12.3 The Paul trap 263Fig. 12.3A linear Paul trap used to store a string of ions. (a) A view looking along the four rods with the end-cap electrode
and ions in the centre. Each of the rods is connected to the one diagonally opposite so that a voltage between the pairs gives
a quadrupole field. (b) A side view of the rod and end-cap electrodes which have a.c. and positive d.c. voltages, respectively.
A string of trapped ions is indicated.
one diagonally opposite and the a.c. voltageV =V 0 cos (Ωt) is applied
between the two pairs. Despite the fact that the voltages vary with time,
we first find the potential by the usual method for electrostatic prob-
lems. The electrostatic potentialφsatisfies Laplace’s equation∇^2 φ=0
(because divE=0andE=−∇φ). A suitable solution for the potential
close to thez-axis, that matches the symmetry of the voltages on the
electrodes, has the form of a quadrupole potential
φ=a 0 +a 2 (x^2 −y^2 ). (12.7)The coefficientsa 0 anda 2 are determined from the boundary conditions.
There are no terms linear inxorybecause of the symmetry under
reflection inx=0andy=0. Thetermsinx^2 andy^2 have opposite
signs, and the variation withzis negligible for rods much longer than
their separation 2r 0. The potential must match the boundary conditions
φ=φ 0 +V 0
2
cos (Ωt)atx=±r 0 ,y=0,φ=φ 0 −V 0
2
cos (Ωt)atx=0,y=±r 0.(12.8)
These conditions are satisfied by the potential^99 This ignores the finite size of the elec-
trodes and that the inner surfaces of the
electrodes would need to be hyperbolic,
e.g. a surface given byx^2 −y^2 =const.,
to match the equipotentials. However,
by symmetry this potential has the cor-
rect form forrr 0 , no matter what
happens near to the electrodes.
φ=φ 0 +V 0
2 r^20cos (Ωt)(
x^2 −y^2)
. (12.9)
To solve Laplace’s equation we have simply made a reasonable guess,
taking into account the symmetry. This is perfectly justified since the
uniqueness theorem says that a solution that fits the boundary condi-
tions is the only valid solution (see electromagnetism texts). The usual