A1
^{ ??}
From the Poisson equation, obtain an expression which establishes a relationship between the coefficients $\alpha$, $\beta$ and $\gamma$.

Let's calculate the derivative straight forward:

\[

\frac{\partial^2 \Phi}{\partial x^2} = \frac{U}{2r_0^2} \frac{\partial^2}{\partial x^2} \left( \alpha x^2 + \beta y^2 + \gamma z^2 \right) = \frac{U}{2r_0^2} \frac{\partial}{\partial x} \left( 2 \alpha x \right) = \frac{U \alpha}{r_0^2}.\]

By analogy we have this relation:

\[ \Delta \Phi = \frac{U}{r_0^2} \left( \alpha + \beta + \gamma \right),\]

so the only way to satisfy $\Delta \Phi = 0$ is to choose coefficients according to

\[ \alpha + \beta + \gamma = 0\]

Answer:
\[\alpha + \beta + \gamma = 0\]

A2
^{ ??}
Calculate the maximum frequency $\omega_\text{fr,max}$ for which the wavelength $\lambda$ is $n=100$ times greater than $r_0$.

The relation which connects the wavelength with the frequency is $2\pi /\lambda = \omega / c$. So, the answer is

\[\omega_\text{fr,max}=\frac{2\pi c}{n r_0}\]

Answer:
\[\omega_\text{fr,max}=\frac{2\pi c}{n r_0} = 19 \cdot 10^9 ~\text{s}^{-1}\]

A3
^{ ??}
Using the equation $\DeclareMathOperator{\Grad}{grad} \vec{E} = - \Grad \Phi$, find the $x$-component of the electric field $\vec{E}$, in which the motion of the ion occurs.

Answer:
$$ E_x = -\cfrac{\partial \Phi}{\partial x} = -\cfrac{x}{r_0^2}\Big( \alpha U + \tilde{\alpha} V \cos (\omega_{rf} t )\Big) $$

A4
^{ ??}
Write an expression for $\xi$, $a_x$ and $q_x$ using time $t$ and parameters of the system: voltage $U$, radio frequency $\omega_\text{rf}$, the amplitude $V$ of the AC voltage, mass $m$ of the ion, it's charge $Ze$, typical size of the setup $r_0$ and coefficients $\alpha$ and $\tilde{\alpha}$.

The projection of the second Newton's law for the ion onto the $x$-axis is

$$ m \cfrac{d^2x}{dt^2} = ZeE_x = -\cfrac{Zex}{r_0^2}\Big( \alpha U + \tilde{\alpha} V \cos (\omega_{rf} t )\Big)$$

Using the subsistution $\xi = \omega_\text{rf} t /2$ we can obtain

$$ \cfrac{d^2x}{d\xi^2}+ x\Bigg( \frac{4 Ze \alpha U}{m r_0^2 \omega_\text{rf}^2}+\frac{4 Ze \tilde{\alpha} V}{m r_0^2 \omega_\text{rf}^2} \cos 2\xi \Bigg) = 0,$$

so the answer is

\[\xi = \frac{\omega_\text{rf}t}{2}, \quad a_x = \frac{4 Ze \alpha U}{m r_0^2 \omega_\text{rf}^2}, \quad q_x = -\frac{2 Ze \tilde{\alpha} V}{m r_0^2 \omega_\text{rf}^2}\]

Answer:
\[\xi = \frac{\omega_\text{rf}t}{2}, \quad a_x = \frac{4 Ze \alpha U}{m r_0^2 \omega_\text{rf}^2}, \quad q_x = -\frac{2 Ze \tilde{\alpha} V}{m r_0^2 \omega_\text{rf}^2}\]

A5
^{ ??}
Qualitatively plot the subset of $a_x$ and $q_x$ parameter values that result in the stable motion along the $x$-axis. Consider the range $a_x \in [-3;\: 7]$ and $q_x \in [0; \:5]$.

By running the Mathieu-stab.py we can obtain a graph that contains some artifacts in the middle of stability region.

Answer:

The region of stability is colored by yellow.

B1
^{ ??}
Qualitatively plot the subset of $U$ and $V$ parameter values that result in stable ion motion in the 3D RF trap. Calculate the values of the characteristic points of the stability region for $\omega_\text{rf} = 3\cdot 10^7~\text{s}^{-1}$, $r_0=1~\text{mm}$ and the ions of rubidium $\rm ^{87}Rb^+$. Consider that the generator in the lab allows you to get this range of voltage values $|U| \in [0; \: 200]~\text{V}$ and $|V| \in [0; \: 400]~\text{V}$.

Stable motion in the trap means that the motion along each axis is stable. We know that the equations of motion along the $x$- and the $y$-axes are the same but for the $z$-axis we have $a_z = -2a_x$ and $q_z = -2q_z$. This results that motion in the trap is stable when points $(a_x,q_x)$ and $(-2a_x,2q_x)$ are in stability region in the same time.

Answer:

The region of overall motion stability is colored by yellow.
The stability region of the motion along only $x$- and $y$-axes is colored by green.
The region of the motion stability along the $z$-axis only is colored by blue.
The region where motion along any axis is unstable is colored by dark blue.

B2
^{ ??}
Qualitatively plot the subset of $U$ and $V$ parameter values that result in different stability for $\rm ^{87}Rb^+$ and $\rm ^{86}Rb^+$ ions. Consider the case where the $\rm ^{87}Rb^+$ ions fly through the trap and the $\rm ^{86}Rb^+$ ions are ejected from the trap. Calculate the values of the characteristic points of the stability region for $\omega_\text{rf} = 3\cdot 10^7~\text{s}^{-1}$ and $r_0=1~\text{mm}$.

If the motion along both $x$- and $y$-axes is stable, the ion will pass through the trap, since $E_z=0$. Otherwise the ion will be ejected. So we have to find the intersection of the regions where the motion of the $\rm ^{87}Rm^+$ ion is stable and the motion of the $\rm ^{86}Rb^+$ ion is unstable.

First, let's find the region where solutions of the Mathieu equation for the parameters $(a_x, q_x)$ and $(-a_x, -q_x)$ are stable at the same time. This follows from $a_y = -a_x$ and $q_y = -q_x$.

The region where the solution of the Mathieu equation for the parameters $(a_x, q_x)$ and $(-a_x, -q_x)$ is simultaneously stable is colored yellow.

Now we scale the coordinates $a_x$ and $q_x$ for both isotopes with masses $m_{86} = 86~\text{Da}$ and $m_{87} = 87~\text{Da}$ and draw their stability regions on the same axes.

Answer:

The region where the ions can be separated is colored yellow.