# What is the electrical potential of a quadrupole ion trap?

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

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.

• So $\phi_0$ is actually the external voltage supply and it has nothing to do with the quadrupole trap itself. And the form of $\phi$(the position depending part) is determined by the structure of this quadrupole trap? –  Hou Apr 1 at 11:08
• @Hou Yes, $\phi_0$ is given by the external voltage supply. And the constants in the position-dependent part ($\lambda, \sigma, \gamma$) are determined by the geometric shape of the metal electrodes. – Thomas Fritsch Apr 1 at 11:15
• Thanks. But can you offer more details about how to determine equation(1)? I know the Laplace’s equation of the potential, but the structure of a quadrupole trap is complex, I want to totally understand the form of the potential $\phi$. –  Hou Apr 1 at 11:20
• @Hou I have added a paragraph at the end of my answer. – Thomas Fritsch Apr 1 at 11:40
• Thanks. So anyway it is a computing problem, and the potential field that the ion feels is $\phi$, which is a function of time and position. –  Hou Apr 1 at 11:50