In this laboratory you will solve the one-dimensional, time-independent Schroedinger equation numerically and find the energy eigenvalues of an electron trapped in a finite square well, a harmonic well, and a triangular well. You will use an Excel spreadsheet for the numerical work.
Theory:
We want to solve the eigenvalue equation
Hψ(x) = Eψ(x),
or
(-ħ2/(2m))∂2ψ(x)/∂x2
+ U(x)ψ(x) = Eψ(x),
or
∂2ψ(x)/∂x2 - k(x)2ψ(x) = 0,
with k2(x) = (2m/ħ2)(E - U(x)).
We want to solve it in a finite region between x = -L and x = L. We assume that in the middle of this region there exists a potential well of width a, with a/2 << L. Because we cannot extend our numerical solution to infinity, we assume that U(x) = ∞ for x < -L and x > L.
We want to find the energies for which the wave function ψ(x) and its derivative ∂ψ(x)/∂x are continuous and the boundary conditions ψ(0) = ψ(L) = 0 are satisfied. These are the energies of the bound states of the system.
We start by expanding ψ(x)
in a Taylor series expansion.
ψ(x +
∆x) = ψ(x) +
∆x*∂ψ(x)/∂x + [(∆x)2/2]*∂2ψ(x)/∂x2 + ...
.
ψ(x -
∆x) = ψ(x) -
∆x*∂ψ(x)/∂x + [(∆x)2/2]*∂2ψ(x)/∂x2 - ...
.
Combining the two equations above yields
[ψ(x + ∆x) + ψ(x -
∆x) - 2ψ(x)]/(∆x)2
=
∂2ψ(x)/∂x2,
or
[ψ(x +
∆x) + ψ(x -
∂x) -2ψ(x)]/(∆x)2 = -k2(x)ψ(x).
ψ(x + ∆x) = (2 - (∆x)2k2(x))ψ(x) - ψ(x - ∆x).
For our numerical work, let {xn} denote points on a grid defined in the region from x = -L to x = L, n = 0, 1, …, N. Define ψn = ψ(xn), kn = k(xn), ∆x = xn+1 - xn. Then
ψn+1 = (2 - (∆x)2kn2ψn - ψn-1.
We have a recipe for finding ψn+1, given ψn and ψn-1. Integrating the wave function using this algorithm, i.e. finding its value on the next grid point from its value at the previous two grid points, is called integrating using the Numerov method.
To solve for the bound states of the system, we pick ψ0 = 0, ψ1 = c1, where c1 is some small number. This number c1 determines the overall normalization of the wave function. We now calculate ψ2, ψ3, etc. We start integrating in the classically forbidden region, where the magnitude of the wave function increases approximately exponentially. As we reach the classically allowed region, the wave function becomes oscillatory at the classical turning point. As we pass through the second turning point and enter again into the classically forbidden region, the integration becomes numerically unstable, because it can contain an admixture of the exponentially growing solution. Integration into a classically forbidden region is likely to be inaccurate. We therefore generate a second solution, picking ψN = 0, ψN-1 = c2, and calculating ψN-2, ψN-3, etc.
ψn-1 = (2 - (∆x)2kn2ψn - ψn+1.
For both integrations we pick the same value for the energy E. To determine whether this energy E is an eigenvalue, we compare the results of our integrations at a matching point xm in the classically allowed region. The constant c2 is chosen so that both integrations yield the same value ψ(xm) at the matching point and we examine the slope ∂ψ(x)/∂x near xm. The values of E for which the slope is continuous across the matching point are the energy eigenvalues, because for these values of E the wave function is a solution to the Schroedinger equation which satisfies all boundary conditions. We search for the eigenvalues by picking different values for the energy E.
Procedure:
Open the linked Excel workbook. It can be used to solve the one-dimensional, time-independent Schroedinger equation for an electron trapped in various potential wells. Sheet 1 contains the data and sheet 2 the user interface.
(a) Examine sheet 1.
Column A contains the position variable x in units of 10-10 m. We are going to solve the Schroedinger equation in the finite region between x = -5 and x = 5.
Column B contains the potential energy function U(x) in units of eV. We are starting with a "symmetric finite square well" extending from x = -1*10-10 m to x = 1*10-10 m with a depth of 500 eV. The user interface on sheet 2 provides us with tools to change U(x).
Cell E2 contains a trial value for the energy E. This value can be changes with a slider located on sheet 2.
Colum C contains k2(x) = (2m/ħ2)(E
- U(x))in units of 1/(10-10 m)2.
With E and U(x) measured in units of eV, (2m/ħ2) has the
value
2*9.11*10-31kg/(1.05*10-34 Js)2 =
1.65*10-38 J-1m-2 *(1.6*10-19
J/eV)
= 2.64 *1019 eV-1m-2 = 0.264 eV-1(10-10
m)-2.
Column D contains the wave function y. It is calculated using ψn+1 = (2 - (∆x)2kn2ψn - ψn-1 for x = -5 to x = 0.2 and using ψn-1 = (2 - (∆x)2kn2ψn - ψn+1 for x = 5 to x = 0.2.
Columns I, J, and K contain different potentials that can be copied into column B with the tools provided on sheet 2.
(b) Switch to sheet 2. Fill in the tables.
(i) Start
with the symmetric square well with width a = 2*10-10 m
and a depth of 500 eV.
Find the allowed energies for an electron
trapped in this potential. How many bound states do exist?
Note the shape of the eigenfunctions. Determine the parity of
each eigenfunction.
[In a symmetric well, i.e. a well that looks the
same when reflected about a line through its center, bound-state
wave functions are either symmetric or anti-symmetric when reflected
about the same line. We say that the symmetric wave functions
have even parity and the anti-symmetric wave functions have odd
parity. In an
asymmetric well bound-state wave functions do not have a well
defined parity.].
Compare the eigenvalues you found with
the energy eigenvalues of the infinite square well.
For the infinite square well we have En = n2π2ħ2/(2ma2).
If we measure a in units of 10-10 m and En in
units of eV, then En = 3.78*n2π2/a2.
Fill in the as many columns in the table below
as you need to in order to see the trends. How many bound
states exist?
|
|
Energy (eV) |
|||||
well
with width |
n |
infinite
well |
symmetric
well |
parity |
symmetric
well |
parity |
asymmetric well |
ground
|
1 |
9.327 |
|
|
|
|
|
first
excited |
2 |
37.307 |
|
|
|
|
|
second excited state |
3 |
|
|
|
|
|
|
... |
|
|
|
|
|
|
|
Comment on your results. What have you learned about the stationary states of an electron in a finite square well?
|
|
Energy (eV) |
||||
spring
constant |
n |
harmonic
well |
harmonic
well |
parity |
triangular
well |
parity |
ground
|
0 |
30.86 |
|
|
|
|
first
excited |
1 |
92.57 |
|
|
|
|
second
excited |
2 |
|
|
|
|
|
... |
|
|
|
|
|
|
Comment on your results. What have you
learned about the stationary states of an electron in a harmonic well
and in a triangular well?