A1  ^?? Find the energy difference $\Delta E = E_{4^2P_{1/2}} - E_{4^2S_{1/2}}$ between the states $4^2S_{1/2}$ and $4^2P_{1/2}$ in electron volts.

The energy difference $\Delta E$ is equal to the energy of a photon of wavelength $\lambda=397~\text{nm}$:
\[\Delta E = \hbar \omega = \frac{2\pi \hbar c}{\lambda} \]

A2  ^?? Using the program Mathieu.py, find the ratio $A=x_\text{max}/(dx / d \xi)_\text{max}$ of the maximum value of $x$ to the maximum value of $dx / d \xi$ during the oscillations. Show that this quantity does not depend on the initial conditions $x(0)$ and $dx/d\xi(0)$ which are defined in the 12th and 13th lines of the program.

For the initial conditions $x(0) = 0.0$, $dx/d\xi(0)=2.0$ the maximum $x_\text{max}=4.20$ and the maximum $(dx/d \xi)_\text{max}=2.87$. For the initial conditions $x(0) = 0.0$, $dx/d\xi(0)=4.0$ the maximum $x_\text{max}=8.34$ and the maximum $(dx/d \xi)_\text{max}=5.87$.

A3  ^?? Estimate the minimum possible uncertainty $\sigma_x$ in the position of the $\rm ^{41}Ca^+$ ion in the trap, assuming that $\sigma_x/\sigma_{dx/d\xi} = A$. Express the answer in terms of the ion's mass $m$, $\omega$ and $A$. Calculate the value of $\sigma_x$ for $\omega_0=9.4 \cdot 10^9~\text{s}^{-1}$. Compare $\sigma_x$ with the wavelength $\lambda$.

Let's express $\sigma_p$ from $\sigma_x$:
$$\sigma_p = m \sigma_{dx/dt} = m\omega \sigma_{dx/d\xi}/2 = m\omega \sigma_x/2A,$$thus
\[ \sigma_x = \sqrt{\frac{A \hbar}{\omega m}}\]

A4  ^?? Let the ion move with velocity $v \ll c$ along the $x$-axis. A photon with frequency $\omega + \delta \omega$ travels along the same axis. At what value of $\delta \omega=\delta\omega_0$ does the energy of the photon in the rest frame of the ion coincide with the energy difference $\hbar \omega$ between the levels of the ion?

The energy of the photon in the lab frame is $E_p=\hbar(\omega + \delta \omega)$ and the projection of the momentum onto the $x$-axis is $p_{x} = \hbar(\omega + \delta \omega) /c$. Using the Lorentz transformation, we can obtain the energy of the photon in the rest frame of the ion:
$$ E_p' = \gamma (E_p - vp_{px}) =\gamma \hbar (\omega + \delta \omega_0) \left( 1-\frac{v}{c} \right)= \hbar (\omega + \delta \omega_0) \sqrt{\cfrac{1-\frac{v}{c}}{1+\frac{v}{c}}}$$On the other hand the energy of the photon $E_p'$ in the rest frame of the ion is coincide with the transition eneregy $\hbar \omega$. So we have
$$ \delta \omega_0 = \omega \left( \sqrt{\cfrac{1+\frac{v}{c}}{1-\frac{v}{c}}} - 1 \right). $$
Using the approximation $v/c \ll 1$ we have:
\[ \delta \omega_0 \approx \omega \frac{v}{c} \]

A5  ^?? What is the photon flux $\Phi=dN/dt$ per unit area in a laser beam of intensity $I$ and photon frequency $\omega$?

Since a photon carries $\hbar \omega$ energy and its flux density is $\Phi$, we have $I = \hbar \omega \Phi$.

A6  ^?? Find the average change in the ion's momentum $\delta p$ after absorbing and emitting a photon of frequency $\omega + \delta \omega$ that has traveled in the direction of the $x$-axis. Assume that $\delta \omega \ll \omega$ and the velocity of the ion is $v \ll c$.

Consider a photon emission in the rest frame of the ion. Let $\theta$ be the angle between the momentum of the emitted photon and the positive direction of the $x$-axis. In this frame all direction of the emission are eqiuprobable, so the probability that $\theta$ will be in the range $[\theta, \theta + d\theta]$ is proportional to the solid angel in equal to the $dw = \cfrac{2\pi \sin \theta d \theta }{4 \pi} = \dfrac{\sin \theta d \theta}{2}$.

Using the Lorentz transformation, we find the momentum projection on the $x$-axis of the emitted photon in the lab frame.
\[ p_{2x} = \cfrac{\gamma \hbar \omega(\cos \theta + \beta)}{c} \]Hence, mean value $\langle p_{2x} \rangle $ is
\[ \langle p_{2x} \rangle = \int\limits_0^\pi \cfrac{\gamma \hbar \omega(\cos \theta + \beta)}{c} \cdot \cfrac{\sin \theta d \theta}{2} = \cfrac{\gamma \beta \hbar \omega}{c}. \]The projection on the $x$-axis of the ion's momentum change $\delta p_x$ is equal to the difference between absorbed and emitted photons:
\[ \delta p_x = \cfrac{\hbar (\omega + \delta \omega)}{c} - \langle p_{2x} \rangle = \cfrac{\hbar \omega}{c} \Big( 1+ \cfrac{\delta \omega}{\omega} - \cfrac{\beta}{\sqrt{1-\beta^2}} \Big) \]According to the approximations $\beta \ll 1$ and $\delta \omega \ll \omega$ we have
\[ \delta p_x \simeq \frac{\hbar \omega}{c}.\]It is following from the symmetry that mean projections of the momentum of emitted photon on the $y$- and $z$-axes are also equal to zero, so $\delta p = \delta p_x$.

A7  ^?? Estimate the average lifetime of the ion's excited state, denoted as $\tau_1'$, measured in its own rest frame. Express the answer in terms of $\Gamma$.

Let's assume that the ion has transitioned to the excited state at $t'=0$.

A simple estimation comes from the idea that if $\Gamma$ is a probability of emission per unit time, then at $t' = 1/\Gamma$ the probability of emission is equal to one.

A more precise approach can be applied. Let $P_1(t')$ be the probability that the ion is still in the excited state at $t'$. We also know that the probability of emission in the time span $t' \in [t'; t'+dt']$ (if the photon hasn't been emitted yet) is equal to $\Gamma dt'$. Thus
$$P_1(t'+dt') = P(t') \cdot (1-\Gamma dt').$$
Hence, we obtain a differential equation for $P_1(t')$:
$$ \cfrac{dP_1}{P_1} = - \Gamma dt'.$$With the initial condition $P_1(0) = 1$ we have
$$ P_1(t') = \exp(-\Gamma t') .$$
And finally
$$ \tau_1' = \int_0^{+\infty} t' \cdot \Gamma \exp(-\Gamma t') dt' = \cfrac{1}{\Gamma}.$$

A8  ^?? Find the average force $F$ acting on the ion moving at velocity $v$ in a laser beam of intensity $I$ and frequency $\omega + \delta \omega$ (this frequency is given in the laboratory frame).

In the resulting expression, the constant component is of no interest. So extract the variable component $F_\text{var}$, which depends on the velocity $v$. Assuming that the time the ion spends in the excited state is much shorter than in the unexcited state, and also assuming that $\Gamma \gg \delta \omega v/c$ and $v/c \ll 1$, write the expression for $F_\text{var}$ in terms of $v$, $\omega$, $\delta \omega$, $\Gamma$, $I$, and $\sigma$, $c$.

Let's find a mean time $\tau_2'$ between the photon emission and absorption in the rest frame of the ion.

In this frame $\Phi' = \gamma \Phi$ and the energy of the incident photon is $\gamma \hbar (\omega + \delta \omega)\left( 1- v/c\right) = \hbar(\omega + \Delta \omega)$. Thus, the mean lifetime of the ion in the unexcited state is
$$\tau_2' = \cfrac{1}{P\Phi' \sigma} = \cfrac{\left[ \gamma \left(1-\frac{v}{c}\right)(\omega + \delta \omega) -\omega \right]^2 + \Gamma^2}{\Gamma^2 \gamma \Phi \sigma}.$$
The mean time $\tau_2$ between photon emission and absorption in the laboratory frame, taking into account $\beta \ll 1$ and $\delta \omega \ll \omega$, is
$$ \tau_2 = \gamma \tau_2' \approx \cfrac{\Gamma^2 + (\delta \omega - v \omega / c)^2}{\Gamma^2 \Phi \sigma} $$
In accordance with $\Gamma \gg \Phi \sigma$, the force acting onto the ion is
$$ F_x \simeq \cfrac{\delta p_x}{\tau_2} = \frac{I\sigma}{c} \frac{\Gamma^2}{\left(\delta \omega - \omega \frac{v}{c}\right)^2 + \Gamma^2 }.$$

Let's separate the constant and the variable components of the force:
$$ F \simeq \frac{I\sigma}{c} + \frac{I\sigma}{c^2} \frac{2 \Gamma^2 \delta \omega \, \omega v }{(\delta \omega^2 + \Gamma^2)^2} $$

A9  ^?? What sign of $\delta \omega$ should be chosen to cool the ion, i.e. to reduce its kinetic energy over time?

If $F=-kv$, where $k$ is a constant, the force $F$ acts like a viscous friction, so it cools the ion. This case is realized when $\delta \omega < 0$.

A10  ^?? What is the value of $\lambda_2$?

The energy difference between the states $4^2S_{1/2}$ and $4^2P_{1/2}$ could be written as the energy of one transition or as the sum of the energies of two transitions:
$$ \Delta E = \cfrac{2\pi \hbar}{\lambda} = \cfrac{2\pi \hbar}{\lambda_1} + \cfrac{2\pi \hbar}{\lambda_2} $$
Thus,
\[\lambda_2 = \frac{\lambda_1 \lambda}{\lambda_1 - \lambda}\]

A11  ^?? At what distance $b$ from the plane of the lens to the photosensitive matrix does the clearest image of the ion form on it? Estimate the size of the clearest image of the ion for a lens with $f=0.6~\text{nm}$, $D=0.2~\text{nm}$.

From the thin lens equation we have
$$ \cfrac{1}{a} + \cfrac{1}{b} = \cfrac{1}{f}, $$so
$$b = \frac{af}{a-f}.$$

The size of the clearest image could be estimated as the size of Airy disk
$$ s = \cfrac{1.22 \lambda f}{D}.$$