Skip to content
Snippets Groups Projects
Commit 2ac69001 authored by T. van der Sar's avatar T. van der Sar
Browse files

Update 11_nearly_free_electron_model.md - typo

parent f12a49dc
No related branches found
No related tags found
No related merge requests found
Pipeline #31111 passed
......@@ -113,7 +113,7 @@ $$ E(\delta k) = E_0 \pm \sqrt{v^2\hbar^2\delta k^2 + |W|^2}$$
that $\psi(x) = u(x) \exp(ikx)$.
#### Physical meaning of $W$
#### Calculating the size of the gaps $W$
To calculate $W=\langle k | V |k' \rangle$, we first express the lattice potential, which is periodic as $V(x)=V(x+a)$, as a Fourier series
......@@ -126,9 +126,9 @@ V_n = \frac{1}{a}\int_0^a e^{- i n 2\pi x /a} V(x) dx
$$
We now calculate the matrix element that couples our basis states $|k\rangle$ and $k'\rangle = |k-2\pi/a '\rangle$
$$W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} dx \left[e^{i k x}\right]^* V(x) \left[e^{-i k'x}\right] = \frac{1}{a}\int_0^a e^{-2\pi i x /a} V(x) dx = V_1$$
$$W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} e^{i k x} V(x) e^{-i k'x} dx = \frac{1}{a}\int_0^a e^{-i 2\pi x /a} V(x) dx = V_1$$
where we have used that $k-k'=2\pi/a$ for the first crossing. We see that the first component of the Fourier-series representation of $V(x)$ determines the strength of the coupling between the two states near the first crossing.
where we have used that $k'-k =2\pi/a$ because we are analyzing the first crossing. We see that the first component of the Fourier-series representation of $V(x)$ determines the strength of the coupling between the two states near the first crossing.
#### Crossings between the higher bands
......
0% Loading or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment