公式速查 + 必背结论 + 重点例题清单

\[ h=6.63\times10^{-34}\,\text{J·s},\quad \hbar=\frac{h}{2\pi}=1.05\times10^{-34}\,\text{J·s},\quad hc=1240\,\text{eV·nm} \] \[ m_e=9.11\times10^{-31}\,\text{kg},\quad m_ec^2=0.511\,\text{MeV},\quad a_0=0.529\,\text{Å},\quad 1\,\text{eV}=1.6\times10^{-19}\,\text{J} \]

必背 \[ E=h\nu=\hbar\omega,\qquad p=\frac{h}{\lambda}=\hbar k \qquad(\text{光子}) \] \[ \tfrac12 mv_{\max}^2=h\nu-W,\quad \nu_0=\frac{W}{h}\qquad(\text{光电效应}) \] \[ \Delta\lambda=\frac{h}{m_ec}(1-\cos\theta),\quad \lambda_C=\frac{h}{m_ec}=2.43\,\text{pm}\qquad(\text{康普顿}) \] \[ L=n\hbar,\quad E_n=-\frac{13.6}{n^2}\,\text{eV},\quad \frac1\lambda=R\Big(\frac1{n_1^2}-\frac1{n_2^2}\Big)\qquad(\text{玻尔}) \] \[ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE_k}}\qquad(\text{德布罗意}),\qquad \lambda_{\max}T=b\qquad(\text{维恩}) \]

必背 \[ i\hbar\frac{\partial\Psi}{\partial t}=\hat H\Psi,\qquad \hat H=-\frac{\hbar^2}{2m}\nabla^2+V \] \[ \hat H\psi=E\psi\ (\text{定态}),\qquad \Psi(\vec r,t)=\psi(\vec r)\,e^{-iEt/\hbar} \] \[ \vec j=\frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*),\qquad \frac{\partial|\Psi|^2}{\partial t}+\nabla\cdot\vec j=0 \] \[ \text{无限深势阱：}\ E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\quad \psi_n=\sqrt{\frac2a}\sin\frac{n\pi x}{a}\ (\text{节点 }n{-}1) \] \[ \text{谐振子：}\ E_n=\Big(n+\tfrac12\Big)\hbar\omega\ (\text{等间距，零点能 }\tfrac12\hbar\omega) \]

必背 \[ [\hat x,\hat p_x]=i\hbar,\qquad [\hat L_x,\hat L_y]=i\hbar\hat L_z,\qquad [\hat L^2,\hat L_z]=0 \] \[ \hat L^2 Y_{lm}=l(l+1)\hbar^2 Y_{lm},\qquad \hat L_z Y_{lm}=m\hbar\,Y_{lm}\ (m=-l,\dots,l) \] \[ \langle\hat F\rangle=\int\psi^*\hat F\psi\,d\tau,\qquad \Delta A\,\Delta B\ge\tfrac12\big|\langle[\hat A,\hat B]\rangle\big| \] \[ \Delta x\,\Delta p\ge\frac{\hbar}{2},\qquad \text{守恒量：}[\hat A,\hat H]=0\ \text{且}\ \partial\hat A/\partial t=0 \]

必背 \[ F_{mn}=\langle\phi_m|\hat F|\phi_n\rangle=\int\phi_m^*\hat F\phi_n\,d\tau\quad(\text{自身表象中为对角阵}) \] \[ \text{幺正变换 }U^\dagger U=1:\ \text{不改变本征值、平均值、迹、内积} \] \[ \sum_n|n\rangle\langle n|=1\ (\text{封闭性}),\qquad \langle\phi|\psi\rangle\ (\text{内积}) \] \[ [\hat a,\hat a^\dagger]=1,\quad \hat a|n\rangle=\sqrt n\,|n-1\rangle,\quad \hat a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle \] \[ \hat N=\hat a^\dagger\hat a,\quad \hat N|n\rangle=n|n\rangle,\quad \hat H=\hbar\omega\big(\hat N+\tfrac12\big) \]

必背 \[ E_n^{(1)}=H'_{nn}=\langle n^{(0)}|\hat H'|n^{(0)}\rangle \] \[ E_n^{(2)}=\sum_{m\ne n}\frac{|H'_{mn}|^2}{E_n^{(0)}-E_m^{(0)}} \] \[ \text{简并：解久期方程}\ \det\big|H'-E^{(1)}I\big|=0\ \Rightarrow\ E^{(1)}\text{ 是 }H'\text{ 的本征值} \] \[ \text{费米黄金定则：}\ w=\frac{2\pi}{\hbar}|H'_{fi}|^2\rho(E_f);\quad \text{共振：}\ \hbar\omega\approx E_m-E_k \]

必背 \[ S=\sqrt{s(s+1)}\,\hbar=\frac{\sqrt3}{2}\hbar,\quad S_z=\pm\frac\hbar2,\quad \hat{\vec S}=\frac\hbar2\vec\sigma \] \[ \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\ \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \] \[ \sigma_i^2=I,\quad \{\sigma_i,\sigma_j\}=2\delta_{ij},\quad [\sigma_x,\sigma_y]=2i\sigma_z \] \[ \hat{\vec J}=\hat{\vec L}+\hat{\vec S},\quad j=l\pm\tfrac12\ (l\ne0) \]