## Envelope function and effective mass approximation Here we mainly consider the behavior of WF when the periodicity of lattice potential gets perturbed, especially by a hydrogen-like impurity. Finally, we'll obtain results as an envelope function. The focus is quantum mechanical motion in a parabolic band. First consider an unperturbed problem, which we have solved, and with a perturbation potential added, as the form. $[H_0 + V(\mathbf{r})] \Psi(\mathbf{r}) = E \Psi(\mathbf{r})$Consider an expansion on solved Bloch states, we have $\Psi(\mathbf{r}) = \sum_{n \mathbf{k}} F_{n \mathbf{k}} \, \psi_{n \mathbf{k}}(\mathbf{r})$ and above equation becomes $\sum_{n \mathbf{k}} \psi_{n \mathbf{k}}(\mathbf{r}) \left[ E_{n}(\mathbf{k}) - E + U(\mathbf{r}) \right] F_{n} (\mathbf{k}) = 0$ Multiply by $\psi_{n' \mathbf{k}'}^{*}$, namely another state, and integrate over $\mathbf{r}$, we will get the following equation, $\sum_{n', \mathbf{k}'} \left[ (E_{n} (\mathbf{k}) - E) \delta_{n \mathbf{k}, n' \mathbf{k}'} + U_{n \mathbf{k}, n' \mathbf{k}'} \right] F_{n'} (\mathbf{k}') = 0$ This equation is $\mathbf{r}$ independent after integration. The expression for $U_{n \mathbf{k}, n' \mathbf{k}'}$ would be, $U_{n' \mathbf{k}', n \mathbf{k}} = \int \mathrm{d}^3r \, \psi_{n' \mathbf{k}'}^*(\mathbf{r}) \, U(\mathbf{r}) \, \psi_{n \mathbf{k}}(\mathbf{r})$By several more transformations between real and $\mathbf{k}$ space, and doing variable transformation, we'll have, $U_{n' \mathbf{k}', n \mathbf{k}} = (2\pi)^3 \sum_{\mathbf{K}} U(\mathbf{k}' - \mathbf{k} - \mathbf{K}) \, C_{n\mathbf{k}}^{n' \mathbf{k}'}(\mathbf{K})$ $\mathbf{K}$ is a translational vector in BZ, and $C_{n\mathbf{k}}^{n' \mathbf{k}'}(\mathbf{K})$ is $C_{n\mathbf{k}}^{n' \mathbf{k}'}(\mathbf{K}) = \frac{1}{V_0} \int_{\text{EZ}} \mathrm{d}^3\mathbf{r} \, e^{-i \mathbf{K} \cdot \mathbf{r}} \, u_{n' \mathbf{k}'}^*(\mathbf{r}) \, u_{n \mathbf{k}}(\mathbf{r})$ This $EZ$ is the elementary cell. Till here we didn't apply any simplification, and now we propose the following approximation: 1. The perturbation potential change slowly on the scale of the lattice constant, i.e. $U(q)$ only significant for $q \ll \frac{\pi}{a}$. (Long range on the scale of lattice constant) 2. Perturbation small compared to typical energy separation. (Weak on the scale of $E_G$) 3. The coefficients $F_n(\mathbf{k})$ have significant value only for small $\mathbf{k}$. With these assumption, we have, - From (3), we only consider $\Gamma$ point. - From (1), only $\mathbf{K}=0$ get into the sum. $U_{n' \mathbf{k}', n \mathbf{k}} = (2\pi)^3 \, U(\mathbf{k}' - \mathbf{k}) \, C_{n\mathbf{k}}^{n' \mathbf{k}'}(0)$ And by simplifying the Bloch integral, we have $U_{n' \mathbf{k}', n \mathbf{k}} \approx U(\mathbf{k}' - \mathbf{k}) \, \delta_{n n'}$ So the equation of motion at very beginning becomes, $\sum_{\mathbf{k}} \left[ (E_{n}(\mathbf{k}) - E) \delta_{\mathbf{k}, \mathbf{k}'} + U(\mathbf{k}' - \mathbf{k}) \right] F_{n}(\mathbf{k}) = 0$ and the wavefunction in real space now becomes $\sum_{\mathbf{k}'} \left[ (E_{n}(\mathbf{k}) - E) \delta_{\mathbf{k}, \mathbf{k}'} + U(\mathbf{k}' - \mathbf{k}) \right] F_{n}(\mathbf{k}') = 0$ Consider the long range nature of $U(\mathbf{r})$, only small $\mathbf{k}$ would be important, so $\Psi(\mathbf{r}) = u_{n0}(\mathbf{r}) \sum_{\mathbf{k}} F_n(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} = u_{n0}(\mathbf{r}) \, F_n(\mathbf{r})$ This is by approximating $U_{nk}(\mathbf{r}) \approx U_{n0}(\mathbf{r})$, and interpret the sum over $\mathbf{R}$ as the Fourier series of a real-space function $F_n(\mathbf{r})$. Consider the dispersion relation for the conduction band: $E_c(k) = {E}_c + \frac{\hbar^2 k^2}{2m^*}$ If we put it into the previous equation of motion, we'll have, $\frac{\hbar^2}{2m^*} k^2 F_c(\mathbf{k}) + \sum_{\mathbf{k}'} U(\mathbf{k} - \mathbf{k}') F_c(\mathbf{k}') = (E - E_c) F_c(\mathbf{k})$ Interpret the summation over $\mathbf{k}'$ as a Fourier series, we'll get the real space $F_c(\mathbf{r})$, which would follow, $\left( -\frac{\hbar^2}{2m^*} \Delta + E_c + U(\mathbf{r}) \right) F_c(\mathbf{r}) = E F_c(\mathbf{r})$ This $E_c + U(\mathbf{r})$ is $E_{c}(\mathbf{r})$, effective potential, and $U(\mathbf{r})$ is the real space, external potential. **This is exactly the Schrödinger equation, where the periodic lattice potential hidden in $H_0$ has disappeared, but the free electron mass in $H_0$ has been replaced by the $m^*$ in the conduction band, while the newly introduced $E_c(\mathbf{r})$ acts as an effective potential in which the conduction band electrons move.** The $F_{c}(\mathbf{r})$ is an envelope function, and under such simplification we do not have to consider oscillatory potentials, but focus on the "perturbation". ![[Drawing 2024-09-01 14.16.18.excalidraw.svg]] With this real space envelope function, we can find the electron density in the same way solely using $F_n(\mathbf{r})$. $\begin{aligned} n(\mathbf{r}) &= |u_{c0}(\mathbf{r})|^2 \sum_i |F_i(\mathbf{r})|^2 f(E_i) \\ &= \sum_i |F_i(\mathbf{R})|^2 f(E_i) \end{aligned}$ if we consider only $\mathbf{R}$ in a primitive cell. >[!Notice] >The idea of filter out oscillatory term and keep only the envelope is kind of similar to [[Pseudopotential]] and [[Slowly varying envelope approximation]]. It's a widely applied technique. >[!Example] >Hydrogen like impurity: >$\left[ -\frac{\hbar^2}{2m^*} \Delta - \frac{e^2}{4\pi \varepsilon_0 \varepsilon_r |\mathbf{r}|} \right] F_c(\mathbf{r}) = (E - E_c) F_c(\mathbf{r})$ >This form is a hydrogen like problem, and the result have the same form, so we have the pre-obtained results, >$\begin{aligned} {E}^{*}_{Ry} &= \frac{e^4 m^*}{2 (4 \pi \varepsilon_0 \varepsilon_r)^2 \hbar^2} = E_{Ry} \frac{m^*}{m_e} \frac{1}{\varepsilon_r^2}, \\ {a}^{*}_B &= \frac{4 \pi \varepsilon_0 \varepsilon_r \hbar^2}{m^* e^2} = a_B \frac{m_e}{m^*} \varepsilon_r. \end{aligned}$ >${E}_{Ry}=13.6\ eV$, ${E}^{*}_{Ry}\ll E_{Ry}$ if $m^{*}\ll m_{e}$ and $\varepsilon_{r}>1$, and that's often the case in the crystals. >$a_{B}=0.53\ \mathrm{\mathring{A}}$, $a_{B}^{*}\gg a_{B}$ in crystals. >This provides some insights on how these impurity states looks like. For GaAs, $\varepsilon_{r}=11.7$, $m^{*}=0.067 m_{e}$, the calculated ${E}^{*}_{Ry}\approx 6\ meV$, $a_{B}^{*} \approx 100\ \mathrm{\mathring{A}}$. And the energy level of the impurities are >$E_{n}=E_{c}-\frac{E_{Ry}^{*}}{n^{2}}$ >Following the convention of hydrogen atom, we all the excitation energy $E_{Ry}^{*}$ "the binding energy". >![[Drawing 2024-09-01 16.06.42.excalidraw.svg]] ^fb8dcb The envelope function may be combined with a magnetic field, and, spin included, the form reads: $\left[ \frac{1}{2m^*} \left( \frac{\hbar}{i} \nabla + |e| \mathbf{A}(\mathbf{r}) \right)^2 + U(\mathbf{r}) + \frac{1}{2} g^* \mu_B \boldsymbol{\sigma} \cdot \mathbf{B} \right] F_c(\mathbf{r}) = (E - E_c) F_c(\mathbf{r})$ In the expression, we have $\frac{\hbar}{i} \nabla$: momentum (of the particle); $|e| \mathbf{A}(\mathbf{r})$: momentum of the field (or say it's a modification of kinetic energy of the particle inside the field); $\frac{1}{2} g^* \mu_B \boldsymbol{\sigma} \cdot \mathbf{B}$: accounts for the external magnetic field. Effective mass $m^{*}$ and effective $g$ factor $g^{*}$ could be calculated from the band edge parameters. $g^{*}$ value could be greatly different from 2, being $10-50$ or negative is common.