What is the electrical potential of a quadrupole ion trap?

Question

I learned from Wikipedia that the quadrupolar potential $\phi$ of a quadrupole ion trap is \begin{equation} \phi = \frac{\phi_0}{r_0^2}(\lambda x^2 + \sigma y^2 + \gamma z^2) \end{equation} where $r_0$ is a size parameter constant and $\lambda, \sigma, \gamma$ are weighting factors for the three coordinates, it also says that $\phi_0$ is the applied electrical potential which is a combination of AC and DC \begin{equation} \phi_0 = U + V\cos(\Omega t) \end{equation} so my question is that what is the difference between $\phi$ and $\phi_0$? Which is the actual potential field that the ion feels? And how to derive the first equation(the expression of $\phi$)?

Answer 1

so my question is that what is the difference between $\phi$ and $\phi_0$?

May be part of your confusion is caused by omitting the dependency on position and time in your equations.

The potential $\phi(x,y,z,t)$ depends on position ($x,y,z$) and time ($t$). It can be decomposed into two factors: One factor depending only on time ($t$), and another factor depending only on position ($x,y,z$). $$\phi(x,y,z,t)=\frac{\phi_0(t)}{r_0^2}(\lambda x^2+\sigma y^2+\gamma z^2) \tag{1}$$ where $$\phi_0(t)=U+V\cos(\Omega t) \tag{2}$$ is the electric potential applied by the external voltage supply.

And how to derive the first equation (the expression of $\phi$)?

Equation (1) can be derived from Laplace's equation for the potential field $\phi(x,y,z,t)$ $$\Delta\phi= 0 \tag{3a}$$ which short-hand notation for $$\frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial y^2}+ \frac{\partial^2\phi}{\partial z^2}=0. \tag{3b}$$

Laplace's equation (3) in turn can be derived from Gauss's law $$\vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_0} \tag{4a}$$ which is short-hand notation for $$\frac{\partial E_x}{\partial x}+ \frac{\partial E_y}{\partial y}+ \frac{\partial E_z}{\partial z} =\frac{\rho}{\epsilon_0} \tag{4b}$$ and the definition of the electric potential $\phi$ $$\vec{E}=-\vec\nabla\phi \tag{5}$$ where $\vec{E}(x,y,z,t)$ is the electric field strength and $\rho(x,y,z,t)$ is the charge density.

The electric charges of the trapped ions are much smaller than the charges on the external metal electrodes. And therefore, in the space between the electrodes, $\rho(x,y,z,t)$ can be neglected and replaced by $0$ in equation (4).

So you need to solve Laplace's equation (3) with the boundary conditions (given by the position-independent potential on your metal electrodes). But at the end you only want to know the potential near the center (small $x$, $y$, $z$) because that's where your ions are located. Therefore you can neglect any higher-order terms, like octupole terms ($\propto x^3,x^2y,xy^2,...$) which are much smaller than the quadrupole terms. And because of the geometric form of the electrodes you don't have any dipole terms ($\propto x,y,z$) in the first place.