Ions inside ion traps interact with the light shining on them, allowing us to obtain their "photographs". This can be achieved using fluorescence: a process of spontaneous transition between two states of a system, followed by the emission of light.
Considering any system from the perspective of quantum mechanics leads to the fact that this system has discrete energy levels and their corresponding states. In other words, the stationary state of a quantum system (as opposed to a classical one) cannot be any state whatever you want. There is a certain discrete set of stationary states for each quantum system.
To describe a huge number of properties of quantum objects, it is sufficient to use a two-level system (TLS), i.e. to consider transitions of the system only between two of its levels.
In particular, let's consider two levels in the $\rm ^{41}Ca^+$ ion. To distinguish them, we will call them $4^2S_{1/2}$ and $4^2P_{1/2}$. The energy difference between them corresponds to the energy of a photon with a wavelength of $\lambda=397~\text{nm}$. The energy of a photon is given by the equation $\hbar \omega$, where $\hbar=1.05 \cdot 10^{-34}~\text{J}\cdot\text{s}$ is the Planck constant, and $\omega = 2\pi c / \lambda$ is the photon frequency. The speed of light is $c=2.98 \cdot 10^8~\text{m}/\text{s}$, and the elementary charge is $e=1.602 \cdot 10^{-19}~\text{C}$.
Every two-level system (TLS) doesn't exist in a pure vacuum but in a "bath" of electromagnetic radiation, with which it can exchange energy. This leads to the fact that if a TLS is not in the lowest energy state, it can spontaneously emit a photon with a certain probability. In our case, if the ion $\rm ^{41}Ca^+$ is in the state $4^2P_{1/2}$, then with a probability per unit time $\Gamma = dP/dt$, it will spontaneously relax to the state $4^2S_{1/2}$ while emitting a photon with a wavelength $\lambda=397~\text{nm}$. This phenomenon is called fluorescence.
Similarly, an ion in the state $4^2S_{1/2}$ can be excited by shining light of wavelength of $\lambda=397~\text{nm}$. With some probability, it will absorb a photon and change its state to the $4^2P_{1/2}$.
Let's consider an ion in the 3D RF Paul's ion trap. Its motion along each axis is described by the Mathieu equation:\[\begin{cases}\dfrac{d^2x}{d\xi^2} + (a - 2q \cos 2 \xi ) x = 0 \\[2pt]\dfrac{d^2y}{d\xi^2} + (a - 2q \cos 2 \xi) y = 0 \\[2pt]\dfrac{d^2z}{d\xi^2} -2 (a - 2q \cos 2 \xi ) z = 0\end{cases},\]
where $\xi=\omega_0 t/2$ is a dimensionless variable proportional to time, $\omega_0$ is the characteristic frequency of the electric field oscillations in the ion trap, $a$ and $q$ are dimensionless constants.
Perform further studies for the following constant values: $a_x=-0.1$, $q_x=0.8$.
The Heisenberg's indeterminacy principle states that for any quantum object, there is an inherently unavoidable uncertainty $\sigma_p$ in the value of the momentum $p$ and an uncertainty $\sigma_x$ in the value of the coordinate $x$, which are related by the expression
\[ \sigma_p \cdot \sigma_x \geq \frac{\hbar}{2}.\]
The absorption of light by ions and fluorescence is used to reduce their kinetic energy, i.e., for cooling. This method is called Doppler cooling. The idea is that a moving ion absorbs light not at a wavelength $\lambda$ but of a wavelength $\lambda + \delta \lambda$ due to the Doppler effect.
In the laboratory frame, the four-momentum of a photon moving along the $x$-axis is given by $p^\mu = (E/c, p_x, p_y, p_z) = (\hbar \omega/c, \hbar \omega/c, 0, 0)$. In a frame moving with velocity $V$ along the $x$-axis, the four-momentum of the photon changes according to the Lorentz transformations:
\[ p'^\mu =
\begin{pmatrix}
\hbar \omega'/c\\
\hbar \omega'/c \\
0\\
0
\end{pmatrix} =
\begin{pmatrix}
\gamma & -\gamma V/c & 0 & 0 \\
-\gamma V/c & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} \begin{pmatrix}
\hbar \omega/c\\
\hbar \omega/c\\
0\\
0
\end{pmatrix} = \Lambda^\mu\!_\nu p^\nu,
\]
where $\gamma = (1 - V^2/c^2)^{-1/2}$ is the Lorentz factor, and $\Lambda^\mu\!_\nu$ is the Lorentz transformation matrix.
When an ion absorbs a photon, it gains a momentum of $\hbar \omega/c$. Conversely, during fluorescence, the ion gives the emitted photon a momentum of $\hbar \omega/c$. However, the momentums of the absorbed photons are directed in one direction, while the momentums of the emitted photons are directed randomly. Therefore, by illuminating the ion with light of frequency $\omega + \delta \omega$, we act on it with a non-zero force on average.
To find the change in the ion's momentum, an energy analysis is also possible, but it requires a more detailed study of the phenomenon: most of the energy $\hbar \omega$ goes into changing the energy of a specific electron, and a small part of $\hbar \omega$ goes into changing the kinetic energy of the entire ion. The relationship between these quantities can be estimated using the conservation of momentum law.
What's more important for us is that in reality, the absorption spectrum of an ion has a certain width. This means that the ion absorbs photons not only with a frequency (in the ion's rest frame) exactly equal to the transition frequency $\omega$. The nature of the absorption spectrum is directly related to the probability per unit time $\Gamma $ of spontaneous transition. If the photon frequency (in the ion's rest frame) is equal to $\omega + \Delta \omega$, then the absorption probability is given by:
\[ P = \frac{\Gamma^2}{\Delta \omega^2 + \Gamma^2}.\]Note that $P$ is not the probability per unit time, but simply the probability of absorption when a photon passes through the effective cross-sectional area $\sigma$ of the ion.
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 $\sigma$, $c$.
After cooling the ion, we can proceed to study its interaction with various physical objects. In the process, we can obtain a "photo" of the ion. As discussed earlier, the ion interacts with light. However, it is inconvenient to use the light from the cooling laser to take an image of the ion, and an alternative method is usually used.
Let's add a 3rd level $3^2D_{3/2}$ to our model. This way the ion in the excited state will not only spontaneously relax during the transitions $4^2P_{1/2} \to 4^2S_{1/2}$ with emitting a photon with a wavelength $\lambda = 397~\text{nm}$ but also spontaneously relax through $4^2P_{1/2} \to 3^2D_{3/2}$ with emitting a photon with a wavelength $\lambda_1=866~\text{nm}$ and $3^2D_{3/2} \to 4^2S_{1/2}$ with emitting a photon with a wavelength $\lambda_2$.
The fluorescent light with a wavelength of $\lambda_1$ could be used to obtain an image of the ion. To do this, we place an optical filter that is nontransparent to $\lambda$ and $\lambda_2$, followed by a flat lens with a diameter $D$ and a focal length $f$, and then a photosensitive matrix. The distance from the plane of the lens to the ion is $a$.