Most textbooks make it clearly that crystal momentum is not true momentum. However, in a lot of literature, crystal momentum is treated as true momentum. Here's two examples:

  1. Rashba spin splitting. It is known that spin orbit coupling in solids can be mathematically expressed as $\nabla V\times \vec{p}\cdot \vec{v}$. If $\nabla V$ is along $\vec{z}$ direction, it can be simplified to $\vec{z}\times \vec{p}\cdot \vec{v}$. Then, in most literature, it is directly converted to $\vec{z}\times \vec{k}\cdot \vec{v}$ and $\vec{k}$ here is the crystal momentum.

  2. $k\cdot p$ method in heterostructures. In heterostructures, a common method to derive the Hamiltonian is to replace crystal momentum in original $k\cdot p$ Hamiltonian by $i\hbar\nabla$, which is the true momentum operator. For example, if some system is described by a $k \cdot p$ Hamiltonian $H(k_x, k_y)$. If we cut a strip along $y$ direction and break translational periodicity in $x$ direction, It is common to use $H(-i\hbar\partial_x, k_y)$ to describe this strip.

To summarize, it seems crystal momentum can sometimes be treated as true momentum. Why is that?


2 Answers 2


I have asked a professor about this and he gave me the answer.

After replacing $\mathbf{k}$ by $-i\nabla$ in $H(h\mathbf{k})$, we are actually getting a new Hamiltonian that acts on envelop of wave functions.

To make this answer relatively complete, I will briefly introduce the main steps focusing on only one band.

Suppose the band we are interested in has a dispersion $E(\mathbf{k})$, we can write its wave function as $\psi_{\mathbf{k}}=e^{i\mathbf{k}\cdot r}u_{\mathbf{k}}(r)$, where $u_{n\mathbf{k}}$ is periodic in space. A wave package can therefore be constructed from Bloch states as follow: $$\psi'(r,t)=\sum_{\mathbf{k}~\text{near}~0}c(\mathbf{k},t)e^{i\mathbf{k}\cdot r}\psi_{\mathbf{0}}(r)$$

We can define $F(r,t)=\sum_{\mathbf{k}~\text{near}~0}c(\mathbf{k},t)e^{i\mathbf{k}\cdot r}$ as the envelop function of the wave packet. Since $\mathbf{k}$ is near $\mathbf{0}$, $F(r,t)$ is slow varying with respect to $r$.

It can be shown that by substituting $\mathbf{k}$ by $-i\nabla$ in $E_n(\mathbf{k})$, we can get a dynamic equation of $F(r,t)$:

$$E(-i\nabla)F(r,t)=i\hbar\frac{\partial F(r,t)}{\partial t}$$

Generally, substituting $\mathbf{k}$ with $-i\nabla$ in a $\mathbf{k}\cdot \mathbf{p}$ Hamiltonian yields an "effective" Hamiltonian that acts on envelop functions.

For more reference, here's a book: http://link.springer.com/book/10.1007%2Fb13586


A eigenstate of a crystal hamiltonian can be written as a Bloch function in space representation $$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) $$ $u$ is periodic with respect to the unit cell. The momentum is now given by $$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{r}) \nabla_r\; e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) \;\text{d}^3r\\ =\hbar \mathbf{k} - i\hbar \int u_\mathbf{k}^*(\mathbf{r}) \nabla_ru_\mathbf{k}(\mathbf{r}) \;\text{d}^3r $$ If we now assume that the second term is small or vanishes due to symmetry reasons we can set $\langle\hat{p}\rangle = \hbar \mathbf{k}$

  • 3
    $\begingroup$ This is where it get's interesting: Under what circumstances vanishes the second term by symmetry reasons? $\endgroup$ Jun 29, 2016 at 10:49
  • 1
    $\begingroup$ One possibility would be inversion symmetry $\endgroup$
    – Jannick
    Jun 29, 2016 at 11:25
  • $\begingroup$ What I mentioned is a general practice and no such symmetry is guaranteed $\endgroup$
    – atbug
    Jun 29, 2016 at 11:40
  • 1
    $\begingroup$ Then it's just an approximation $\endgroup$
    – Jannick
    Jun 29, 2016 at 12:20
  • 2
    $\begingroup$ I get what you mean, but I would suggest replacing your expectation value "$\langle u_{\bf k}| \hat{p}|u_{\bf k} \rangle$" with the explicit expression $-i \hbar \int d^3r\, u_{\bf k}^*({\bf r}) {\bf \nabla} u_{\bf k}({\bf r})$, since "$| u_{\bf k} \rangle$" (as opposed to $| \psi \rangle$) isn't actually a quantum state. $\endgroup$
    – tparker
    Jul 4, 2016 at 17:53

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