Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Schroedinger Wave Equation

Date: 09/14/98 at 04:36:22
From: Adam Chowanetz
Subject: Schroedinger's wave equation

I was wondering if you could help me understand the concepts behind 
Schrodinger's wave equation. I really want to understand the way that 
the equation models the atomic shells and sub shell orbitals. 

I do have a reasonably good background in calculus and the knowledge 
for solving differential equations and partial fractions, but the 
references I have read do not describe the variables very well. 

Is there any way you could explain to me these variables and how they 
are involved in the model of the atom? My chemistry classes have
inspired an interest in the topic.

Thank you so much for your time, and I do understand that it must take 
an enormous amount of time trying to answer mathematical questions 
from people all over the world.

Thanks again,

Date: 09/14/98 at 06:20:20
From: Doctor Mitteldorf
Subject: Re: Schroedinger's wave equation

Dear Adam,

Long ago, I learned this subject from a classic text by Pauling and 
Wilson, _Introduction to Quantum Mechanics_. Its approach is 
straightforward but rather old-fashioned. A somewhat more modern 
approach is in Shankar's _Principles of Quantum Mechanics_.

Here are the basic steps:

The first and perhaps most difficult thing is to grasp what the 
Schroedinger equation is about. It doesn't tell you where the electron 
is at a given time, or how it moves. This is the big shocker, because 
those are questions you'd want to ask, the more so in 1925. The 
solution of the equation is a "wave function" but it's not at all clear 
what it has to do with waves. In other circumstances, it looks like a 
sine wave, but here in the atom it has little in common with anything 
you'd call a wave. The wave function isn't measurable directly by any 
means. It is a complex number attached to every point in space and 
time, which must be multiplied by its complex conjugate to get its 
absolute square. The absolute square is proportional to the probability 
of finding the electron at a given spot.

The equation itself is:

   -h^2/(2m) del^2 psi + V*psi = E*psi

In this equation, h is Planck's constant over 2 pi, m is the mass of 
the electron, del^2 is the Laplacian operator, which I'll come back to, 
and V is the potential energy, a function of where you are in space.  
For the Hydrogen atom, V is (-e^2/r) where r is distance from the 
origin. E is just a constant number, equal to the energy of the 
electron. In principle, E can be anything, but you find when you solve 
the equation that there are only scattered values of E for which the 
equation can be solved at all. This corresponds to the fact that energy 
of the hydrogen atom is quantized, and can only take on certain values. 
(These energy values were known early on from spectroscopic 
experiments, and the fact that the Schroedinger equation reproduced 
them accurately was the first thing that gave people confidence in such 
a crazy system as quantum mechanics.) 

Back to the Laplacian. It is the second partial derivative with respect 
to x, plus the second partial derivative with respect to y plus the 
second partial derivative with respect to z:

   del^2 psi = d^2 psi /dx^2 + d^2 psi /dy^2 + d^2 psi /dz^2 
This works for a psi that's a function of (x,y,z) coordinates in space. 
But the V part of the equation is a function of r = sqrt(x^2+y^2+z^2). 
This makes the equation very messy to solve. A standard trick for 
dealing with it is to transform variables, to write the Laplacian 
operator as an equivalent combination of partial derivatives in the 
spherical coordinate system (r,theta,phi). This looks like:

 del^2 psi = (1/r^2)d/dr (r^2*d psi/dr) 
            + 1/(r^2sin^2(theta)) d^2 psi / d phi^2 
            + 1/(r^2*sin(theta)) d/dtheta (sin(theta)*d psi / d theta)

This looks awfully messy, but it turns out to be easier to solve than 
if you wrote the whole thing out in terms of x, y and z, where V is 
proportional to 1/sqrt(x^2+y^2+z^2).  

The next step is to write the solution as a product of three functions, 
one a function of r alone, one a function of phi alone, and one a 
function of theta alone. There's no guarantee that this will work, but 
it simplifies the equation enormously, getting rid of all the problems 
associated with partial derivatives; so if you can find a solution this 
way it saves much work. Focus first on the r part of the equation, and 
notice that the partial derivatives with respect to r must be dealt 
with, but the partial derivatives with respect to theta and phi (2d and 
3d terms) are zero by definition.  

I'll leave you to work through the solution to the three equations 
given in Pauling and Wilson. The solution to the (theta,phi) part of 
the equation is a standard set of polynomials in cos(theta) and 
cos(phi) called "spherical harmonics" that turn out to be useful in 
many contexts. The solution to the r part is a polynomial times a 
Gaussian e^(-(r/a)^2), where a is the "Bohr radius" of the 
atom = h^/(m*e^2), about half an angstrom.

Even once you have the complete set of solutions, it's a long way from 
here to explaining chemistry. For one thing, the equation is for a 
single electron. If there are 2 electrons, the equations become so much 
more complicated that solutions weren't even attempted until recently, 
using numerical methods on a supercomputer. If there are 3 electrons, 
the equations are hopeless. So a standard thing to do is to make the 
very rough approximation that the electrons "don't see each other" and 
just form independent clouds of charge about the nucleus. It's a bad 
approximation, but it's very difficult to approach the equations 
without this assumption.
The actual Schroedinger equation for two electrons is a partial 
differential equation in 6 dimensions, since the wave function depends 
on where both electrons are: (x1, y1, z1, x2, y2, z2). This is the one 
that's solved on a supercomputer. The wave function for 3 electrons 
depends on 9 variables and is intractable. A simple oxygen atom has 
8 electrons and is hopeless.

But it turns out that much about chemistry can be related in a general 
way to the solutions of the Schroedinger equation for a hydrogen atom.  
The electrons in more complex atoms behave as if they were roughly 
independent of each other just enough that the hydrogen solutions are 
recognizable in complex atoms.

In hydrogen, the electron seeks the lowest energy solution to the 
Schroedinger equation, and once it gets there will remain stable for a 
long time. But if you have more than one electron, the "exclusion 
principle" sets in. It turns out that electrons are forbidden by 
quantum mechanical rules (that have nothing to do with the Schroedinger 
equation) from occupying the same energy state as one another. So in an 
oxygen atom with 8 electrons, the "real" wave function is a function of 
24 different coordinates, describing where all 8 electrons are. But 
this is so difficult to deal with that we settle for the rough 
approximation of 8 different functions of 3 dimensional coordinates. 
And the 8 functions are different because the electrons refuse to 
occupy the same state. So only the first lucky 2 get to occupy the 
lowest energy state. (Two, not one, because they can have different 
spins. We haven't even talked about spin yet...) After that, the next 
2 electrons occupy a higher energy level, and they keep piling up in 
higher and higher energy solutions to the Hydrogen Schroedinger 
equation. Some basic properties of the Periodic Table can be explained 
by the pattern of energies in the solutions to the r part of the 
Schroedinger equation.

I'm going to stop here. I hope I've given you a taste of the journey 
that you're embarking on.

- Doctor Mitteldorf, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
College Physics
High School Physics/Chemistry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum