Nope!So when I said thermal noise earlier, I meant thermal noise due to the Johnson-Nyquist noise of the tank circuit, which gets applied to the electrodes and accelerates the ion. If you disconnect the tank circuit, it doesn't add any noise to the ion. There might be other sources of thermal noise, but they small relative to the tank circuit's contribution.
Also, for others following this thread and confused what a Penning trap is, check out
this article, sections 3.1 and 3.2.It's both! (Although I think the amplitude of the jitter should be *inverse* proportional to the Q-factor.) Damping and thermal noise are intrinsically related, and this is the content of the fluctuation-dissipation theorem (FDT). Honestly, I wouldn't bother remembering the mathematical statement for FDT. I've wasted so many hours on that, only to end up back where I started. The juicy part of the theorem is that the noise power (noise amplitude squared) is proportional to the damping coefficient. This fact shows up everywhere: Brownian motion, electrical thermal noise, thermally-driven electromagnetic fields, even thermal jitter of mechanical resonators.
P.S., Technically, the "juicy part" of the FDT states that the autocorrelation of the thermal noise (as opposed to the noise power) is proportional to the damping coefficient. Again, the exact math isn't terribly useful since it's usually easier to derive the FDT for a given system from scratch than to use a generalized FDT. Just my two cents.
Specifically, for the Penning trap, here's how you get the FDT:
First you need to find an expression for ##\gamma## in terms of the resonator circuit resistance ##R##. You know that the induced charge on the electrode is given by ##q' = q \frac{z}{D}## for some length ##D## that depends on the trap geometry. Thus, the current going through the resonator circuit is ##\dot{q'} = q \frac{\dot{z}}{D}##, and the power dissipated is $$\begin{equation} \tag{1}P = I^2 R = q^2 \frac{\dot{z}^2}{D^2} R \end{equation}$$. Likewise, you know the energy dissipated in the ion's motion is ##F_{damp} \cdot \dot{z}##, so $$\begin{equation} \tag{2} P = \gamma \dot{z}^2 \end{equation}$$ Combining equations (1) and (2) yields $$\begin{equation} \tag{3} \gamma = \frac{q^2}{D^2}R \end{equation}$$
Next, you need to evaluate the spectral density of the thermal noise. Let ##\tilde{h}(\omega)## be the Fourier transform of ##h(t)##, the force on the ion due to thermal disturbances divided by mass. Then we know that $$\begin{equation} \tag{4} \tilde{h}(\omega) = \frac{q}{m} \tilde{E_t}(\omega) = \frac{q}{m} \frac{\tilde{V_t} (\omega)}{L} \end{equation}$$ where ##E_t## is the electric field at the ion's position due to thermal noise, ##V_t## is the voltage at the electrode due to thermal noise, and ##L## is another geometry-dependent length scale that converts electrode voltage to electric field at the trap center. We know that the voltage on the electrode ##V_t## will be given by
Johnson-Nyquist noise: $$\begin{equation} \tag{5} |\tilde{V_t}(\omega)|^2 = 4 k_B T R \end{equation}$$. So, combining equations (4) and (5), $$\begin{equation} \tag{6} |\tilde{h} (\omega)|^2 = \left(\frac{q}{m}\right)^2 L^{-2} \times 4k_B T \end{equation}$$
Combining equations (3) and (6) gives us the FDT: $$|\tilde{h} (\omega)|^2 = \left( \frac{D}{L} \right)^2 \frac{4 k_B T}{m^2} \gamma$$ As was promised, ##|h|^2 \propto \gamma##.