13
$\begingroup$

I'm currently taking a Solid State Physics class, and is currently reading about the quantum mechanical description of solids.

I then came across the following figure:

Band structure for Al

It's supposed to be the band structure for aluminium. My question is basically: How do I interpret these band structures ? I can't even see why they cut of the region from $X$ to $\Gamma$ at the dotted line in the first figure, and not in the second :/ Why is the last bit of $X$ to $\Gamma$ that you see in figure 2, not important in figure 1 ?

I understand that, the $X$, $\Gamma$, $K$, $L$ and $W$ are points of symmetry on the brillioun zone. But what does it mean for the electron(s) ? Only that they have higher energies at some points ?

I'm just having some trouble understanding what this band structure tells me, and what I'm able to see/do with it ?

So can anyone maybe explain it, maybe even easily, what I actually have to understand when I see this ?

Thanks in advance.

$\endgroup$

2 Answers 2

6
$\begingroup$

As you have already stated, the polyeder in the upper part of the figure is plotted in 3D k-space and visualizes the first Brillouin zone. The dotted line nicely visualizes a closed (one-dimensional) path in k-space that -- by convention -- runs through the special points you already mentioned. Now, this closed path is precisely the horizontal axis in the lower part of the figure (this is why it starts and ends with $\Gamma$) and what is plotted in this figure is the energy dispersion $E(\vec{k})$.

As stated in the answer of user D-K, this is done to capture some important features in a 2D plot because we don’t know how to conveniently draw a 4D plot on a sheet of paper ;) .

$\endgroup$
2
$\begingroup$

To fully present the dispersion relation E(k) needs a 4D space, which cannot be intuitive. Instead, we can use a 2D band diagram to show the most important features (along high-symmetry directions) of the band structure. Another way is to give the constant-E isosurfaces, such as Fermi surface. Many such isosurfaces will give you a 3D feeling of what the band structure is like.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.