# Exercise 1： Unitary and Hermitian

$\quad$ Consider the following matrices

$$U_1=\left(\begin{matrix} e^{-i\delta}\cos\theta & -e^{i\delta}\sin\theta \\\\ e^{i\delta}\sin\theta & e^{i\delta}\cos\theta \end{matrix}\right),\delta\ne0；U_2=\left(\begin{matrix} -1 & e^{-i\pi/2} \\\\ e^{i\pi/2} & 1 \end{matrix}\right) ；U_3=\frac{1}{\sqrt{2}}\left(\begin{matrix} 1 & e^{-i\theta} \\\\ e^{i\theta} & -1 \end{matrix}\right)$$

Which of these matrices is unitary? Hermitian? Explain why.

Solution:
$\quad$For $U_1$

$$U_1^\dagger = (U_1^*)^T = \left(\begin{matrix} e^{i\delta}\cos\theta & e^{-i\delta}\sin\theta \\\\ -e^{-i\delta}\sin\theta & e^{-i\delta}\cos\theta \end{matrix}\right) \ne U_1$$

then

$$U_1U_1^\dagger = \left(\begin{matrix} e^{-i\delta}\cos\theta & -e^{i\delta}\sin\theta \\\\ e^{i\delta}\sin\theta & e^{i\delta}\cos\theta \end{matrix}\right) \left(\begin{matrix} e^{i\delta}\cos\theta & e^{-i\delta}\sin\theta \\\\ -e^{-i\delta}\sin\theta & e^{-i\delta}\cos\theta \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 0 &1 \end{matrix}\right) = U_1^\dagger U_1$$

So $U_1$ is Unitary, but is not hermitian.
$\quad$For $U_2$

$$U_2^\dagger = (U_2^*)^T = \left(\begin{matrix} -1 & e^{-i\pi/2} \\\\ e^{i\pi/2} & 1 \end{matrix}\right) = U_2$$

then

$$U_2U_2^\dagger = \left(\begin{matrix} -1 & e^{-i\pi/2} \\\\ e^{i\pi/2} & 1 \end{matrix}\right) \left(\begin{matrix} -1 & e^{i\pi/2} \\\\ e^{-i\pi/2} & 1 \end{matrix}\right) = \left(\begin{matrix} 0 & -2i \\ -2i & 0 \end{matrix}\right)$$

So $U_2$ is hermitian, but is not unitary.
$\quad$For $U_3$

$$U_3^\dagger = (U_3^*)^T = \frac{1}{\sqrt{2}}\left(\begin{matrix} 1 & e^{-i\theta} \\\\ e^{i\theta} & -1 \end{matrix}\right) = U_3$$

then

$$U_3U_3^\dagger = \frac{1}{2}\left(\begin{matrix} 1 & e^{-i\theta} \\\\ e^{i\theta} & -1 \end{matrix}\right) \left(\begin{matrix} 1 & e^{-i\theta} \\\\ e^{i\theta} & -1 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) = U_3^\dagger U_3$$

So $U_3$ is hermitian and unitary.

# Exercise 2: Density matrix

$\quad$Consider the two following matrices $\rho_1=\left(\begin{matrix} 1/2 & 1/4 \\ 1/4 & 1/2 \end{matrix}\right)$ and $\rho_2=\left(\begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix}\right)$. Which one represent a mixed state? Explain why? For the other one, write down the corresponding pure state.

Solution:
$\quad$ For pure state, it has the following property

$$tr(\rho^2) = 1$$

For mixed state, it has the following property

$$tr(\rho^2) < 1$$

We can judge pure state and mixed state by their different properties.
$\quad$ For $\rho_1$

$$tr(\rho_1^2) = 5/8 < 1$$

So $\rho_1$ represent a mixed state. For $\rho_2$

$$tr(\rho_2) = 1$$

So $\rho_2$ represent a pure state. Write down the corresponding pure state, as

$$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{1} + \frac{1}{\sqrt{2}} \ket{2}$$

$\ket{1}$ and $\ket{2}$ are normalized and orthogonal to each other.

# Exercise 3: Atomic transitions

Which of the following transitions of the cesium atom are electric dipole-allowed? Which transitions can be considered as cycling (closed transitions)?

$$\begin{aligned} \ket{6S_{\frac{1}{2}},F=3,m_F=2} &\rightarrow \ket{5D_{\frac{3}{2}},F=4,m_F=2} \\ \ket{6S_{\frac{1}{2}},F=4,m_F=4} &\rightarrow \ket{6P_{\frac{3}{2}},F=5,m_F=5} \\ \ket{6P_{\frac{1}{2}},F=3,m_F=2} &\rightarrow \ket{7D_{\frac{5}{2}},F=3,m_F=2} \\ \ket{6S_{\frac{1}{2}},F=3,m_F=0} &\rightarrow \ket{6P_{\frac{1}{2}},F=3,m_F=0} \\ \ket{6P_{\frac{1}{2}},F=3,m_F=2} &\rightarrow \ket{5D_{\frac{3}{2}},F=2,m_F=1} \\ \ket{6S_{\frac{1}{2}},F=4,m_F=-4} &\rightarrow \ket{6P_{\frac{3}{2}},F=5,m_F=-5} \end{aligned}$$

Solution：
$\quad$ Electric dipole-allowed transitions satisfy the selection rules, as

$$\Delta L = \pm1 \ ;\ \Delta J=0,\pm1\ ;\ \Delta F=0,\pm1\ ;\ \Delta m_F=0,\pm 1$$

Based on the selection rules, the following transitions are electric dipole-allowed:

$$\begin{aligned} \ket{6S_{\frac{1}{2}},F=4,m_F=4} &\rightarrow \ket{6P_{\frac{3}{2}},F=5,m_F=5} \\ \ket{6S_{\frac{1}{2}},F=3,m_F=0} &\rightarrow \ket{6P_{\frac{1}{2}},F=3,m_F=0} \\ \ket{6P_{\frac{1}{2}},F=3,m_F=2} &\rightarrow \ket{5D_{\frac{3}{2}},F=2,m_F=1} \\ \ket{6S_{\frac{1}{2}},F=4,m_F=-4} &\rightarrow \ket{6P_{\frac{3}{2}},F=5,m_F=-5} \end{aligned}$$

Cycling (closed) transitions are those that start and end in the same hyperfine level. From the provided transitions, the following transitions can be considered cycling (closed):

$$\ket{6S_{\frac{1}{2}},F=3,m_F=0} \rightarrow \ket{6P_{\frac{1}{2}},F=3,m_F=0}$$

# Exercise 4: Intensity and Rabi frequency

$\quad$ Calculate the Rabi frequency of an atomic transition driven by a light field of intensity $1W/cm^2$. The relevant dipole matrix element is $d=4.2a.u.$ (the dipole is given in atomic units $=qa_0$).
$\quad$ This level of intensity $(1W/cm^2)$ is obtained at the focus of a Gaussian laser beam where the waist is $w_0=60\mu m$. What is the total power of the laser beam?

Solution：
$\quad$ For Rabi frequency, we can use the formula:

$$\Omega_R = \frac{dE}{\hbar}$$

where $\Omega_R$ is the Rabi frequency, $d$ is the dipole matrix element, $E$ is the electric field amplitude, and $\hbar$ is the reduced Planck's constant.
$\quad$ Given that the intensity of the light field is $1W/cm^2$, we can calculate the electric field amplitude using the formula:

$$I = \frac{1}{2} c \epsilon_0 E^2$$

where $I$ is the intensity of the light field, $c$ is the speed of light, $\epsilon_0$ is the vacuum permittivity.
$\quad$ We can solve for $E$ :

$$E = \sqrt{\frac{2I}{c\epsilon_0}}$$

And then we can calculate the Rabi frequency:

$$\Omega_R = \frac{d}{\hbar} \sqrt{\frac{2I}{c\epsilon_0}}$$

$\quad$ To calculate the total power of the laser beam, we can use the formula:

$$P = I \pi w_0^2 = (1W/cm^2) \times \pi \times 60\mu m \approx 1.13\times 10^{-4} W$$

where $P$ is the total power of the laser beam.

# Exercise 5: Light-atom interaction

$\quad$ We consider a 2-level atom interacting with a linearly polarized light field. The levels are denoted as $\ket{g},\ket{e}$. The light field is $\vec{E}_L=E_0 \cos(\omega_Lt)\vec{e}_z$. The detuning of the field from the respective atomic transition is $\Delta=\omega_L-\omega_{eg}$.
$\quad$ Apply the rotating frame transformation $U=\left(\begin{matrix}1 & 0 \\ 0 & e^{i\omega_Lt} \end{matrix}\right)$ to the system and simplify the Schrödinger equation using the so-called rotating wave approximation. The rotating wave approximation consists in neglecting the fast-oscillating terms in the Hamiltonian written in the rotating frame. Show that an effective Hamiltonian very similar to that described in the lectures for a two-level atom interacting with a circularly polarized light can be obtained, expressed with the relevant Rabi coupling $\Omega_L$ and detuning $\Delta$. Please pay attention to provide the details of the derivation.

Solution:
$\quad$ we have the following two types of Schrödinger equations in Schrödinger picture and rotating frame picture:

$$i\hbar\frac{d}{dt}\ket{\psi}^\mathcal{S} = \hat{H}^\mathcal{S}\ket{\psi}^\mathcal{S}\ ;\ i\hbar\frac{d}{dt}\ket{\psi}^\mathcal{R} = \hat{H}^\mathcal{R}\ket{\psi}^\mathcal{R}$$

where $\mathcal{S}$ represents in Schrödinger picture, $\mathcal{R}$ represents in rotating frame picture.
$\quad$ We can use the rotating frame transformation operator $\hat{U}$ to transform $\ket{\psi}^\mathcal{S}$ into $\ket{\psi}^\mathcal{R}$. The corresponding equation as follows:

$$\begin{aligned} \ket{\psi}^\mathcal{R} &= \hat{U} \ket{\psi}^\mathcal{S} \\ \ket{\psi}^\mathcal{S} &= \hat{U}^\dagger \ket{\psi}^\mathcal{R} \end{aligned}$$

Then, we apply the transformation to the Schrödinger equations:

$$\begin{aligned} i\hbar\frac{d}{dt}\ket{\psi}^\mathcal{R} &= i\hbar\frac{d}{dt} \left(\hat{U}\ket{\psi}^\mathcal{S}\right) \\&= i\hbar\left(\frac{d\hat{U}}{dt}\ket{\psi}^\mathcal{S}+\hat{U}\frac{d\ket{\psi}^\mathcal{S}}{dt}\right) \\&= i\hbar\frac{d\hat{U}}{dt}\ket{\psi}^\mathcal{S} + \hat{U}\hat{H}^\mathcal{S}\ket{\psi}^\mathcal{S} \\&= i\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger\ket{\psi}^\mathcal{R} + \hat{U}\hat{H}^\mathcal{S}\hat{U}^\dagger\ket{\psi}^\mathcal{R} \\&= \left(i\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger+\hat{U}\hat{H}^\mathcal{S}\hat{U}^\dagger\right)\ket{\psi}^\mathcal{R} \\&= \hat{H}^\mathcal{R}\ket{\psi}^\mathcal{R} \end{aligned}$$

So

$$\hat{H}^\mathcal{R} = i\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger+\hat{U}\hat{H}^\mathcal{S}\hat{U}^\dagger$$

We have known the matrix form of $\hat{U}$:

$$\hat{U}=\left(\begin{matrix}1 & 0 \\ 0 & e^{i\omega_Lt} \end{matrix}\right)$$

And we know the matrix form of $\hat{H}^\mathcal{S}$:

$$\begin{aligned} \hat{H}^\mathcal{S} &= \hat{H}_0^\mathcal{S} + \hat{V}^\mathcal{S} = \left(\begin{matrix}0 & 0 \\ \\ 0 & \hbar\omega_{eg}\end{matrix}\right) + \left(\begin{matrix}0 & \hbar\Omega_L\cos(\omega_L t) \\ \\ \hbar\Omega_L\cos(\omega_L t) & 0\end{matrix}\right) \\\\&= \left(\begin{matrix}0 & \hbar\Omega_L\cos(\omega_L t) \\ \\ \hbar\Omega_L\cos(\omega_L t) & \hbar\omega_{eg}\end{matrix}\right) = \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2}(e^{i\omega_Lt}+e^{-i\omega_Lt}) \\ \\ \frac{\hbar\Omega_L}{2}(e^{i\omega_Lt}+e^{-i\omega_Lt}) & \hbar\omega_{eg}\end{matrix}\right) \end{aligned}$$

Then we can calculate the matrix form of $\hat{H}^\mathcal{R}$:

$$\begin{aligned} \hat{H}^\mathcal{R} =& i\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger+\hat{U}\hat{H}^\mathcal{S}\hat{U}^\dagger \\\\=& i\hbar\left(\begin{matrix}0 & 0 \\ 0 & i\omega_Le^{i\omega_Lt} \end{matrix}\right)\left(\begin{matrix}1 & 0 \\ 0 & e^{-i\omega_Lt} \end{matrix}\right) \\&\qquad+ \left(\begin{matrix}1 & 0 \\ 0 & e^{i\omega_Lt} \end{matrix}\right) \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2}(e^{i\omega_Lt}+e^{-i\omega_Lt}) \\ \\ \frac{\hbar\Omega_L}{2}(e^{i\omega_Lt}+e^{-i\omega_Lt}) & \hbar\omega_{eg}\end{matrix}\right) \left(\begin{matrix}1 & 0 \\ 0 & e^{-i\omega_Lt} \end{matrix}\right) \\\\=& i\hbar\left(\begin{matrix}0 & 0 \\ 0 & i\omega_L \end{matrix}\right) + \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2} \\ \frac{\hbar\Omega_L}{2} & \hbar\omega_{eg} \end{matrix}\right) + \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2}e^{-i2\omega_Lt} \\ \frac{\hbar\Omega_L}{2}e^{i2\omega_Lt} & 0 \end{matrix}\right) \\\\=& \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2} \\ \frac{\hbar\Omega_L}{2} & -\hbar\Delta \end{matrix}\right) + \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2}e^{-i2\omega_Lt} \\ \frac{\hbar\Omega_L}{2}e^{i2\omega_Lt} & 0 \end{matrix}\right) \end{aligned}$$

We can use the rotating wave approximation. In this approximation, we neglect the fast-oscillating terms in the Hamiltonian written in the rotating frame. Then we can obtain the following:

$$\hat{H}^\mathcal{R} \approx \left(\begin{matrix}0 & \frac{\hbar\Omega_L}{2} \\ \frac{\hbar\Omega_L}{2} & -\hbar\Delta \end{matrix}\right)$$

(By the way, in the derivation above, we assumed an equation $\Omega_L^*=\Omega_L$)
$\quad$ As you can see, the effective Hamiltonian in the rotating frame is very similar to the one described in the lectures for a two-level atom interacting with a circularly polarized light. The diagonal terms represent the detuning, and the off-diagonal terms represent the Rabi coupling.

# Exercise 6: definitions for Rb

$\quad$ Calculate the recoil energy $E_R$. Give the result in units of Hz ($E_R/h$ where $h$ is the Planck constant). Calculate the recoil velocity $V_R$ and finally the Doppler temperature.

Solution:

$$v_R = \frac{hk_L}{m}$$

$$E_R = \frac{1}{2}mv_R^2$$

$$T_D = \frac{\hbar\Gamma}{2k_B}$$

# Exercise 7: Zeeman slower

$\quad$ Consider a $\mathrm{Cs}$ atom moving at a velocity of $350\mathrm{~m/s}$ and being slowed in a Zeeman slower by a laser driving the $\ket{6S_{1/2},F=4,m_F=4} \rightarrow \ket{6S_{3/2},F=5,m_F=5}$ atomic transition. For this problem, consider that the detuning as seen by the atom is constant equal to $-\Gamma/2$, where $\Gamma$ is the excited state lifetime, and the saturation parameter is $s=2$.

(a) How many cycles of absorption and emission are necessary for slowing down an atom completely?

(b) What is the duration of such a slowing?

(c) Consider that the Bias magnetic field of the Zeeman slower is 0 G. What is the magnetic field at the entrance of the Zeeman slower? Give the result in Gauss (G).

(d) What is the light field frequency detuning $\omega_L-\omega_A'$ at the entrance into the Zeeman slower? Note that $\omega_A'$ is the frequency of the transition including the Zeeman effect and excluding the Doppler shift.

(e) What is the stopping distance? Plot the strength of the magnetic field as a function of the slowing distance.

# Exercise 7: Dipole (trapping) force

$\quad$ Consider a standing wave formed by the superposition of two counter-propagating plane waves along $x$ with identical frequency $\omega_L$ and a two-level Rb atom at rest.

(a) Give the expression of the total light intensity due to the coherent superposition of two counter-propagating plane waves.

(b) Express the total dipole and scattering forces as a function of $\Delta$ and position (2-level atom system). We will assume that the total intensity is much smaller than the saturation intensity.