# 物理代写|固体物理代写Solid-state physics代考|PHYSICS3544

## 物理代写|固体物理代写Solid-state physics代考|Electrons and holes

The case of semiconductors, characterised by two partially filled topmost bands, is especially intriguing and deserves more attention.

Let us consider a semiconductor with $N_{\text {val }}$ valence electrons and suppose that $N_{\mathrm{CB}}$ ones have been promoted to the conduction band by thermal excitation, while $N_{\mathrm{VB}}=N_{\mathrm{val}}-N_{\mathrm{CB}}$ electrons are left in the valence band. If an electric field is applied, the $N_{\mathrm{CB}}$ electrons in $\mathrm{CB}$ provide a contribution $J_{\mathrm{q}, \mathrm{CB}}$ to the total charge current density as large as
$$J_{\mathrm{q}, \mathrm{CB}} \sim \sum_{k_{\mathrm{occ}} \in \mathrm{CB}} v_{\mathrm{CB}}\left(k_{\mathrm{occ}}\right),$$
where the sum runs over the wavevectors corresponding to the occupied $\mathrm{CB}$ states. Similarly, the VB electrons provide a contribution
$$J_{\mathrm{q}, \mathrm{VB}} \sim \sum_{k_{\mathrm{oxs}} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\mathrm{oce}}\right),$$
where in this case the sum runs over the wavevectors of the VB states still occupied by the unexcited electrons. It is useful to rewrite this current density contribution in a different way
\begin{aligned} J_{\mathrm{q}, \mathrm{VB}} & \sim \sum_{k_{\mathrm{ocR}} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\mathrm{occ}}\right)+\sum_{k_{\text {cmply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right)-\sum_{k_{\text {emply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right) \ & \sim \sum_{k_{\text {cumply }} \in \mathrm{VB}} v_{\mathrm{VB}}(k)-v_{\mathrm{VB}}\left(k_{\text {empty }}\right) \end{aligned}
where the $\sum_{k_{\text {eaply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {emply }}\right)$ term is obtained by summing over the wavevectors corresponding to the emptied states in the valence band. By applying the result $8.34$, we understand that the first term on the right-hand side of the above equation is zero since it contains a sum over all the states of a totally filled band. This leads to the remarkable result
$$J_{\mathrm{q}, \mathrm{VB}} \sim-\sum_{k_{\text {emply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right),$$
which in physical terms is interpreted as if the $J_{\mathrm{q}, \mathrm{VB}}$ contribution is given by a new type of charge carriers which occupy the VB states left empty by excited electrons.

## 物理代写|固体物理代写Solid-state physics代考|Effective mass

We now readdress the electron dynamics under the action of an external electric field $\mathbf{E}$, preliminarily treated in section 8.3.3 where, however, we only considered the effect of the external force $-e|\mathbf{E}|$. Even if we still assume for the moment an idealised picture where no defects are present and ions are clamped at their ideal positions, we know that an electron travelling within a crystal experiences as well the interactions with the lattice and with all the remaining electrons. In short, we are faced with the problem of inserting the fundamental notion that we are dealing with band electrons, rather than free electrons.

By considering an infinitesimal time interval $d t$, the work done by external force $-e|\mathbf{E}|$ changes the electron energy by the amount
\begin{aligned} d E_n(k) &=-e|\mathbf{E}| v_n(k) d t \ &=\frac{-e|\mathbf{E}|}{\hbar} \frac{d E_n(k)}{d k} d t, \end{aligned}

$$\frac{d k}{d t}=\frac{-e|\mathbf{E}|}{\hbar} .$$
The rate of change of the electron velocity is
$$\frac{d v_n}{d t}=\frac{d v_n}{d k} \frac{d k}{d t}=\frac{1}{\hbar} \frac{d^2 E_n(k)}{d k^2} \frac{-e|\mathbf{E}|}{\hbar},$$
so that its equation of motion is conveniently written in the form
$$-e|\mathbf{E}|=\left(\frac{1}{\hbar^2} \frac{d^2 E_n(k)}{d k^2}\right)^{-1} \frac{d v_n}{d t},$$
which indeed represents a very suggestive way of putting things: the quantity
$$m_e^*=\left(\frac{1}{\hbar^2} \frac{d^2 E_n(k)}{d k^2}\right)^{-1},$$
plays the role of an electron effective mass, fully incorporating any band feature through the second derivative of the $E_n(k)$ dispersion.

## 物理代写|固体物理代写固态物理代考|电子与空穴

.电子与空穴 .电子与空穴

$$J_{\mathrm{q}, \mathrm{CB}} \sim \sum_{k_{\mathrm{occ}} \in \mathrm{CB}} v_{\mathrm{CB}}\left(k_{\mathrm{occ}}\right),$$
，其中总和经过与被占据的$\mathrm{CB}$状态对应的波矢。类似地，VB电子提供了一个
$$J_{\mathrm{q}, \mathrm{VB}} \sim \sum_{k_{\mathrm{oxs}} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\mathrm{oce}}\right),$$

\begin{aligned} J_{\mathrm{q}, \mathrm{VB}} & \sim \sum_{k_{\mathrm{ocR}} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\mathrm{occ}}\right)+\sum_{k_{\text {cmply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right)-\sum_{k_{\text {emply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right) \ & \sim \sum_{k_{\text {cumply }} \in \mathrm{VB}} v_{\mathrm{VB}}(k)-v_{\mathrm{VB}}\left(k_{\text {empty }}\right) \end{aligned}
，其中$\sum_{k_{\text {eaply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {emply }}\right)$项是通过对价带中空态对应的波矢求和得到的。通过应用结果$8.34$，我们知道上面方程右边的第一项是零，因为它包含一个完全充满的带的所有状态的和。这导致了显著的结果
$$J_{\mathrm{q}, \mathrm{VB}} \sim-\sum_{k_{\text {emply }} \in \mathrm{VB}} v_{\mathrm{VB}}\left(k_{\text {empty }}\right),$$
，这在物理上被解释为$J_{\mathrm{q}, \mathrm{VB}}$贡献是由一种新的载流子给出的，这种载流子占据了被激发电子留下的空的VB态。

## 物理代写|固体物理代写固态物理代考|有效质量

\begin{aligned} d E_n(k) &=-e|\mathbf{E}| v_n(k) d t \ &=\frac{-e|\mathbf{E}|}{\hbar} \frac{d E_n(k)}{d k} d t, \end{aligned}

$$\frac{d k}{d t}=\frac{-e|\mathbf{E}|}{\hbar} .$$电子速度的变化率
$$\frac{d v_n}{d t}=\frac{d v_n}{d k} \frac{d k}{d t}=\frac{1}{\hbar} \frac{d^2 E_n(k)}{d k^2} \frac{-e|\mathbf{E}|}{\hbar},$$
，因此它的运动方程可以方便地写成
$$-e|\mathbf{E}|=\left(\frac{1}{\hbar^2} \frac{d^2 E_n(k)}{d k^2}\right)^{-1} \frac{d v_n}{d t},$$

$$m_e^*=\left(\frac{1}{\hbar^2} \frac{d^2 E_n(k)}{d k^2}\right)^{-1},$$

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