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

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• Philethan
In summary, Grosso's Solid State Physics says that:-The wavefunction for a free electron is a plane wave-The wavefunction for a periodic potential is also a plane wave, but with a periodic component added-The wavefunction for a harmonic potential is a sum of plane waves
Philethan
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.

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

Philethan
Nguyen Son said:
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

Philethan
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

Philethan

## 1. What is Bloch's theorem?

Bloch's theorem is a fundamental principle in solid-state physics that describes the behavior of electrons in a periodic potential. It states that the wavefunction of an electron in a periodic potential can be written as a product of a periodic function, known as the Bloch function, and a plane wave. This allows for the simplification of the Schrödinger equation for a periodic potential to only consider a single unit cell of the lattice.

## 2. How does V(x)=V(x+ma) relate to Bloch's theorem?

V(x)=V(x+ma) is a condition of periodicity in the potential energy function, where a is the lattice constant. This condition is necessary for Bloch's theorem to hold, as it ensures that the potential energy experienced by the electron is the same at each point in the lattice, allowing for the simplification of the Schrödinger equation.

## 3. What is the proof for Bloch's theorem with V(x)=V(x+ma)?

The proof for Bloch's theorem with V(x)=V(x+ma) involves using the condition of periodicity in the potential energy function to show that the wavefunction can be written as a product of a periodic function and a plane wave. This simplifies the Schrödinger equation to only consider a single unit cell of the lattice, making it easier to solve.

## 4. Can Bloch's theorem be applied to any periodic potential?

Yes, Bloch's theorem can be applied to any periodic potential as long as the condition V(x)=V(x+ma) is met. This includes crystalline solids, as well as other periodic systems such as superlattices and photonic crystals.

## 5. Are there any limitations to Bloch's theorem?

One limitation of Bloch's theorem is that it only applies to non-interacting particles, as it does not take into account the effects of electron-electron interactions. Additionally, it assumes a perfect periodic potential, which may not be the case in real materials. However, it is still a valuable tool for understanding the electronic properties of periodic systems.

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