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
\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 $\phi$)?
 A: 
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.
