## Crystal momentum

A current will flow if electric and/or magnetic fields are applied to the electrons in a crystal. The total force on an electron is the sum of the externally applied forces and forces exerted by the ions in the crystal lattice. In this section we show that if we construct a superposition of Bloch waves to form a localized wave packet of charge, this wavepacket moves as if only the externally applied fields are acting on it, $\vec{F}_{\text{ext}} = m^*\frac{\vec{v}_g(\vec{k})}{dt}$ where $\vec{F}_{\text{ext}}$ is the external force caused by the electric and magnetic fields, $m^*$ is the effective mass, and $\vec{v}_g(\vec{k})$ is the group velocity of a wave packet constructed from waves that all have wavenumbers close to $\vec{k}$.

We start out by reviewing the concept of Bloch waves and then determine how the Bloch waves respond to an external force. Any wave function that is defined in a crystal with periodic boundary conditions can be written as a Fourier series,

$$\psi (\vec{r}) = \sum\limits_{\vec{k}}c_{\vec{k}}e^{i\vec{k}\cdot\vec{r}}.$$

Here $\vec{k}$ sums over all points in reciprocal space. We can restrict the sum over $\vec{k}$ to the first Brillouin zone and the rest of reciprocal space can be reached by adding reciprocal lattice vectors $\vec{G}$. $$\psi (\vec{r}) = \sum\limits_{\vec{k}\in\text{1BZ}}\sum\limits_{\vec{G}}c_{\vec{k}+\vec{G}}e^{i(\vec{k}+\vec{G})\cdot\vec{r}}$$

We define a Bloch wave as,

$$\psi_\vec{k} (\vec{r}) = e^{i\vec{k}\cdot\vec{r}}\sum\limits_{\vec{G}}c_{\vec{k}+\vec{G}}e^{i\vec{G}\cdot\vec{r}}=e^{i\vec{k}\cdot\vec{r}}u_\vec{k} (\vec{r}).$$

From the definition of a Fourier series we know that,

$$u_\vec{k} (\vec{r}) = \sum\limits_{\vec{G}}c_{\vec{k}+\vec{G}}e^{i\vec{G}\cdot\vec{r}},$$

is a periodic function with the periodicity of the crystal. The arbitrary wave function that we started with can be written as a sum of Bloch waves.

$$\psi (\vec{r}) = \sum\limits_{\vec{k}\in\text{1BZ}}\psi_\vec{k} (\vec{r}).$$

It is convenient to write electron wave functions in terms of Bloch waves because Bloch waves are eigenstates of the Hamiltonian $\textbf{H}_0$. To see this, consider what happens when the translation operator acts on a Bloch wave,

$$\textbf{T} \psi_\vec{k} (\vec{r}) = e^{i\vec{k}\cdot(\vec{r}+\vec{a})}u_\vec{k}(\vec{r}+\vec{a}) = e^{i\vec{k}\cdot\vec{a}}e^{i\vec{k}\cdot\vec{r}}u_\vec{k}(\vec{r}) = e^{i\vec{k}\cdot\vec{a}}\psi_\vec{k} (\vec{r}).$$

Here $\vec{a}$ translates the lattice to an equivalent position. We observe that a Bloch wave is an eigenfunction of the translation operator with an eigenvalue $\langle\textbf{T}\rangle=e^{i\vec{k}\cdot\vec{a}}$. There are many possible translation vectors $\vec{a}$. Since the Hamiltonian commutes with the translation operator, Bloch waves are also eigenstates of the Hamiltonian.

The time derivative of the expectation value of an operator $\langle \textbf{A} \rangle$ is given by the Ehrenfest theorem,

$$i \hbar \frac{d}{dt} \langle \textbf{A} \rangle = \langle [\textbf{A}, \textbf{H} ] \rangle .$$

Since the translation operator and the Hamiltonian of the crystal commute, the expectation value of the translation operator is constant in time,

$$i \hbar \frac{d}{dt} \langle \textbf{T} \rangle = \langle [\textbf{T}, \textbf{H}_0 ] \rangle =0.$$

Consider what happens if an external force $\vec{F}$ is applied. A force is related to a potential energy term by $\vec{F} = -\nabla U$ so a force can be included by adding a potential energy term $U=-F_xx-F_yy-F_zz$ to the Hamiltonian. This force breaks the translational symmetry and the expectation value of the translation operator acquires a time dependence,

$$i \hbar \frac{d}{dt} \langle \textbf{T} \rangle = \langle [\textbf{T}, \textbf{H}_0-F_xx-F_yy-F_zz ] \rangle.$$

Since $[\textbf{T},\textbf{H}_0] = [\textbf{H}_0,\textbf{T}] =0$, this is,

$$i \hbar \frac{d}{dt} \langle \textbf{T} \rangle = \langle -\textbf{T}(F_xx+F_yy+F_zz)+(F_xx+F_yy+F_zz)\textbf{T} \rangle.$$

Allow $\textbf{T}$ to act on $(F_xx+F_yy+F_zz)$,

$$i \hbar \frac{d}{dt} \langle \textbf{T} \rangle = \langle -(F_x(x+a_x)+F_y(y+a_y)+F_z(z+a_z))\textbf{T}+(F_xx+F_yy+F_zz)\textbf{T} \rangle = \langle -(F_xa_x+F_ya_y+F_za_z)\textbf{T}\rangle.$$

Here $a_x,\,a_y,\,a_z$ are integer steps in the periodic lattice. As $(F_xa_x+F_ya_y+F_za_z)$ is a constant,

$$i \hbar \frac{d}{dt} \langle \textbf{T} \rangle = -(F_xa_x+F_ya_y+F_za_z)\langle\textbf{T}\rangle.$$

For a Bloch wave $\langle\textbf{T}\rangle = \langle e^{-i\vec{k}\cdot\vec{r}}u_{\vec{k}}^*|\textbf{T}|e^{i\vec{k}\cdot\vec{r}}u_{\vec{k}}\rangle = e^{i\vec{k}\cdot\vec{a}}\langle e^{-i\vec{k}\cdot\vec{r}}u_{\vec{k}}^*|e^{i\vec{k}\cdot\vec{r}}u_{\vec{k}}\rangle = e^{i\vec{k}\cdot\vec{a}}$,

$$i \hbar \frac{d}{dt} e^{i\vec{k}\cdot\vec{a}} = -(F_xa_x+F_ya_y+F_za_z)e^{i\vec{k}\cdot\vec{a}}.$$

The external force will cause the momentum of the electron to change so $\vec{k}$ will be time dependent. Performing the differentiation with respect to time,

$$- \hbar \left(\frac{dk_x}{dt} a_x + \frac{dk_y}{dt} a_y +\frac{dk_z}{dt} a_z\right) e^{i\vec{k}\cdot\vec{a}} = -(F_xa_x+F_ya_y+F_za_z)e^{i\vec{k}\cdot\vec{a}}.$$ $$\label{cm} \hbar \frac{d\vec{k}}{dt} = \vec{F}_{\text{ext}}.$$

This result says that applying an external force to an electron in a Bloch wave causes the wavevector to change like $\hbar \frac{d\vec{k}}{dt} = \vec{F}_{\text{ext}}$. Since this looks a lot like the relationship between the momentum and the total force $\frac{d\vec{p}}{dt} = \vec{F}_{\text{total}}$, $\hbar \vec{k}$ is called the crystal momentum.

Bloch waves oscillate with a frequency $\hbar\omega (\vec{k}) = E(\vec{k})$ where $E(\vec{k})$ is the energy of the Bloch wave. The waves move with a phase velocity $v_p = \omega/|\vec{k}|$. We now consider building a wave packet out of Bloch waves. If this wave packet is to be well defined for a long time it shoud be contructed from waves that all have about the same phase velocity. A wave packet that is built out of waves with wavenumbers close to $\vec{k}$ moves with a group velocity $\vec{v}_g(\vec{k}) = \nabla_{\vec{k}} \omega (\vec{k}) =\nabla_{\vec{k}} E(\vec{k})/\hbar$. Differentiating the group velocity with respect to time,

$$\frac{d\vec{v}_g(\vec{k})}{dt} = \frac{1}{\hbar}\frac{d}{dt}\left(\frac{\partial E}{\partial k_x}\hat{k}_x+\frac{\partial E}{\partial k_y}\hat{k}_y+\frac{\partial E}{\partial k_z}\hat{k}_z\right)$$

The time dependence of $E(\vec{k})$ is due to the fact that $\vec{k}$ changes when an external force is applied,

$$\frac{d\vec{v}_g(\vec{k})}{dt} = \frac{1}{\hbar}\left(\frac{\partial^2 E}{\partial k_x^2}\frac{dk_x}{dt}\hat{k}_x+\frac{\partial^2 E}{\partial k_y^2}\frac{dk_y}{dt}\hat{k}_y+\frac{\partial^2 E}{\partial k_z^2}\frac{dk_z}{dt}\hat{k}_z\right)=\frac{1}{\hbar}\nabla_{\vec{k}}^2E\frac{d\vec{k}}{dt}.$$

The expression for crystal momentum can be used to write this as,

$$\vec{F}_{\text{ext}} = m^*\frac{d\vec{v}_g(\vec{k})}{dt},$$

where the effective mass is, $$m^*(\vec{k}) = \frac{\hbar^2}{\nabla_{\vec{k}}^2E(\vec{k})}.$$

In summary, particle-like wavepackets moving in a crystal constructed out of electron wave functions respond to external fields as if they are free particles with an effective mass $m^*$.