- #1

Philethan

- 35

- 4

$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\:\psi(x)=E\:\psi(x)\tag{1}$$

$$V(x)=V(x+ma)\tag{2}$$

$$V(x)=\sum_{-\infty}^{+\infty}V_n\:e^{ih_nx},\tag{3}$$where ##h_n=n\cdot2\pi/a##.

$$W_k(x)=\frac{1}{\sqrt{L}}e^{ikx}\tag{4}$$

If we apply the operator ##H=(p^{2}/2m)+V(x)## to the plane wave ##W_{k}(x)##, we see that ##H\left|W_{k}(x)\right\rangle## belongs to the subspace ##\mathbf{S}_{k}## of plane waves of wavenumbers ##k+h_{n}##:

$$\mathbf{S}_{k}\equiv\left\{W_k(x),W_{k+h_1}(x),W_{k-h_1}(x),W_{k+h_2}(x),W_{k-h_2}(x),\cdots\right\}$$

Therefore, I plug (4) into (1), and I expect that I can get the following relationship, which proves that ##H\left|W_{k}(x)\right\rangle## belongs to the subspace ##\mathbf{S}_{k}## of plane waves of wavenumbers ##k+h_{n}##:

$$\left[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x) \right]W_k(x)\propto W_{k+h_n}(x)=\frac{1}{\sqrt{L}}e^{i(k+h_n)x}$$

Here's my derivation:

$$\begin{align}\left[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x) \right]W_k(x)&=\left[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x) \right]\frac{1}{\sqrt{L}}e^{ikx}\\[4ex]&=\frac{\hbar^2k^2}{2m}\left(\frac{1}{\sqrt{L}}e^{ikx}\right)+\sum_{-\infty}^{+\infty}V_ne^{ih_nx}\left(\frac{1}{\sqrt{L}}e^{ikx}\right)\\[4ex]&=\frac{\hbar^2k^2}{2m}\left(\frac{1}{\sqrt{L}}e^{ikx}\right)+\sum_{-\infty}^{+\infty}V_ne^{ih_n(x+ma)}\left(\frac{1}{\sqrt{L}}e^{ikx}\right)\\[4ex]&=\frac{\hbar^2k^2}{2m}\left(\frac{1}{\sqrt{L}}e^{ikx}\right)+\sum_{-\infty}^{+\infty}V_ne^{ih_n}\left(\frac{1}{\sqrt{L}}e^{i(kx+h_nma)}\right)\end{align}$$

Then, I just don't know how to do now. I have no idea how to simply and rewrite this result to prove that it really is proportional to ##W_{k+h_n}(x)=e^{i(k+h_n)x}/\sqrt{L}##. Could you please help me? I'll really appreciate that.