(Check out section 15.1.1 of the book for the details of this calculation.) This equations describes the avoided crossings observed in the plot above.
(Check out section 15.1.1 of the book for the details of this calculation.) This equation describes the avoided crossings observed in the plot above. We observe that the gap at $\delta k=0$ is equal to $2W$.
??? question "Does our solution $\psi(x)$ satisfy the Bloch theorem? What is $u(x)$ in this case?"
The wave function has a form $\psi(x) = \alpha \exp[ikx] + \beta \exp[i(k - 2\pi/a)x]$
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
We will now show that $W=\langle k | V |k' \rangle$ represents a Fourier component of the lattice potential. To see this, we express the lattice potential (which is periodic with $V(x)=V(x+a)$) as a Fourier series
$$ V(x) = \sum_{n=-\infty}^{\infty} V_n e^{2\pi i n x/a}$$
...
...
@@ -125,10 +125,10 @@ $$
V_n = \frac{1}{a}\int_0^a e^{- i n 2\pi x /a} V(x) dx
$$
We now calculate the matrix element that couples our basis states $|k\rangle$ and $k'\rangle = |k-2\pi/a '\rangle$
Calculating $W$, we find
$$W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} e^{i k x} V(x) e^{-i k'x} dx = \frac{1}{a}\int_0^a e^{-i 2\pi x /a} V(x) dx = V_1$$
where we have used that $k'-k =2\pi/a$ 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.
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 gap near the first crossing.
