## Information

• 官方介绍: https://ocw.mit.edu/courses/physics/8-421-atomic-and-optical-physics-i-spring-2014/

Larmor frequency $\Omega_L = \gamma B_{\mathrm{eff}} \gg \dot{\theta}$ . $\theta$ 是 $B_{\mathrm{eff}}$ 与 $z$ 轴的夹角.

\begin{align} B_{z, \mathrm{eff}} = B_0 - \frac{\omega(t)}{\gamma} \end{align} \begin{align} \theta \approx \frac{\pi}{2} - \frac{B_{z, \mathrm{eff}}}{B_1} \end{align} \begin{align} \dot{\theta} = \frac{\dot{\omega}}{\gamma B_1} = \frac{\dot{\omega}}{\omega_R} \ll \omega_R \end{align}

\begin{align} |\dot{\omega}| \ll \omega_R^2 \end{align}

## Quantized spin in a magnetic field

Hamiltonia

\begin{align} H = - \vec{\mu}\cdot \vec{B}_0 = - \gamma \hat{L}_z B_0 \end{align}

\begin{align} \frac{\mathrm{d}}{\mathrm{d}t} \vec{\hat{\mu}} = \frac{i}{\hbar}[\hat{H}, \vec{\hat{\mu}}] = \frac{i}{\hbar}\gamma[\hat{H}, \vec{\hat{L}}] = \gamma \vec{\hat{\mu}} \times \vec{\hat{B}} \end{align}

\begin{align} \frac{\mathrm{d}}{\mathrm{d}t} \langle\vec{\hat{\mu}}\rangle = \gamma \langle\vec{\hat{\mu}}\rangle \times \vec{\hat{B}} \end{align}
• valid for any spin
• valid for spin $1/2$ $\Rightarrow$ any two level system
• composite angular momentum (比如原子的总角动量) (unless $B$ field is so strong that it breaks up the coupling of the angular momentum)
• valid for a system of $N$ two-level systems (e.g. Dicke superradiance).

## Two level system spin $1/2$

What is the level structure of the electron in magnetic field?

Spin down is lowest state for the electron!

The magnetic moment has to point in the pulse direction( $\vec{B}$ ), which up, but the electron has negtive charge, the $\gamma$ factor is *negtive* and that's why for electron, the vector of spin and the vector of the magnetic momentum are oppsite. (差一负号)

\begin{align} \langle \mu_z \rangle = \frac{\gamma\hbar}{2}(P_{\downarrow} - P_{\uparrow}) = - \frac{\gamma\hbar}{2}(2P_e - 1) \end{align} \begin{align} P_e(t) = \frac{1}{2} - \frac{1}{\hbar\gamma} \langle\mu_z\rangle = \frac{1}{2} - \frac{1}{2} \left( 1 - 2 \frac{\omega_R^2}{\Omega_R^2}\sin^2 \frac{\Omega_Rt}{2} \right) \end{align}

\begin{align} P_e(t) = \frac{\omega_R^2}{\Omega_R^2} \sin^2 \frac{\Omega_Rt}{2} \end{align}

## Spin $\frac{1}{2}$ Hamiltonian

\begin{align} H = \frac{\hbar}{2} \begin{pmatrix} \omega_0 & \omega_Re^{-\mathrm{i}\omega t}\\ \omega_Re^{\mathrm{i}\omega t} & -\omega_0 \end{pmatrix} \end{align}

Question: Can be exactly realized in nature? Or it is an approximation?

Can be exactly realized in nature!

\begin{align} H_0 =& - \vec{\mu}\cdot \vec{B}_0 = - \gamma \hat{S}_z B_0 \\ =&\frac{\hbar}{2}\omega_0 \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} = \frac{\hbar\omega_0}{2}\sigma_z \end{align}

$|e\rangle, |g\rangle$ eigenstates with eigenvalues $\pm \frac{\hbar\omega_0}{2}$

Now add rotating fields

\begin{align} H_1 =& -\vec{\mu} \cdot\vec{B}_1(t) = -\vec{\mu}\cdot \frac{\omega_R}{\gamma}(-\hat{e}_{x}\cos\omega t - \hat{e}_y\sin\omega t) \\ =& \omega_R (\hat{S}_x\cos\omega t + \hat{S}_y \sin\omega t)\\ =& \frac{\hbar\omega_R}{2}\begin{pmatrix} 0 & e^{-\mathrm{i}\omega t} \\ e^{\mathrm{i}\omega t} & 0 \end{pmatrix} \end{align}

So $H = H_0 + H_1$ .

We go complex in our Hamiltonian, not because we have approximated a real field $\cos \omega t$ by $e^{\mathrm{i}\omega t}$ , but because when we have a rotating field and we write down it in Pauli spin matrix, we get imaginary units for the $\sigma_y$ spin matrix.

Famous dressed atom Hamiltonian describes two-level system *+* one mode of the EM field (with arbitrary amplitudes). (8.422 课程会细讲)

Since dressed atom picture in the standard way assumes that the photon number $N$ is large. There is a correspondance that in the limit of $N$ being large, the Fock state descriptin and the coherent state description fully agree.

## Solution of the Hamiltonian

Transform to rotating frame. Unitary transofrmation (Rotation operation around $\hat{e}_z$ )

\begin{align} U = \left.e^{\mathrm{i}\frac{S_z}{\hbar}\theta}\right|_{\theta = \omega t} = \begin{pmatrix} e^{\mathrm{i} \omega t/2} & 0\\ 0 & e^{-\mathrm{i}\omega t/2} \end{pmatrix} \end{align}

Then

\begin{align} H' = \frac{\hbar}{2} \begin{pmatrix} \delta & \omega_R \\ \omega_R & -\delta \end{pmatrix} \end{align}

is time independent, where $\delta = \omega - \omega_0$ . Care about $U \mathrm{i}\hbar\partial_t U^{\dagger}$ result a term $\sim \omega \sigma_z$ .

Solve it (homework # 1). A special case amplitudes

\begin{align} a_{g} (0) =& 1 \\ |a_e(t)|^2 =& \frac{\omega_R^2}{\Omega_R^2}\sin^2 \frac{\Omega_Rt}{2} \end{align}

as derived before.

## Rapid adiabatic passage --- Quantum treatment

\begin{align} H_0 = \frac{\hbar}{2} \begin{pmatrix} \delta & 0 \\ 0 & -\delta \end{pmatrix} \end{align}

\begin{align} H_0 = \frac{\hbar}{2} \begin{pmatrix} \delta & \omega_R \\ \omega_R & -\delta \end{pmatrix} \end{align}

\begin{align} P_{\mathrm{na}} = e^{- 2\pi \Gamma} \end{align}

where $\mathrm{na}$ means non-adiabatic, $\Gamma$ is landau-Zener parameter given as

\begin{align} \Gamma = \left(\frac{\omega_R}{2}\right)^2 \left[ \frac{\mathrm{d}\omega}{\mathrm{d}t} \right]^{-1} = \frac{1}{4}\frac{\omega_R^2}{\dot{\omega}} \end{align}

## Fast Sweep (没有完全理解)

Let's estimate the result in perturbation. Should we calculate that transition probability by using perturbation theory for an incoherent transition or for coherent transition?

Coherent!

• Coherent: population $\dot{a}_2  -\mathrm{i}\frac{H_{12}}{\hbar}$ with $H_{12}  \frac{\hbar}{2}\omega_R$ , 那么短时的演化为 $a_2  \frac{1}{2}\omega_R t$ , so $P_2\sim |a_2|^2 \omega_R^2 t_{\mathrm{eff}}^2$ , effective time $t_{\mathrm{eff}}$ (when wave function really changes) of driving the system. Coherent process are always quadratic in time.
• Incoherent: Fermi's Golden Rule $P \sim \frac{\omega_R^2 t_{\mathrm{eff}}}{\mathrm{density\, of\, states}}$

Q: What is $t_{\mathrm{eff}}$ in LZ?

• A. $t_{\mathrm{eff}} \sim \frac{\omega_R}{\dot{\omega}}$
• B. $t_{\mathrm{eff}} \sim \sqrt{\frac{1}{\dot{\omega}}}$
• C. $t_{\mathrm{eff}} \sim \frac{1}{\omega_R}$

the answer is B. $t_{\mathrm{eff}}$ is the "dephasing" time $\Delta t$ , the time during which everything is coherent

\begin{align} \Delta\omega \cdot \Delta t = \dot{\omega} \Delta t \cdot\Delta t \sim \pi \Rightarrow \Delta t \sim \sqrt{\frac{1}{\dot{\omega}}} \end{align}

Expand $e^{- 2\pi \Gamma}$ (small $\Gamma$ )

\begin{align} 1 - P_{\mathrm{na}} = 2\pi\Gamma \propto \frac{\omega_R^2}{\dot{\omega}} \end{align}