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I found it very instructive to see how the $E$-$k$ relationship of a free particle can be roughly identified from the extended band structure of a solid: The following is the outcome of the one dimensional Kronig–Penney model.

enter image description here

For three dimensions things get more complicated. Lets have a look on Si:

enter image description here

Taking only the 100-Direction and putting the slices together, I come to this "extended" zone scheme:

enter image description here

The red and blue branches are what I expect from the one-dimensional picture. But, to put an example, where is the yellow branch coming from? It doesn't fit in to the "simple" picture.

I always thought, that that there is a unique k(E) Relationship for each Brillouin-Zone and the red, blue and green branches are from first, second and third Zone, respectively.

Now I see, that my picture cannot be completely true, because the yellow branch doesn't fit into my model in mind. Can the yellow branch assigned to a particular zone or is my idea of a one-to-one relationship Branch<-->Zone not reasonable at all?

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  • $\begingroup$ One thing to keep in mind is that your top picture, the K-P model, shows a free (or nearly-free) electron in a periodic potential. Because of the interactions of an electron with the lattice, most band structures do not look like free electrons. Well, most of the ones in textbooks do, but there are a lot of really weird ones out there. (See, for example, the Fermi surface of beryllium: physics.stackexchange.com/q/538629) $\endgroup$ – Jon Custer Sep 29 '20 at 12:59
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First, it's not correct to call the shifted first Brillouin zone (as in the extended zones scheme) "second" BZ. The higher-order Brillouin zones are still $\Gamma$-centered, with the shape becoming more complicated as the zone order increases. See e.g. the pictures here.

Next, your 1-dimensional band structure is deceiving you, making you forget that in 2 dimensions the band structure will have additional branches for the added dimension. E.g. for the 1D free-electron dispersion relation

$$E_n(k)=\left(k+\frac{2\pi}a n\right)^2$$

we have the following band diagram:

1D E(k)

But if we now consider the 2D dispersion relation for an electron in a square lattice,

$$E_{n_x,n_y}(k_x,k_y)=\left(k_x+\frac{2\pi}a n_x\right)^2+\left(k_y+\frac{2\pi}a n_y\right)^2,$$

we'll get the following, considerably more complicated, chart:

2D E(kx,ky)

Now you can see the same situation for the Si lattice: just plot the empty-lattice bands to get the following chart on the LHS (red curves), and compare them with the actual Si bands on the RHS (black curves):

empty lattice FCC bands Si bands

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The red and blue branches are what I expect from the one-dimensional picture. But, to put an example, where is the yellow branch coming from? It doesn't fit in to the "simple" picture.

It is easier to start with phonon band structures. With two atoms per unit cell in one dimension, one then gets optical branches that look like the ones that you marked yellow.

Silicon has two atoms per primitive unit cell. A band structure must also consider different wave functions. The tight-binding approximation is a better starting point and then one must consider different atomic orbitals, not just $2p$. Sodium has a band structure that is easier to understand in a nearly-free electron model with one atom per primitive unit cell.

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