## Effective mass (10 minutes)

Let's first consider the mass of a classical object. Some important characteristics:

1. Tells us how much inertia an object has.
2. Equivalently, tells us the force needed to accelerate an object.
3. Equivalently, tells us how much kientic energy increases as momentum increases.

Now, for a classical object, the kinetic energy of the object is equal to

$$E \; = \; \frac{1}{2}mv^{2} \; = \; \frac{p^{2}}{2m} \; \; .$$

So if we take the derivative of the energy with respect to momentum twice, we find that

$$\frac{d^{2}E}{dp^{2}} \; = \; \frac{1}{m} \; \; .$$

• Using the de Broglie Wavelength of the electron, we can find that the electron's momentum in a crystal is related to it's wave vector by

$$p \; = \; \hbar k \; \; .$$

If we insert this into our classical notion of mass, we can find that the effective mass of the electron is

$$\frac{1}{\hbar^{2}}\frac{d^{2}E}{dk^{2}} \; = \; \frac{1}{m_{eff}} \; \; .$$

• What does this tell us about an electron in an LCAO? Let's assume we have small values of $k$ and expand the energy equation

$$E \; = \; \alpha + 2 \beta\, \cos{(ka)} \; \; ,$$

$$E \; \simeq \alpha +2 \beta \left[1-\frac{1}{2}(ka)^{2}\right] \; \; ,$$

$$E \; \simeq \alpha + 2\beta - \beta a^{2}k^{2} \; \; .$$

Plugging this into our effective mass equation, we find that

$$m_{eff} \; = \; \frac{\hbar^{2}}{-2\beta\, a^{2} \, } \; \; .$$

How can we interpret this? This expression tells us that if the $\beta$ value for an LCAO is small, the electrons behave as if they are heavier and will require more energy to achieve higher velocities. This does make mathematical sense; a small $\beta$ value means that the overlap between the atomic wave functions is small and that the electron will find it more challenging to move across the crystal.

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