2}{\hbar^2} 2m (E+V_o-E)[/eq]

[eq]R^2 = \frac{2ma^2}{\hbar^2} V_o[/eq]

From the results, the allowable states only occurs when [eq]k_1[/eq] and [eq]k_2[/eq] are positive. That argumnet would only be possible if the constants are located in the first quadrant. The solutions thus can only be found at [eq]R<\frac{\pi}{2}[/eq].

[eq]\frac{2ma^2}{\hbar^2}V_o\;\;<\;\;\frac{\pi^2}{4}[/eq]

[eq]V_o a^2\;\;<\;\;\frac{\hbar^2}{2m}\frac{\pi^2}{4}[/eq]

[eq]V_o a^2\;\;<\;\;\frac{\pi^2 \hbar^2}{8m}.[/eq]

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Questions About The Wavefunction of Quantum Mechanics

How to understand the wavefunction in quantum mechanics better? This perhaps is the most popular question asked by students of quantum mechanics in our part of the country this year.

These students have previously studied about a year or more of quantum concepts through formal classes in modern physics and quantum mechanics with different instructors.

In a 10-minute short exercise, they wrote questions they have about the concept of wavefunction in quantum mechanics. Some of the more common questions are the subject of this post.

Why is it called the wavefunction? Why is it important in quantum mechanics?

The wavefunction describes the state of the system completely. Is the wavefunction an observable? If not, what’s the sense of showing a graphical representation of it?

How important is the wavefunction in the real world? How was it created without knowing its physical interpretation?

Is there another way of obtaining the wavefunction apart from solving Schrodinger equation?

How do we know or prove that the wavefunction obtained from a system is true and correct?

Are there classical and quantum properties of ?

What is the difference if the wavefunction is expressed as or ? How to separate the time-independent part?

What does it mean that is square-integrable? What is the physical meaning of the norm of ?

What is the significance of in Hilbert’s space?

Is probability the only meaning that can be attached to ?

These are just a few of the lingering questions students would have about learning the basic concepts of quantum mechanics. These questions show the many interesting aspects of quantum mechanics; in part showing its reputation as a tough course to master for young students.

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Further Use of Quantum Ladder Operators

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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Simple Quantum System:Infinite Square Well Potential

by JUN BONITA

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.

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Taylor Series Expansion of Operators

by BIENVENIDO M. BUTANAS JR.

       
Some properties and examples of Hermitian and unitary operators were discussed in previous posts given by Mary Jane Madulara and Bebelyn Rosales. Hermitian operators are such that its matrix elements are equal to the elements of its corresponding adjoint matrix. Also that a matrix is classified as a unitary if its adjoint is equal to its inverse, among other properties.

Given the matrix representation of an operator, the procedure in extracting the eigenvalues and corresponding eigenvectors of this operator was shown.

An example of a unitary transformation which allows a Hermitian matrix to become diagonal by constructing a unitary matrix from its eigenvectors is also shown in the procedure involving the simultaneous diagonalization of two Hermitian matrices.

Here we will show another commonly used mathematical expression in the form of Taylor Series. Taylor series are used to define functions and operators in diverse areas of mathematics. In particular, this is true in areas where the classical definitions of functions break down. Using Taylor series, one may define analytical functions of matrices and operators such as matrix exponential or matrix algorithm (See for example http://en.wikipedia.org/wiki/Taylor_expansion#Taylor_series_as_definitions).
     

Here we show a few calculations involving the Taylor series expansions of two matrices.

To do these calculations, we use the fact that functions of matrices are defined by their Taylor expansions, i.e.:

a. We can then find the expression exp(M) if given that

by substituting into the Taylor series expansion and doing the necessary matrix multiplications, we get

.

In the last expression above, it is easily seen that the fourth term yields zero. Therefore, the subsequent terms which are multiples of this are also equal to zero. Hence, we arrive at the result,

For the next example, we take the matrix M to be

Again we apply Taylor series expansion to get

re-grouping these terms we obtain

We can now see that each expression in the braces can be represented in compact form as

Next we show that

.

We first take the first given matrix Mand note that

we get the determinant also and substituting these calculated values we arrive at the following results

.

For the next sample matrix, we have

so we have

evaluating the right hand side, this becomes

Functions of Hermitian and Unitary Operators

Lastly, we show that if H is hermitian, show that is unitary.

The proof of the above statement, starts with the definition of hermitian operators, that is;
and that of unitary operators which is: . We let then we can show that .

Simplifying the last equation, we have
but and simplifying the left hand side of the equation, we get
so and . Therefore, is unitary.

ABOUT THE AUTHOR:

BIENVENIDO BUTANAS JR. is a graduate student in physics at the MSU-Iligan Institute of Technology. He is fascinated with quantum mechanics in particular and would like to do finish more advanced studies in the near future.

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