In general, the total momentum of a system is conserved. Nevertheless, the momentum of a single electron or phonon is not conserved in a crystal. In fact, for an electron there are two types of momentum:

  1. Crystal momentum $k$ which is conserved up until a reciprocal lattice vector $G$. This is due to the fact that we can convert a $k$ outside the first Brillouin zone (BZ) into the first BZ by subtracting or adding $G$. Also the Bloch function is invariant under addition of $G$.
  2. The real momentum of the electron i.e, the momentum of a bloch wave given by $$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{r}) \nabla_r\; e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) \;\text{d}^3r\\ =\hbar \mathbf{k} - i\hbar \int u_\mathbf{k}^*(\mathbf{r}) \nabla_ru_\mathbf{k}(\mathbf{r}) \;\text{d}^3r $$

My question is:

Let’s imagine three cases:

  1. An electron scatters with another electron. Momentum of the electron is not necessarily conserved, but total momentum is. Where does the missing momentum come from?
  2. An electron scatters with a phonon and the phonon loses all its momentum to the electron. The momentum of the electron, however, has not to be equal to the momentum of the two separate particles (for example if the phonon transfers enough momentum to put the electron out of the first BZ). Where does the missing momentum appear to conserve the total momentum?
  3. Two phonons interact with each other to generate a new phonon. The resulting phonon can have less crystal momentum then the two separate phonons (umklapp process). How is total momentum conserved in this case?

Proof that electron momentum is not conserved: The momentum operator has not the same eigenfunctions as the Hamilton operator (Bloch waves):

$$\begin{eqnarray} \left.\langle x \right| \hat{p}\left|\Psi \rangle\right. &=& -i\hbar \left(\frac{d}{dx}\right) u_k(x) e^{i k x} \\ &=& -i \hbar \left( i k u_k(x) e^{ikx} + u_k'(x)e^{ikx}\right) \end{eqnarray}$$

Since the eigenfunctions are different, both operators do not commute. This means that the momentum of the Bloch function is not a conserved quantity.


1 Answer 1


How can the total momentum of a crystal be conserved?

The total Hamiltonian for a crystal looks like: $$ H = K_{\text ions} + K_{\text electrons} + \sum_{\text electrons; r_i\neq r_j}\frac{e^2}{2}\frac{1}{|\vec r_i-\vec r_j|} +\sum_{\text ions; R_i\neq R_j}\frac{Z^2e^2}{2}\frac{1}{|\vec R_i-\vec R_j|} +\sum_{\text electrons, ions; r_i,R_j}\frac{Ze^2}{|\vec r_i-\vec R_j|}\;. $$

If the entire crystal, including all electrons and ions is translated by a fixed distance $\vec a$: $$ \vec r_i \to \vec r_i + \vec a $$ and $$ \vec R_i \to \vec R_i + \vec a $$ the Hamiltonian is unchanged.

On the other hand, if we treat the ions as fixed and only translate the electronic variables then the Hamiltonian changes. So we see that the total electronic momentum is not conserved, but the total momentum of all electrons and ions is conserved.

  • $\begingroup$ So if two electrons scatter, some momentum will go into the crystal lattice? What if two phonons scatter…is momentum transferred to the electrons? What if a phonon and electron scatter, where is the momentum going? $\endgroup$
    – Lockhart
    Jan 27, 2023 at 6:50
  • $\begingroup$ I didn't attempt to answer these additional questions because I don't think they are sufficiently clear. The question in the title of your post is clear and answerable. $\endgroup$
    – hft
    Jan 27, 2023 at 15:06
  • $\begingroup$ But roughly speaking, yes. If it appears that the momentum is not conserved for two scattering particles, then that momentum must have been transferred to other components of the system that have been hidden via the formalism. $\endgroup$
    – hft
    Jan 27, 2023 at 15:09
  • $\begingroup$ Yes maybe the title question was a bit to general. $\endgroup$
    – Lockhart
    Jan 27, 2023 at 18:45

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.