Next: 4 A Sample Problem
Up: Visual Schrödinger: A VisualizerSolver
Previous: 2 Documentation
Subsections
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:
An advantage of using the solver in a separate window is that it can be
resized.
When first loaded or when ``1d Harmonic Oscillator'' is
selected, a simple harmonic oscillator is set up where natural
units are chosen
so that
= m = = 1 , and the energies are
in units of
. 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
V_{scale} 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 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
X_{min} and X_{max}
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.
The radial Schrödinger equation for central potentials has the
same form as the onedimensional Schrödinger equation. The boundary
condition is that the radial wave function goes to zero at r = 0 , so
we need to choose X_{min}
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 X_{min} out from the origin to avoid overflows and
underflows.
Choosing ``H atom 1s State'' sets up the usual hydrogen atom bound
state problem.
The units chosen for the Hydrogen atom are
= m = e = 1 which
give results in atomic units. Lengths are in Bohr radii (0.529 Angstrom),
and energies are in Hartree = 2 Rydbergs 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 X_{max}
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 X_{max}
value is extended to 30 Bohr. You can see the standard degeneracies
by integrating the equations with the sum of L and Nodes constant.
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 BornOppenheimer
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} .
Correct models of the deuteron require a tensor force. A
central potential that gives some of the correct features is the
MalflietTjon 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.
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.
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((xa)^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.
This is simply a square well of depth 1 running from x =  1 to
x = 1 . It shows how to use the pulse function.
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.
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 V_{scale} 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 X_{min} and X_{max} values are changed to 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.
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
where
ln()' is the logarithmic derivative of the ground
state wave function.
This is in the form

 A^{ }A = (E_{n}  E_{0})
 (13)

Multiplying through by A gives the equation,

 AA^{ }(A) = (E_{n}  E_{0})(A).
 (14)

which shows that A is the eigenvector of the Hamiltonian

H_{S} =  AA^{ } + E_{0}
 (15)

with the same eigenvalues as the original Hamiltonian, with the
exception that since A = 0 , the eigenvalue E_{0} does not
occur. For the infinite square well running from 1 to 1,
with
/(2m) = 0.5 , the potential for H_{S} is easily
verified to be
(/(2cos(x/2)))^{2}. You can verify
that the energy eigenvalues of this potential are
E_{n} = (n + 2)^{2}/8 where n is the selected value of Nodes.
Next: 4 A Sample Problem
Up: Visual Schrödinger: A VisualizerSolver
Previous: 2 Documentation