# E vs. k diagram in semiconductor physics?

Can someone please explain why the E vs. k diagram forms a parabola for a free electron?

In my book, it calls "k" the "crystal momentum, and goes on to show the E vs. k diagrams for different crystallographic directions....which don't really make sense to me either.

### 1 Answer

- Steve4PhysicsLv 78 years agoFavorite Answer
DeBroglie wavelength is give by λ = h/p where p = momentum (=mv)

The wave number, k =2π/λ. Since p = h/λ, p = hk/(2π).

This is often written using the symbol 'h_bar' where h_bar = h/(2π), so p = h_bar.k.

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Kinetic energy (E = mv^2/2) and momentum (p = mv) are algebraically related by E = p^2/(2m) (check the algebra for yourself if you are not familiar with this handy formula).

For a free electron, the electron's energy is all kinetic so:

E = p^2/(2m)

= (h_bar.k)^2 / (2m)

= (h_bar^2/(2m)).k^2

= (constant) x k^2

So a graph of E vs k is parabolic (like y=Ax^2) for a free electron.

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For an electron in a crystal, the potential energy must be included in the calculation, of E as a function of k. The maths gets very messy and I'm not familiar with the details. However, the potential energy varies with position due to the periodic lattice. Along any line, the potential energy will vary depending on the line's direction/orientation through the lattice; this gives different E-k graphs for different directions relative to the lattice.