Rabi's Formula (update 15/Apr/2020)

Table of Contents

1. Statement of the Problem

二能级系统的 Hamiltonian

\begin{align*} H = H_0 + V \end{align*}

\(H_0\) 的两个本征态为 \(|1\rangle ,\,|2\rangle\) . 在这两个 本征态组成的 Hilbert 空间写出 Hamiltonian 的矩阵形式

\begin{align*} H = \pmatrix{E_1 & 0 \\0 & E_2} + \pmatrix{0 & V_{12} \\ V_{21} & 0} \end{align*}

其中 \(V_{12}=V_{21}^{*} = \gamma e^{\mathrm{i}\omega t}\) 在 Dirac Representation 中, \(t\) 时刻处于态 \(|\psi(t) \rangle_D = c_1(t) |1\rangle +c_2(t)|2\rangle\) ,且 \(c_1(0) = 1, \quad c_2(0) = 0\)

其中\(|\psi(t) \rangle_D = e^{-\frac{1}{\mathrm{i}\hbar}H_0t}|\psi(t)\rangle\) , 求 \(c_1(t),\quad c_2(t)\)

2. Dirac Representation

\begin{align*} \mathrm{i}\hbar\frac{\partial}{\partial t}c_1(t) &= \mathrm{i}\hbar \frac{\partial}{\partial t} \langle 1 |\psi(t)\rangle_D \\ &= \langle 1|\cdot\mathrm{i}\hbar \frac{\partial}{\partial t} |\psi(t)\rangle_D \end{align*}

而 Dirac Representation 中的运动方程

\begin{align*} \mathrm{i}\hbar \frac{\partial}{\partial t} |\psi(t)\rangle _D = V_D |\psi(t)\rangle_D \end{align*}

所以

\begin{align*} \mathrm{i}\hbar\frac{\partial}{\partial t}c_1(t) &= \langle 1 |V_D |\psi(t)\rangle_D \end{align*}

插入一组完备基

\begin{align*} \mathrm{i}\hbar\frac{\partial}{\partial t}c_1(t) =& \langle 1 |V_D \cdot \left( |1 \rangle\langle 1| + |2\rangle\langle 2| \right) \cdot |\psi(t)\rangle_D \\ =&e^{\frac{\mathrm{i}}{\hbar}(E_1-E_1)t}V_{11} c_1(t) +e^{\frac{\mathrm{i}}{\hbar}(E_1-E_2)t}V_{12} c_2(t) \end{align*}

其中

\begin{align*} V_{11}=& \langle 1| V_D |1\rangle \\ V_{12}=& \langle 1 |V_D | 2\rangle \end{align*}

同理可得 \(c_2(t)\) 的微分方程

若记 \((E_n-E_m)/\hbar =\omega_{nm}\) , 则最终得到一个微分方程组

\begin{align*} \mathrm{i}\hbar\dot{c}_1 = V_{11} c_1 + e^{\mathrm{i}\omega_{12}t}V_{12} c_2 \\ \mathrm{i}\hbar \dot{c}_2 =e^{\mathrm{i}\omega_{21}t} V_{21} c_1 + V_{22} c_2 \end{align*}

3. 求解微分方程组

3.1. 化简

对于这个问题, \(V_{11} = V_{22} =0\) , \(V_{12}=V_{21}^{*} = \gamma e^{\mathrm{i}\omega t}\) ,所以

\begin{align*} \mathrm{i}\hbar\dot{c}_1 = e^{\mathrm{i}\omega_{12}t}V_{12} c_2 \\ \mathrm{i}\hbar\dot{c}_2 =e^{\mathrm{i}\omega_{21}t} V_{21} c_1 \end{align*}

\begin{align*} \mathrm{i}\hbar\dot{c}_1 =& \gamma e^{\mathrm{i}(\omega-\omega_{21})t} c_2 \tag{1}\\ \mathrm{i}\hbar\dot{c}_2 =&\gamma e^{-\mathrm{i}(\omega-\omega_{21})t} c_1\tag{2} \end{align*}

3.2. 求解

\((2)\) 移项得

\begin{align*} c_1 =\frac{\mathrm{i}\hbar}{\gamma}e^{\mathrm{i}(\omega-\omega_{21})t} \dot{c}_2 \end{align*}

两边对 \(t\) 求导

\begin{align*} \dot{c}_1 =\frac{\mathrm{i}\hbar}{\gamma}e^{\mathrm{i}(\omega-\omega_{21})t} [\mathrm{i}(\omega-\omega_{21})\dot{c}_2 +\ddot{c}_2] \tag{3} \end{align*}

将 \((1)\) 式代入 \((3)\) 式得

\begin{align*} \frac{\gamma}{\mathrm{i}\hbar} e^{\mathrm{i}(\omega-\omega_{21})t} c_2 =\frac{\mathrm{i}\hbar}{\gamma}e^{\mathrm{i}(\omega-\omega_{21})} [\mathrm{i}(\omega-\omega_{21})\dot{c}_2 +\ddot{c}_2] \end{align*}

整理成标准形式

\begin{align*} \ddot{c}_2 + \mathrm{i}(\omega-\omega_{21})\dot{c}_2 + \frac{\gamma^2}{\hbar^2}c_2 =0 \end{align*}

其解的形式为

\begin{align*} c_2(t) = e^{-\frac{\mathrm{i}}{2}(\omega-\omega_{21})t} \left(Ae^{\mathrm{i}\Omega t} + B e^{-\mathrm{i}\Omega t} \right) \tag{4} \end{align*}

其中

\begin{align*} \Omega = \sqrt{\frac{(\omega-\omega_{21})^2}{4}+\frac{\gamma^2}{\hbar^2}} \end{align*}

\((4)\) 式代入 \((2)\) 式得

\begin{align*} c_1(t)=\frac{\mathrm{i}\hbar}{\gamma}e^{\mathrm{i}(\omega-\omega_{21})t} \cdot e^{-\frac{\mathrm{i}}{2}(\omega-\omega_{21})t}\left[ -\frac{i}{2}(\omega-\omega_{21})\left(Ae^{\mathrm{i}\Omega t} + B e^{-\mathrm{i}\Omega t} \right) + \mathrm{i}\Omega\left(Ae^{\mathrm{i}\Omega t} - B e^{-\mathrm{i}\Omega t} \right)\right]\tag{5} \end{align*}

将初值条件 \(c_1(0) = 1, \quad c_2(0) = 0\) 代入 \((4), (5)\) 式得

\begin{align*} A+B =& 0 \\ \frac{\mathrm{i}\hbar}{\gamma}\cdot \mathrm{i}\Omega(A-B) =&1 \end{align*}

解得

\begin{align*} A =& -\frac{\gamma}{2\hbar\Omega} \\ B =& \frac{\gamma}{2\hbar\Omega} \end{align*}

将系数 \(A,\,B\) 的结果代入 \((4) ,\,(5)\) 式得

\begin{align*} c_2(t) =& -\frac{\mathrm{i}\gamma}{\hbar\Omega} e^{-\frac{\mathrm{i}}{2}(\omega-\omega_{21})t} \sin(\Omega t) \\ c_1(t) =&-\frac{\mathrm{i}}{2\Omega}(\omega-\omega_{21})e^{\frac{\mathrm{i}}{2}(\omega-\omega_{21})t}\sin (\Omega t) + e^{\frac{\mathrm{i}}{2}(\omega-\omega_{21})t}\cos (\Omega t) \end{align*}

取模方有

\begin{align*} |c_2(t)|^2 = \frac{1}{1+\frac{\hbar^2(\omega-\omega_{21})^2}{4\gamma^2}}\sin^2\left( \Omega t \right) \end{align*}

4. Results

4.1. code

import numpy as np
from matplotlib import pyplot as plt
from matplotlib import animation

N = 1000

gamma = 1

t = np.linspace(0,16,N)

def c2(omega):
    Omega = np.sqrt( (omega-1)**2/4 + gamma )
    c2 = 1/( 1+(omega-1)**2/(4*gamma**2) ) \
	*np.sin(Omega*t)**2
    ax.set_xticks([np.pi/Omega, 2*np.pi/Omega, \
		   3*np.pi/Omega, 4*np.pi/Omega, \
		   5*np.pi/Omega])
    ax.set_xlim(0,4*np.pi/Omega)
    return c2

fig, ax = plt.subplots()
line, = ax.plot(t,c2(1))

#ax3 = fig.add_axes([0.1, 0.1, 0.8, 0.8])



def init():
    line.set_ydata(c2(1))
    return line

def animata(i):
    line.set_ydata(c2(1+.01*i))
#    ax.text(1+0.1*i,1,'x')
    return line

ani = animation.FuncAnimation(fig=fig, func=animata, \
			      frames=400, interval=20)

ax.set_xlabel('$t$')
ax.set_ylabel('$|c_2(t)|^2$')
ax.set_title('Plot of $|c_2(t)|^2$ at $\omega = \omega_{21}$ to $\omega =5 \omega_{21}$')

ax.set_yticks([0,0.2,0.5,1])

ax.set_xticklabels(['$\pi/\Omega$', '$2\pi/\Omega$', \
		    '$3\pi/\Omega$', '$4\pi/\Omega$', \
		    '$5\pi/\Omega$'])
ani.save('fig.gif',writer='imagemagick')
plt.show()

5. 总结

如果取 \(\hbar = 1\) , 记 detuning \(\Delta = \omega-\omega_{21}\) , 那么对于一个二能级系统的(含时) 微扰

\begin{align} V_{12} = \gamma e^{\mathrm{i}\omega t} \end{align}

在 Dirac 表象(态矢中去除了 \(H_0\) 的演化)中

\begin{align} c_2(t) =& -\frac{\mathrm{i}\gamma}{\Omega} e^{-\frac{\mathrm{i}}{2}\Delta t} \sin(\Omega t) \end{align} \begin{align} |c_2(t)|^2 = \left(\frac{\gamma}{\Omega}\right)^2\sin^2\left( \Omega t \right) \end{align}

其中 Rabi frequency \(\Omega\) ( \(|c_2(t)|^2\) 以 \(2\Omega\) 为角频率振荡) 为

\begin{align} 2\Omega = \sqrt{\Delta^2+(2\gamma)^2} \end{align}

可以看出, detuning 越大, Rabi frequency 越大, \(|c_2(t)|^2\) 振幅越小.这是当然的, 失 谐大了, 耦合就弱, 因为初态取了 \(c_2(0)=0\) .

6. Reference

http://farside.ph.utexas.edu/teaching/qmech/Quantum/node113.html

J. J. Sakurai, Jim Napolitano, Modern Quantum Mechanics 2nd

Date: 2019-03-06 Wed 00:00

Created: 2022-04-18 Mon 20:38

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