Electronic Properties of Materials/Quantum Mechanics for Engineers/Hydrogen

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Electronic Properties of Materials/Quantum Mechanics for Engineers
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This is the seventh chapter of the first section of the book Electronic Properties of Materials.


Imagine a "simple" Hydrogen atom.

<FIGURE "Title" (Description)

Two particles with unique . This is a many-body problem and quite complicated, but the many-body accounts for translation of the hydrogen atom which we're not really interested in. If is the couplumb potential then the Hamiltonian classically looks like:

We want to change to a more natural coordinate system which will allow us to distinguish hydrogen atom translation from <FIGURE> interaction. We also want a simplified version of .

<FIGURE> "Title" (Description)

Identify the center of mass:

There are two approaches to this transformation:

The first approach is to use relative and total momentum. (classical picture)

<FIGURE> "Title" (Description)

The relative momentum is: Which m? Use the "Reduced Mass":

Solve for in terms of and . Substitute into , and rearrange...

This is classical, let's turn it into a quantum Hamiltonian.

This approach is physically intuitive, but not purely quantum. Also, it is not generalizable.

The second, better, approach is to use quantum .

Remember the Laplacian operator:

Using the relations , and to perform the coordinate transformation, . Remember that the chain rule for the transformation of differential operators.

With some algebra, it results in the same .

We now have:

This means the solution is separable:

Look at . This is just a free particle, . We've already solved this problem and found is planewaves.

Look at . This is an , with mass correction in the kinetic energy term, moving around a central potential:

How big is the reduced mass correction? Tiny.

Experimentally, it is possible to distinguish hydrogen and deuterium by spectral shifts. As more particles are added this method can be extended. For three particles:

<FIGURE> "Title" (Description)

<MATHY STUFF> (related to figure)

The problem we want to solve is:

But this is a spherically symmetric potential. Cartesian coordinates aren't the best choice. Switch to spherical coordinates.

<FIGURE> "Spherical Axis" (Description)

Spherical coordinates are highly relevant to the following mathematical calculations, now is a good time to pause and familiarize yourself if needed.

In Spherical Coordinates:

Wow! What a mess! How do we solve this? #SeparationOfVariables


Multiply Left By:

So both sides equal a constant, . Thus:

Let's look closer at the second one:


As it happens, this is an operator. First I'll tell you the answer, and then I'll show you where it originates.

The operator is , where is angular momentum. The eigenequation is:

, where is an integer and .

Angular Momentum[edit | edit source]

Before we can proceed with studying Hydrogen, we need to learn a little about angular momentum. In Quantum Mechanics there are two types of angular momentum. "Orbital" momentum is analogus to the classically understood angular momentum where , and "Spin" momentum which has no classical equivalent. We will talk about spin later, and for now we will focus on orbital angular momentum.

Orbital Angular Momentum[edit | edit source]

Classically: is a vector. Thus:

Consider the commutation of the operator.

Which means that we cannot simultaneously measure all components of . Very odd properties for a vector!

What about the magnitude of ?