🗒️ Note: we will start by considering the NEFM
Electrons group velocity is defined as
$$ v=\frac{1}{\hbar}\frac{\text d E_k}{\text dk} $$
as $k$ $\uparrow$, $\rm KE$ $\uparrow$ then $\downarrow$ as can be seen
🗒️ Note: in NEFM there is both $\rm KE$ and $\rm PE$
Acceleration of the electron in state $k$ can be written
$$ a=\frac{1}{\hbar}\frac{\text d}{\text dt}\frac{\text dE}{\text dk}= \frac 1 \hbar \frac{\text dk}{\text dt} \frac{\text d^2 E}{\text dk^2} $$
Applying force $F$ during $\text dt$: work $\delta E=Fv\delta t$
$$ \begin{aligned} \text{when }\delta t\to 0 \qquad Fv&=\frac{\text dE}{\text dt}=\frac{\text dE}{\text dk}\frac{\text dk}{\text dt} \qquad \\ F&= \frac 1v \frac{\text dE}{\text dk} \frac{\text dk}{\text dt} = \hbar \frac{\text dk}{\text dt} \end{aligned} $$
used group velocity, now using acceleration :
$$ F=\frac{\hbar ^2}{\text d ^2 E/\text d^2 k} a $$
Applying newtons 2 law: effective mass $m^*$
$$ m^* =\frac{\hbar ^2}{\text d ^2 E/\text d^2 k} $$
🗒️ Note: we would have gotten the same result with FEM
💼 Case: consider $m^* <0$
Due to a small number of vacant state (holes), which move opposite of net electron flow direction
🗒️ Note: we can think of negative $m^*$ as electrons which are positively charged
💼 Case: consider a quantum state where a band is almost full of electrons with just one vacant site
applying an electric field $\vec E$, $R \to L$, electrons move $L\to R$ as they are negatively charged
$$ \vec F=-e\vec E=m^* \frac{\text d \vec v}{\text dt}=\hbar \frac{\text d \vec k}{\text dt} \qquad \frac{\text d\vec v}{\text d t}=-\frac e {m^*} \vec E $$
the holes move $R\to L$ as if positive charge so
$$ \begin{aligned} \vec F&=e\vec E=-m^* \frac{\text d \vec v}{\text dt}=-\hbar \frac{\text d \vec k}{\text dt} \\ \frac{\text d\vec v}{\text d t}&=\frac e {m^*} \vec E \end{aligned} $$
💎 Conclusion: A vacancy in a nearly-full electron band behaves like an extra positively charged particle
💼 Case: Consider a sample, a magnetic field is applied $\perp$ to an electric field driving a current
Electrons experience A Lorentz force to the right, curving its trajectory, causing a build-up on the edge of the slab ($y$-axis)
Creates a transverse electric field keeping electrons moving along $x$ with drift velocity
$$ v_x=-\frac{j_x}{ne} \qquad \text{where $j_x$: current density and $n$: electron density} $$
Equilibrium: force magnetic field $B_z$ = opposite transverse electric field (The Hall field $E_H$)
$$ \begin{aligned} F_{y,m}&=-ev_x B_z=\frac{j_x B_z}{n}=-F_{y,e}=-eE_H \\ \Rightarrow \qquad E_H&=-\frac{j_x B_z}{ne}=R_H j_x B_z \qquad R_H=-\frac{1}{ne} \; \text{(Hall coefficient for electrons)} \end{aligned} $$
🗒️ Note: thus for holes we have $R_H =+ 1/(pe)$ where $p$ is the hole concentration
Experimentally the hall voltage is
$$ V_H=E_H w $$
Thus for electrons
$$ \begin{aligned} R_H=-\frac 1 {ne}=\frac{E_H}{j_x B_z} = \frac{V_H/w}{I_x B_z / wt}= \frac{V_H t}{I_x B_z} \end{aligned} $$
🗒️ Note: the Hall coefficient relates mobility to the conductivity
$$ \sigma=ne\mu \qquad \mu=|R_H| \sigma $$
| Metal | Sign of $R_H$ |
|---|---|
| $\rm Na$ | - minus |
| $\rm K$ | - minus |
| $\rm Cu$ | - minus |
| $\rm Be$ | + plus |
| $\rm Mg$ | - minus |
| $\rm Cd$ | + plus |
| $\rm Al$ | - minus |
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/28ebaf8f-7ad0-4ee3-9736-e11c4d348549/Intrinsic_conductor.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/28ebaf8f-7ad0-4ee3-9736-e11c4d348549/Intrinsic_conductor.png" width="40px" /> Intrinsic conductor: Conductivity arises from the material's own properties (no impurities), it possesses a small energy gap allowing some electrons to enter the conduction band at normal temperatures, enabling a slight current. This conductivity, involving both electrons and holes, increases significantly with temperature as more electrons get excited.
</aside>
🗒️ Note: thus density of current carrying electrons $n$ $\uparrow$ with $T$, $n=p$ so concentration of holes $\uparrow$
