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?

  • 2
    $\begingroup$ Good question: the short answer is that $k$ is NOT the momentum of the electron. Momentum is not conserved in the presence of a lattice. You will find more information about that when you search for the term pseudomomentum. Also a good exercise is think how to describe a free electron in terms of Bloch momentum (as Bloch's theorem for sure also applies for free space with $V=0$). $\endgroup$
    – Fabian
    May 5, 2014 at 10:01
  • $\begingroup$ OP should accept an answer or indicate why the existing ones are not acceptable. $\endgroup$
    – DanielSank
    Sep 16, 2014 at 5:16

2 Answers 2


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.

  • $\begingroup$ So, I have the following question that : does the k contain the effect of the periodic lattice or not? if not how do we use it in the semiclassical theory of bloch electron transport and ignore the lattice effect in describing the bloch electrons? (N. Ashcroft- chapter 12) - and so if not why do we call it crystal momentum? $\endgroup$
    – P.A.M
    Nov 1, 2016 at 23:00
  • $\begingroup$ @P.A.M I'm not sure what you mean by "contain the effect". $\endgroup$
    – DanielSank
    Nov 2, 2016 at 17:25

You can't mix the crystal momentum $\vec{k}$ with the actual electron momentum, because in crystal, the actual translation symmetry is broken. That is to say, translate a very small distance, the system is changed, thus the actual momentum is not a good quantum number.

You can check that $\psi_{nk}=\psi_{nk+K}$ and $E_{nk}=E_{nk+K}$, so that $\psi_{nk}$ and $\psi_{nk+K}$ actually describe the same quantum state, therefor one can always translate crystal momentum outside 1BZ into it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.