Dispersion Relations (tight binding energy)

[cobi18]

The nearest neighbour tight-binding energy band for an fcc metal can be written as:
E(kx; ky; kz) = Ei - A - 8t[cos(pi*kx)cos(pi*ky) + cos(pi*ky)cos(pi*kz) + cos(pi*kz)cos(pi*kx)]where kx; ky; kz are express in units of 2*pi/a .
Sketch this band dispersion (a) between the (capital Gamma) point (kx = ky = kz = 0) and the X point
(kx = 1; ky = kz = 0) (located on the square face of the Brillouin zone), and (b) between
the point and the L point (kx = ky = kz = 1=2) (located on the hexagonal face of the
Brillouin zone).I am not sure even how to start the sketch of this, do I need to know Ei and A, or can I be a lot more simple when just sketching something.

I am just really bad a graphs in reciprocal space, and my lecturer wasn't very helpful.

Any help will be greatly appreciated!

 

[mark726]

Hi there! I can definitely help you with this. Don't worry, you don't need to know the specific values of Ei and A to sketch this band dispersion.

To start, let's define some variables for the sake of simplicity. Let's say kx = x, ky = y, and kz = z. This means that the equation for the energy band becomes:

E(x; y; z) = Ei - A - 8t[cos(pi*x)cos(pi*y) + cos(pi*y)cos(pi*z) + cos(pi*z)cos(pi*x)]

Now, let's look at the two points specified in the forum post: the Gamma point (x = y = z = 0) and the X point (x = 1, y = z = 0).

(a) Sketching between the Gamma point and the X point:
At the Gamma point, all three variables are equal to 0. This means that the energy band becomes:

E(0; 0; 0) = Ei - A - 8t[cos(0)cos(0) + cos(0)cos(0) + cos(0)cos(0)]
= Ei - A - 8t[1 + 1 + 1]
= Ei - A - 24t

This gives us the starting point for our sketch. Now, at the X point, x = 1 and y = z = 0. Substituting these values into the equation, we get:

E(1; 0; 0) = Ei - A - 8t[cos(pi*1)cos(pi*0) + cos(pi*0)cos(pi*0) + cos(pi*0)cos(pi*1)]
= Ei - A - 8t[cos(pi) + 1 + cos(pi)]
= Ei - A - 8t[-1 + 1 - 1]
= Ei - A

This gives us the end point for our sketch. Now, we can simply connect these two points with a curve. The shape of the curve will depend on the values of Ei and A, but it should generally look like a downward sloping curve.

(b) Sketching between the Gamma point and the L point:
At the Gamma point, all three variables are equal to 0. This gives us the same