I am studying Bloch's theorem, which can be stated as follows:
The eigenfunctions of the wave equation for a period potential are the product of a plane wave $e^{ik \cdot r}$ times a modulation function $u_{k}(r)$, which has the periodicity of the lattice. In total: $\psi_k (r) = u_k(r)e^{ik\cdot r} $. [Reference: Kittel - Introduction to solid sate physics.]
I have some problems understanding Bloch's theorem in full. Can I view the wavevector $k$ as the actual, physical momentum of the electron, which moves in a periodic potential, i.e., does it define the wavelength via $\lambda = 2\pi/k$? And how does this relate to the fact that all wavevectors can be translated back to the first Brouillon zone?
Answer
Here's a simple-minded answer:
Let's just compute the momentum of a particle with a Bloch wave function
$$\begin{eqnarray} \left.\langle x \right| \hat{p}\left|\Psi \rangle\right. &=& -i\hbar \left(\frac{d}{dx}\right) u_k(x) e^{i k x} \\ &=& -i \hbar \left( i k u_k(x) e^{ikx} + u_k'(x)e^{ikx}\right) \\ &=& \left( pu_k(x) - i\hbar u_k'(x)\right)e^{ikx} \end{eqnarray}$$
where in the last line we defined $p\equiv \hbar k$. This pretty clearly shows that the Bloch wave function is not an eigenfunction of the momentum operator. So, while you can always break the wave function down into plane waves $e^{ikx}$, and each component is a momentum eigenstate with momentum $p=\hbar k$, the Bloch functions are not themselves momentum eigenstates. Therefore, $k$ in $u_k(x)e^{ikx}$ is not the momentum of the Bloch state. Note, however, that if $u_k(x)=\text{constant}$ so that $u_k'(x)=0$, then we get
$$\left.\langle x \right|\hat{p} \left| \Psi \rangle \right. = pu_k(x)e^{ikx}=p\left.\langle x \ |\Psi \rangle \right. = \left.\langle x \right|\ p \left| \Psi \rangle \right. $$ or in other words
$$\hat{p}\left|\Psi\rangle\right. = p\left|\Psi \rangle\right. .$$
Please make a separate question for the Brillouin zone thing. I would like to answer this, but it belongs in a separate question.
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