Numerical solution of the Schrödinger equation

UMEÅ UNIVERSITY December 5, 2011
Department of Physics Atomic and Molecular Physics, 5p
Peter Olsson
Numerical solution of the Schrödinger
equation

1 Background
The starting point is the Schrödinger equation in a spherically symmetric
potential, V(r) = −Ze 2 /4πǫ 0 r:
{ −¯h
2
2m e
∇ 2 +V(r)
In spherical coordinates we have
}
ψ(r,θ,φ) = Eψ(r,θ,φ).
∇ 2 = 1 (
∂
r 2 ∂ )
− 1 ,
r 2 ∂r ∂r r 2l2
where the operator l 2 contains the terms that depend on θ and φ. If one
assumes that the solution may be written as a product
ψ(r,θ,φ) = R(r)Y(θ,φ),
the equation separates into radial and angular parts,
( )
1 ∂
r 2∂R
R∂r
∂r
− 2m er 2
¯h 2 {V(r)−E} = 1 Y l2 Y. (1)
Since the respective sides depend on different variables the equation is only
satisfied is both sides equal a constant, b. For the angular equation we have
l 2 Y = bY,
which is an eigenvalue equation described in detail in section 2.1.1 of C. J.
Foot, Atomic Physics. The eigenvalues are b = l(l+1), where l is the orbital
angular momentum quantum number.
To proceed we introduce the dimensionless variable x = r/a 0 , a dimensionless
energy Ẽ = E/(e2 /8πǫ 0 a 0 ), and the effective potential
In these units Eq. (1) becomes
Ṽ(x) = l(l+1)
x 2 − 2Z x . (2)
d 2 R
dx + 2 dR
2 x dx + (Ẽ −Ṽ(x)) R(x) = 0.
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2 The task
The purpose with this lab is:
• To get (more) experience of solving differential equations on the computer.
• To get familiar with the “quantum defect” and to get a better understanding
of the effect of the orbital angular momentum quantum
number l.
This computer lab is to be solved individually. Go through the steps below
and fill in the tables. Also make some quick sketches of the various wave
functions with a pen on a single sheet of paper (don’t use the printer). There
is no need to write a report, but you have to show your results to the lab
supervisor and discuss them shortly.
2.1 Test the method
Before doing any serious calculations one has to make sure that the method
works. The first step is therefore to use Matlab to solve the Schrödinger
equation for the hydrogen atom. We will be using the function ode45. Note
the following:
• This function can solve a system of first order differential equations;
we therefore need to rewrite our second order differential equation as
two first order ones. Use f(1) = R and f(2) = dR/dx. Complete the
expressions for the derivatives of f(1) and f(2), which are to be used
below:
df(1)
dx =
df(2)
dx =
• Make a text-file hydrogen.m that contains the function hydrogen with
arguments x and (the vector) f. (Here v is meant to be the effective
potential, Eq. (2).)
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function fprim=hydrogen(x,f)
v=...;
fprim=[... ; ...];
This expression will of course depend on the trial energy. The simplest
way to put in the trial energy is to edit hydrogen.m before each integration.
Another method is to let the energy be f(3), and specify its
derivative to be zero, fprim=[...;...;0]. The energy is then given
as one of the initial conditions.
• Call ode45 as [x,f]=ode45(’hydrogen’,[0.001 20], [1 -1]);
– The first argument is the name of the routine that ode45 calls
many times with different values of x and f.
– The second argument is the x-interval for the integration. Why
don’t we start at x = 0 (Try that, to see what happens.)
Note that you will need to change the final value for x. In some
cases it should better be smaller than 20, whereas you will sometimes
see that you can get higher precision for the energy if you
integrate out to x much bigger than 20.
– The third argument gives the start values, R(0.001) = 1 and
dR/dx| 0.001 = −1.
• Sincethewavefunctiongivestheprobabilitydensityfortheelectron, we
need wave functions that arenormalizable. This means that ∫ dr|ψ(r)| 2
hastobefinite. Thiswill betrueonlyifR(r)vanishesforlarger, which
will be true only for certain values of the energy, the eigenvalues.
• Our approach, where we start with some arbitrarily chosen initial values
for R(≈ 0) and dR/dx, doesn’t give a properly normalized wave
function. That need not bother us if the goal is only to determine the
energy eigenvalues, since they are independent of the prefactor of R.
(Convince yourself of this fact.)
• To plot the wave function use plot(x,f(:,1)). See the Appendix on
the last page for a note on a more effective use of matlab.
Calculate R(x) for n = 1, n = 2, and n = 3, so far with l = 0. In our reduced
units, the energy is simply E n = −1/n 2 , independent of l. Also examine the
effect on the wave function of anenergy which is slightly off the correct value.
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2.2 The lithium atom
The lithium atom is a three-electron system and is much more difficult to
solve exactly. It is however rather easy to study a rough approximation,
and it turns out that even very simple models captures many features of the
true behavior. We will below calculate the “quantum corrections” δ s and δ p ,
which turn out to be reasonably constant even in our simple approximation.
In our approximation the potential felt by the outermost electron is, at
large distances, given by Z = 1, but close to the core, r < a 0 (i.e. x < 1) the
electron feels the three protons, and the potential is, apart from an additive
constant, given by Z = 3, cf. Fig. 4.7 in Foot, Atomic Physics. (Physically,
this means that we assume that the two core electrons are located at a very
thin shell at r = a 0 — admittedly a gross simplification.) Make a new file
lithium.m with a function that implements this potential. Note that you need
toaddaconstant to thex < 1-partofthepotential, tomake Ṽ(x) continuous
at x = 1.
Results with l = 0: Fix l = 0 and determine the energy eigenvalues for
n = 1, 2, 3, and 4. You now have to guess an initial value of the energy
which then has to be adjusted many times to get an acceptable solution with
lim x→∞ R(x) = 0. (When necessary, integrate the equation out to large x to
get higher precision in the energy). Complete the table below. Calculate also
the quantum correction from the expression for the energy of Alkali metals,
Eq. (4.1) in Foot, Atomic physics,
1
Ẽ(n,l) = −
(n−δ l ) 2,
l = s,p,d,....
For l = 0: energy eigenvalues and the quantum defect δ s .
n Energy, Ẽ(n,0) δ s
1
2
3
4
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The effect of l > 0: The orbital angular momentum quantum number,
l, gives a term in the potential, Eq. (2), which makes the wave function
smaller close to the core. The wave functions are then less affected by the
modified potential close to the core, and therefore become more similar to
the hydrogenic wave functions. We therefore expect the quantum correction
to be smaller for larger l. Fix l = 1 and determine the energy eigenvalues for
n = 2, 3, and 4, and estimate the value of δ p in this model.
For l = 1: energy eigenvalues and the quantum defect δ p .
n Energy, Ẽ(n,1) δ p
2
3
4
A More effective use of Matlab
(This remark is on the last point in Sec. 2.1.)
The plot will appear in a new window. It is now handy to shrink the matlab
window such that the two or three windows you need will fit together on the
screen without overlapping one another. Note also that it is possible to put
both commands you need (calculations and plotting) on a single line.
[x,f]=ode45(’hydrogen’,[0.001 20], [1 -1]); plot(x,f(:,1))
Youcanthenuse the arrow-upkey to get back thecommand lineused before,
and do both calculations and plotting in one go.
With the method described above you need to change the energy value
in hydrogen.m for each calculation. Since that needs to be done very many
times it is much more convenient to instead give the energy as one of the
initial values (as described above). One can then e.g. write
[x,f]=ode45(’hydrogen’,[0.001 20], [1 -1 -0.22]); plot(x,f(:,1))
where “-0.22” is the energy given as an initial value.
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