In this laboratory you will solve the onedimensional, timeindependent 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
(ħ^{2}/(2m))∂^{2}ψ(x)/∂x^{2}
+ U(x)ψ(x)
= Eψ(x),
or
∂^{2}ψ(x)/∂x^{2}
 k^{2}(x)ψ(x) = 0,
with k^{2}(x) = (2m/ħ^{2})(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]*∂^{2}ψ(x)/∂x^{2}
+ ... .
ψ(x 
∆x)
= ψ(x) 
∆x*∂ψ(x)/∂x
+ [(∆x)^{2}/2]*∂^{2}ψ(x)/∂x^{2}
 ... .
Combining the two equations above yields
[ψ(x
+ ∆x)
+ ψ(x 
∆x)
 2ψ(x)]/(∆x)^{2}
=
∂^{2}ψ(x)/∂x^{2},
or
[ψ(x +
∆x)
+ ψ(x 
∂x)
2ψ(x)]/(∆x)^{2}
= k^{2}(x)ψ(x).
ψ(x + ∆x) = (2  (∆x)^{2}k^{2}(x))ψ(x)  ψ(x  ∆x).
For our numerical work, let {x_{n}} denote points on a grid defined in the region from x = L to x = L, n = 0, 1, …, N. Define ψ_{n }= ψ(x_{n}), k_{n }= k(x_{n}), ∆x = x_{n+1 } x_{n}. Then
ψ_{n+1} = (2  (∆x)^{2}k_{n}^{2}ψ_{n}  ψ_{n1}.
We have a recipe for finding ψ_{n+1}, given ψ_{n} and ψ_{n1}. 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 ψ_{0} = 0, ψ_{1} = c_{1}, where c_{1} is some small number. This number c_{1} determines the overall normalization of the wave function. We now calculate ψ_{2,} ψ_{3}, 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 ψ_{N} = 0, ψ_{N1} = c_{2}, and calculating ψ_{N2,} ψ_{N3}, etc.
ψ_{n1} = (2  (∆x)^{2}k_{n}^{2}ψ_{n}  ψ_{n+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 x_{m} in the classically allowed region. The constant c_{2} is chosen so that both integrations yield the same value ψ(x_{m}) at the matching point and we examine the slope ∂ψ(x)/∂x near x_{m}. 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 onedimensional, timeindependent 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 k^{2}(x) = (2m/ħ^{2})(E
 U(x))in units of 1/(10^{10} m)^{2}.
With E and U(x) measured in units of eV, (2m/ħ^{2})
has the value
2*9.11*10^{31}kg/(1.05*10^{34} Js)^{2} = 1.65*10^{38}
J^{1}m^{2} *(1.6*10^{19} J/eV)
= 2.64 *10^{19} eV^{1}m^{2} = 0.264 eV^{1}(10^{10}
m)^{2}.
Column D contains the wave function y. It is calculated using ψ_{n+1} = (2  (∆x)^{2}k_{n}^{2}ψ_{n}  ψ_{n1} for x = 5 to x = 0.2 and using ψ_{n1} = (2  (∆x)^{2}k_{n}^{2}ψ_{n}  ψ_{n+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 and keep a log of your answers to the question in the text below.
(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, boundstate wave functions
are either symmetric or antisymmetric when reflected about the same line.
We say that the symmetric wave functions have even parity and the
antisymmetric wave functions have odd parity. In
an asymmetric
well boundstate 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 E_{n} = n^{2}π^{2}ħ^{2}/(2ma^{2}).
If we measure a in units of 10^{10} m and E_{n} in units of
eV, then E_{n} = 3.78*n^{2}π^{2}/a^{2}.
Fill in the appropriate columns in the table
below. Extend the table downward when necessary.


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?
Use your log to prepare a report.
Summarize the laboratory exercise.
Insert your tables and discuss your results. In the discussion you should answer the following questions.
How do the energy eigenvalues of a finite square well compare to the energy eigenvalues of an infinite square well?
How do the shapes of the energy eigenfunctions of a finite square well compare to the shapes of the energy eigenfunctions of an infinite square well?
For a symmetric square well, what is the parity of the eigenfunctions?
Excluding nodes at infinity, how many nodes does the eigenfunction of a square well corresponding to the eigenvalue E_{n} have?
How do the numerically obtained energy eigenvalues of the harmonic well compare to the analytically obtained energy eigenvalues?
How do the shapes of the energy eigenfunctions of the harmonic well compare to the shapes of the energy eigenfunctions of a finite square well? Describe similarities and differences.
How do the shapes of the energy eigenfunctions of the
triangular well compare to the shapes of the energy eigenfunctions of the
harmonic well? Describe similarities and differences.