# Bloch theorem proof with V(x)=V(x+ma)

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## Main Question or Discussion Point

In Grosso's Solid State Physics, chapter 1, page 2, The author said that:

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

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

$$V(x)=\sum_{-\infty}^{+\infty}V_n\:e^{ih_nx},\tag{3}$$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.

## Answers and Replies

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Hmmm, the author says "in free-electron case $V(x)=0$, the wavefunctions are simply plane waves and can be written in the form $W_{k}(x)=\frac{1}{\sqrt{L}}e^{ikx}$".
Then he talks about periodic potential case. I'm not sure, but do you think that substitute $W_{k}(x)$ in (4) into (1) is correct? Because formula (4) is for free-electron case

Hmmm, the author says "in free-electron case $V(x)=0$, the wavefunctions are simply plane waves and can be written in the form $W_{k}(x)=\frac{1}{\sqrt{L}}e^{ikx}$".
Then he talks about periodic potential case. I'm not sure, but do you think that substitute $W_{k}(x)$ in (4) into (1) is correct? Because formula (4) is for free-electron case
Well....... Hmm... I'm not sure, but he also said "the plane waves (4) constitute a complete set of orthonormal functions, that can be conveniently used as an expansion set." Does that mean substitute $W_{k}(x)$ into (1) is correct? =0=

Wait a minute... suppose that we can use formula (4) for this case and put into (1), it's really simple
\begin{align}
\frac{-\hbar^{2}}{2m} \nabla^2 W_{k}(x)+V(x)W_{k}(x)&=\frac{-\hbar^{2}}{2m} \nabla^2 \frac{1}{\sqrt{L}}e^{ikx}+ \sum_{n=-\infty}^\infty V(h_{n})e^{ih_{n}x}\frac{1}{\sqrt{L}}e^{ikx}
\nonumber \\
&=\frac{\hbar^{2}k^{2}}{2m} W_{k}(x) + \sum_{n=-\infty}^\infty V(h_{n})\frac{1}{\sqrt{L}}e^{i(k+h_{n})x}
\nonumber \\
&=\frac{\hbar^{2}k^{2}}{2m} W_{k}(x) + \sum_{n=-\infty}^\infty V(h_{n})W_{k+h_{n}}(x)
\nonumber
\end{align}
So it's belong to that subspace $S_{k}$ above

Oh!!!!!! Haha. I'm so stupid that I didn't see it. Thank you so much! I understand it now :D

Oh you're welcome but don't say you're stupid. I think it's a conventional situation when we prove a formula/relation, we usually want to expand everything, substitute everything from everywhere to see the expected result at the end of the progression, but the result goes too far away that we couldn't see the relation