Hamiltonian
考虑碱金属原子的基态 \(|^2S_{1/2}\rangle\) 和两个激发态 \(|^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 之间的跃迁.
如果不考虑精细结构, 两个激发态 \(|^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 是简并的. 那么它的 Hamiltonian 在基底 \(|^2S_{1/2}\rangle, |^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 下可以写为
\begin{align} \left( \begin{array}{ccc} E_{\mathrm{ex}} & 0 & 0 \\ 0 & E_{\mathrm{ex}} & 0 \\ 0 & 0 & 0 \end{array} \right) \end{align}其中, 选取了基态 \(|^2S_{1/2}\rangle\) 能量为 \(0\) , 激发态能量为 \(E_{\mathrm{ex}}\) .
两个激发态可以由 spin-orbital coupling 项 \(\frac{A_{\mathrm{fs}}}{\hbar^2}\hat{\vec{L}}\cdot\hat{\vec{S}}\) 解除简并
\begin{align} \hat{\vec{L}}\cdot\hat{\vec{S}} &= \frac{1}{2}\left[(\hat{\vec{L}}+\hat{\vec{S}} )^2 - \hat{\vec{L}}^2 - \hat{\vec{S}}^2\right] \\ &= \frac{1}{2}\left[\hat{\vec{J}} ^2 - \hat{\vec{L}}^2 - \hat{\vec{S}}^2\right] \end{align}所以
\begin{align} \frac{A_{\mathrm{fs}}}{\hbar^2}\hat{\vec{L}}\cdot\hat{\vec{S}} |^2S_{1/2}\rangle =&0|^2S_{1/2}\rangle \\ \frac{A_{\mathrm{fs}}}{\hbar^2}\hat{\vec{L}}\cdot\hat{\vec{S}} |^2P_{1/2}\rangle =& \frac{A_{\mathrm{fs}}}{2} |^2P_{1/2}\rangle\\ \frac{A_{\mathrm{fs}}}{\hbar^2}\hat{\vec{L}}\cdot\hat{\vec{S}} |^2P_{3/2}\rangle =&- A_{\mathrm{fs}}|^2P_{3/2}\rangle \end{align}那么它的 Hamiltonian 在基底 \(|^2S_{1/2}\rangle, |^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 下可以写为
\begin{align} H_{\mathrm{at}} = \left( \begin{array}{ccc} E_{\mathrm{ex}} + \frac{A_{\mathrm{fs}}}{2} & 0 & 0 \\ 0 & E_{\mathrm{ex}} - A_{\mathrm{fs}} & 0 \\ 0 & 0 & 0 \end{array} \right) \end{align}在偶极近似下, 光会通过偶极相互作用 \(\hat{H}_{\mathrm{d}} = \hat{\vec{d}}\cdot\hat{\vec{E}} = d_jE^0_j \cos(\phi_j - \omega t)\) (对相同的指标求和)将基态与激发态耦合在一起. 由于 \(|^2S_{1/2}\rangle\) 是偶宇称的态, \(|^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 是奇宇称的态, 而偶极相互作用只耦合不同宇称的态, 所以它 在基底 \(|^2S_{1/2}\rangle, |^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 下可以写为
\begin{align} H_{\mathrm{d}} = \left( \begin{array}{ccc} 0 & 0 & V \\ 0 & 0 & V \\ V^{*} & V^{*} & 0 \end{array} \right) \end{align}其中 \(V = \langle^2P_{1/2}|d_j|^2S_{1/2}\rangle E^0_j \cos(\phi_j - \omega t) = \langle^2P_{3/2}|d_j|^2S_{1/2}\rangle E^0_j \cos(\phi_j - \omega t) \equiv \langle \mathrm{ex}|d_j|\mathrm{g}\rangle E^0_j \cos(\phi_j - \omega t)\)
所以总的 Hamiltonian 在基底 \(|^2S_{1/2}\rangle, |^2P_{1/2}\rangle, |^2P_{3/2}\rangle\) 下可以写为
\begin{align} H = H_{\mathrm{at}} + H_{\mathrm{d}} = \left( \begin{array}{ccc} E_{\mathrm{ex}} + \frac{A_{\mathrm{fs}}}{2} & 0 & 0 \\ 0 & E_{\mathrm{ex}} - A_{\mathrm{fs}} & 0 \\ 0 & 0 & 0 \end{array} \right) +\left( \begin{array}{ccc} 0 & 0 & V \\ 0 & 0 & V \\ V^{*} & V^{*} & 0 \end{array} \right) \end{align}Rotating Wave Approximation
对 Hamiltonian 作变换
对 Hamiltonian 做一个幺正变换
\begin{align} U = \left( \begin{array}{ccc} e^{-\mathrm{i}\omega t} & 0 & 0 \\ 0 & e^{-\mathrm{i}\omega t} & 0 \\ 0 & 0 & 1 \end{array} \right) \end{align} \begin{align} U^{\dagger}H U = U^{\dagger}H_{\mathrm{at}} U + U^{\dagger}H_{\mathrm{d}} U = H_{\mathrm{at}} + \left( \begin{array}{ccc} 0 & 0 & V e^{\mathrm{i}\omega t} \\ 0 & 0 & V e^{\mathrm{i}\omega t} \\ V^{*}e^{-\mathrm{i}\omega t} & V^{*}e^{-\mathrm{i}\omega t} & 0 \end{array} \right) \end{align}其中
\begin{align} V e^{\mathrm{i}\omega t} = & \langle \mathrm{ex}|d_j|\mathrm{g}\rangle E^0_j \cos(\phi_j - \omega t) e^{\mathrm{i}\omega t}\\ = &\langle \mathrm{ex}|d_j|\mathrm{g}\rangle E^0_j \frac{1}{2}(e^{\mathrm{i}(\phi_j-\omega t)} + e^{-\mathrm{i}(\phi_j-\omega t)}) e^{\mathrm{i}\omega t} \\ = &\langle \mathrm{ex}|d_j|\mathrm{g}\rangle E^0_j \frac{1}{2}e^{\mathrm{i}\phi_j}(1 + e^{2\mathrm{i}\omega t}) \end{align}如果略去高频项 \(e^{2\mathrm{i}\omega t}\) , 并记 \(E_j \equiv E_j^0 e^{\mathrm{i}\phi_j}\) 那么 记
\begin{align} U^{\dagger}H_{\mathrm{d}} U \approx \left( \begin{array}{ccc} 0 & 0 & V_r \\ 0 & 0 & V_r \\ V_r^{*} & V_r^{*} & 0 \end{array} \right) \end{align}对 \(\mathrm{i}\hbar \frac{\partial}{\partial t}\) 作变换
做一个幺正变换相当于选了一组基底, 幺正变换不会改变本征能量. 但是这个幺正变换是含 时的, 会对时间演化产生影响, 它的时间演化为
\begin{align} \left(U^{\dagger} \mathrm{i}\hbar \frac{\partial}{\partial t} U \right) \left(U^{\dagger}|\psi\rangle \right) = \left(U^{\dagger} H U \right) \left(U^{\dagger}|\psi\rangle \right) \end{align}计算时间演化的幺正变换 \(\left(U^{\dagger} \mathrm{i}\hbar \frac{\partial}{\partial t} U \right)\)
\begin{align} \frac{\partial}{\partial t} U \left(U^{\dagger}|\psi\rangle \right) = & \frac{\partial}{\partial t} \left( \begin{array}{ccc} e^{-\mathrm{i}\omega t} & 0 & 0 \\ 0 & e^{-\mathrm{i}\omega t} & 0 \\ 0 & 0 & 1 \end{array} \right)\left(U^{\dagger}|\psi\rangle \right) \\ = & \left[ \begin{array}{c} \\ \\ \\ \end{array} \right. \left( \begin{array}{ccc} -\mathrm{i}\omega & 0 & 0 \\ 0 & -\mathrm{i}\omega & 0 \\ 0 & 0 & 0 \end{array} \right)U + U\frac{\partial}{\partial t} \left. \begin{array}{c} \\ \\ \\ \end{array} \right] \left(U^{\dagger}|\psi\rangle \right) \end{align}所以新的时间演化为
\begin{align} \mathrm{i}\hbar \frac{\partial}{\partial t} \left(U^{\dagger}|\psi\rangle \right) = H_r \left(U^{\dagger}|\psi\rangle \right) \end{align}其中
\begin{align} H_r =& U^{\dagger} H U - \left( \begin{array}{ccc} \hbar\omega & 0 & 0 \\ 0 & \hbar\omega & 0 \\ 0 & 0 & 0 \end{array} \right) \\ =&\left( \begin{array}{ccc} E_{\mathrm{ex}} + \frac{A_{\mathrm{fs}}}{2}- \hbar\omega & 0 & 0 \\ 0 & E_{\mathrm{ex}} - A_{\mathrm{fs}} - \hbar\omega & 0 \\ 0 & 0 & 0 \end{array} \right) +\left( \begin{array}{ccc} 0 & 0 & V_r \\ 0 & 0 & V_r \\ V_r^{*} & V_r^{*} & 0 \end{array} \right) \end{align}如果记 \(\Delta_{\mathrm{ex2}} \equiv E_{\mathrm{ex}} + \frac{A_{\mathrm{fs}}}{2}- \hbar\omega\) , \(\Delta_{\mathrm{ex1}} \equiv E_{\mathrm{ex}} + A_{\mathrm{fs}}- \hbar\omega\) 那么
\begin{align} H_r =\left( \begin{array}{ccc} \Delta_{\mathrm{ex2}} & 0 & 0 \\ 0 & \Delta_{\mathrm{ex1}} & 0 \\ 0 & 0 & 0 \end{array} \right) +\left( \begin{array}{ccc} 0 & 0 & V_r \\ 0 & 0 & V_r \\ V_r^{*} & V_r^{*} & 0 \end{array} \right) \end{align}用 Dirac 表象来理解 Rotating Wave Approximation
回到我们做 Rotating Wave Approximation 之前的 Hamiltonian
\begin{align} H = H_{\mathrm{at}} + H_{\mathrm{d}} = \left( \begin{array}{ccc} E_{\mathrm{ex}} + \frac{A_{\mathrm{fs}}}{2} & 0 & 0 \\ 0 & E_{\mathrm{ex}} - A_{\mathrm{fs}} & 0 \\ 0 & 0 & 0 \end{array} \right) +\left( \begin{array}{ccc} 0 & 0 & V \\ 0 & 0 & V \\ V^{*} & V^{*} & 0 \end{array} \right) \end{align}如果打入的光的频率对应的能量 \(\hbar\omega\) 很接近 \(E_{\mathrm{ex}}\) , 那么 \(\Delta_{\mathrm{ex2}}\) 和 \(\Delta_{\mathrm{ex2}}\) 可以看做是一个微扰, 所以可以将 Hamiltonian 写成 \(H = H_0 + V_\mathrm{p}\) (添加下标 \(p\) 代表 perturbation, 以区别于 \(V = \langle \mathrm{ex}|d_j|\mathrm{g}\rangle E^0_j \cos(\phi_j - \omega t)\) )的形式
\begin{align} H = H_0 + V_{\mathrm{p}} = \left( \begin{array}{ccc} \hbar\omega & 0 & 0 \\ 0 & \hbar\omega & 0 \\ 0 & 0 & 0 \end{array} \right) +\left( \begin{array}{ccc} \Delta_{\mathrm{ex2}} & 0 & V \\ 0 & \Delta_{\mathrm{ex1}} & V \\ V^{*} & V^{*} & 0 \end{array} \right) \end{align}所以在 Dirac 表象中, 它的演化方程为
\begin{align} \mathrm{i}\hbar\frac{\partial}{\partial t}|\psi^D(t) \rangle = V_{\mathrm{p}}^D(t) |\psi^D(t) \rangle \end{align}其中上标 \(D\) 代表 Dirac 表象. \(|\psi^D(t)\rangle = U_0^{\dagger}(t)| \psi(t)\rangle\) 是 Schrodinger 表象中 的态矢去掉 \(H_0\) 的演化. \(V_{\mathrm{p}}^D(t) = U_0^{\dagger}(t) V_p U_0(t)\) 是 Dirac 表象中 的微扰项. 而根据 Dirac 表象的定义 \(U_0(t)\) 为 \(H_0\) 对应的时间演化为
\begin{align} U_0 = e^{-\frac{\mathrm{i}}{\hbar}H_0t} = \left( \begin{array}{ccc} e^{-\mathrm{i}\omega t} & 0 & 0 \\ 0 & e^{-\mathrm{i}\omega t} & 0 \\ 0 & 0 & 1 \end{array} \right) \end{align}就是之前进行用到过的幺正变换. 所以如果我们在这里同样地略掉高频项, 就有 \(V_{\mathrm{p}}^D(t) = H_{r}\) 与之前的结果相符.
Reference
- [1] H. Z. 的 lecture
- [2] 博客中的 Term Symbol, 多体物理读书会:3.1节 Dirac表象中的演化算符-diagram techniques的起点, 定态微扰论总结 (update 18/Apr/2020)