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Subsections

3 Prepared Examples

   The example inputs for many textbook quantum problems are available by clicking the ``Example Inputs'' menu bar on the applet. In this section, each of the examples is described to provide a starting point to explore some solutions of Schrödinger's equation.

One way to use these is to click on the ``Click Here to create a separate Solver Window'' button below the applet, to get a solver Window that you can use while reading about the examples. For convenience, an equivalent button is here:
No Java here
An advantage of using the solver in a separate window is that it can be resized.

3.1 1-D Harmonic Oscillator

    When first loaded or when ``1-d Harmonic Oscillator'' is selected, a simple harmonic oscillator is set up where natural units are chosen so that $\hbar$ = m = $\omega$ = 1 , and the energies are in units of $\hbar$$\omega$ . Clicking Solve will display the ground state energy and the wave function. Try changing the Nodes input variable to calculate higher excited states. For example, you might want to look at states with Nodes = 20 to see how the wave function looks for this excited state. You will notice that the green energy line is off scale unless you change the Vscale variable to something like 20. For Nodes greater than 40 or so, you will see that the wave function is being affected by the boundaries at $\pm$ 10 . To calculate accurately for higher excited states, you will need to change these. You will also need to add more integration points by changing Grid Points input variable to a larger value. Note that if you choose values for Xmin and Xmax too far apart, the integration may fail due to overflows or underflows. The applet does not do much error checking, does no rescaling, and does not use a variable step size. Our philosophy behind this omission is given in section 5.

3.2 L=1, Radial 3-d Harmonic Oscillator

    The radial Schrödinger equation for central potentials has the same form as the one-dimensional Schrödinger equation. The boundary condition is that the radial wave function goes to zero at r = 0 , so we need to choose Xmin to be 0. However, the radial Schrödinger equation also contains the r - 2 centripetal term that blows up at the origin. The usual technique is to expand the solutions at the origin. A sleazier technique, which we adopt, is to simply ignore the potential at the end points. Note the potential is given by
L=1
0.5*(x^2+ L*(L+1)/x^2)
By simply changing L=1 to L=0 or L=5, etc. you can change the orbital angular momentum of the solution. The potential area is scrollable and editable to allow these changes to be made easily. For large L values you will have to move Xmin out from the origin to avoid overflows and underflows.

3.3 H Atom 1s State and H Atom 2p State

        Choosing ``H atom 1s State'' sets up the usual hydrogen atom bound state problem. The units chosen for the Hydrogen atom are $\hbar$ = m = e = 1 which give results in atomic units. Lengths are in Bohr radii (0.529 Angstrom), and energies are in Hartree = 2 Rydbergs $\approx$ 27.2 electron Volts.

The hydrogen atom states have a few extra problems that should be mentioned. For the ground state, the attractive coulomb potential cusp cannot be produced accurately near the origin with a fixed step size Numerov method and our simple boundary conditions. However, the results are probably good enough for our purposes. The default integration parameters given produce an energy that is good to 4 places for the ground state. The potential is given by

L=0
-1/x + L*(L+1)/(2*x^2)
for the s state with Xmax given by 10 Bohr. The only difference in the p state parameters is that L = 1 is substituted for L = 0 in the potential, and the Xmax value is extended to 30 Bohr. You can see the standard degeneracies by integrating the equations with the sum of L and Nodes constant.

3.4 Cl 2 Morse Potential

    To demonstrate a more complicated potential, this example uses the Morse potential given in the introduction for 2 Chlorine atoms. This is a good approximation to the electronic ground state Born-Oppenheimer potential surface. The mass is the reduced mass of two Chlorine atoms. The units are energies in cm - 1 and length in Angstroms. The L=0 state is given, but L is given as a potential parameter so it can be easily changed. You can expand the Morse potential about its minimum and see where it begins to deviate from the harmonic approximation, and calculate the vibrational and rotational spectrum of Cl 2 .

3.5 Malfliet-Tjon Deuteron

    Correct models of the deuteron require a tensor force. A central potential that gives some of the correct features is the Malfliet-Tjon potential. The units are length in Fermis (or femtometers) and energy in MeV. The calculated binding energy of the deuteron is 0.4 MeV which should be compared to the correct value of 2.2 MeV. The weak binding with most of the probability outside the attractive potential region is correct.

3.6 30th Excited of 0.5*x**4

    This is given as another example of a simple one dimensional potential. You can compare the states of this potential to those of the harmonic oscillator.

3.7 The Double Harmonic Oscillator

    The double harmonic oscillator can be studied using analytic methods (see E. Merzbacher, Quantum Mechanics), and is a favorite sort of problem from which to study perturbations. The potential is written as
a = 3 
0.5*min((x-a)^2,(x+a)^2)
where 2a is the separation of the two wells and can be adjusted by simply typing in a new value. You can study the various states of the double well and look at the energy splittings as you increase the number of nodes and vary the spacing.

3.8 Square Well

    This is simply a square well of depth 1 running from x = - 1 to x = 1 . It shows how to use the pulse function.

3.9 Two Narrow Pulses

    While the applet doesn't support true delta functions, they can be approximated by narrow pulses. You should be careful that the grid is fine enough to achieve the desired accuracy. A favorite analytic problem is to calculate the bonding and antibonding (nodes = 0 and nodes = 1) states of two delta functions. This example allows us to solve and visualize an approximation to this problem. The variable a sets the spacing of the pulses, and w is their width.

3.10 ``Atom'' for Solid Slab and Solid Slab

        These two examples can be used to look at what happens when we bring atoms together to make a solid. We choose for the ``atom'' a potential
pi=4.*atan(1.)
b=pi/2.
-10*pulse(x,b)*cos(x)^2
which looks complicated, but the pulse value is simply 0 for x > |b| and 1 for x < |b| . This forms a cosine shaped attractive well. A large Vscale value is chosen to keep the plots from getting too crowded. By incrementing the Nodes value, you will find that this potential has 3 bound states.

Now select ``Solid Slab'' and the only change to the input data is that

b=21*pi/2
and the Xmin and Xmax values are changed to $\pm$ 40 . Calculating, you will see that we now have 21 of these ``atom'' potentials brought close together. The wave function looks like a cosine modulating the ``atomic'' orbitals as expected from tight binding theory. You can increase the number of nodes and you will find the lowest 21 states give the beginnings of an energy band. The next 21 give a second band and the following 21 give a third band. You can calculate an approximate bandwidth from these values. You will also see that the higher energy states are more affected by the surface and tend to stick out beyond the slab as expected. If you increase the number of nodes to 63 you will see that the wave function is heavily damped inside the ``solid.'' For these systems you may wish to use the zoom feature of the plots.

3.11 Infinite Well Super Symmetric Partner

    Most quantum texts include a paragraph or two on the super symmetric partner potentials for one particle in one dimension. See for example, Quantum Mechanics, Richard Liboff, Introductory Quantum Mechanics, third edition, Addison Wesley, 1998, page 343. Write the Schrödinger equation as

$\displaystyle\left \{
-\frac{\hbar^2}{2m}\left [ \frac{d}{dx} + \ln(\psi_0)' \right ]
\left [ \frac{d}{dx} - \ln(\psi_0)' \right] + E_0
\right\}$$\displaystyle\psi_{n}^{}$ = En$\displaystyle\psi_{n}^{}$, (12)

where ln($\psi_{0}^{}$)' is the logarithmic derivative of the ground state wave function. This is in the form

- $\displaystyle{\frac{\hbar^2}{2m}}$A $\scriptstyle\dagger$A$\displaystyle\psi_{n}^{}$ = (En - E0)$\displaystyle\psi_{n}^{}$ (13)

Multiplying through by A gives the equation,

- $\displaystyle{\frac{\hbar^2}{2m}}$AA $\scriptstyle\dagger$(A$\displaystyle\psi_{n}^{}$) = (En - E0)(A$\displaystyle\psi_{n}^{}$). (14)

which shows that A$\psi_{n}^{}$ is the eigenvector of the Hamiltonian

HS = - $\displaystyle{\frac{\hbar^2}{2m}}$AA $\scriptstyle\dagger$ + E0 (15)

with the same eigenvalues as the original Hamiltonian, with the exception that since A$\psi_{0}^{}$ = 0 , the eigenvalue E0 does not occur. For the infinite square well running from -1 to 1, with $\hbar^{2}_{}$/(2m) = 0.5 , the potential for HS is easily verified to be ($\pi$/(2cos($\pi$x/2)))2. You can verify that the energy eigenvalues of this potential are En = $\pi^{2}_{}$(n + 2)2/8 where n is the selected value of Nodes.  
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