Symmetries and solutions of the three-dimensional Paul trap

Using the symmetries of the three-dimensional Paul trap, we derive the solutions of the time-dependent Schr\"odinger equation for this system, in both Cartesian and cylindrical coordinates. Our symmetry calculations provide insights that are not always obvious from the conventional viewpoint.


Introduction
I first got to know of Joe (the speaker was MMN) because of his paper with Singh on the time-energy uncertainty relation [1]. [See the Appendix for this story.] To use the language of the discussions of this conference, the Eberly-Singh paper was "fundamental" though maybe not "useful." But it certainly was fun. That is what we are presenting today. Something that is fun. We are going to discuss the Paul trap from a different point of view. We will discuss the Paul trap's symmetries and use them to obtain the time-dependent solutions, without having to solve the mixed, second-order, partial, differential Schrödinger equation. We will be exemplifying something that Christopher Gerry alluded to yesterday: although the Heisenberg and Schrödinger formulations of quantum mechanics are equivalent, different things can be more transparent in one of them. It was Pauli who taught us this.

The Paul trap
The Paul trap provides a dynamically stable environment for charged particles [2]- [4] and has been widely used in fields from from quantum optics to particle physics.
Paul gives a delightful mechanical analogy [4]. Think of a mechanical ball put at the center of a saddle surface. With no motion of the surface, it will fall off of the saddle. However, if the saddle surface is rotated with an appropriate frequency about the axis normal to the surface at the inflection point, the particle will be stably confined. The particle is oscillatory about the origin in both the x and y directions. But it's oscillation in the z direction is restricted to be bounded from below by some z 0 > 0.
The potential energy can be parametrized as [2] V (x, y, where These potentials can be used to set up the classical motion problem The solutions to these equations are oscillatory Mathieu functions for the bound case [2]. The oscillatory motion is similar in the (x, y) directions but is different in the z direction, since g(t) = g 3 (t).
In the quantum mechanical treatment of the Paul trap, one must solve the Schrödinger equation, which in Cartesian coordinates has the form (h = m = 1) The time-dependent functions g and g 3 in Eq. (8) are where is the "dc" plus "ac time-dependent" electric potential that is applied between the ring and the end caps. Exact solutions for the 1-dimensional, quantum case were first investigated in detail by Combescure [5]. In general, work has concentrated on the z coordinate, but not entirely [6]. Elsewhere [7], stimulated by the work of Ref. [8], we discussed the classical/quantum theory of the Paul trap in the z-direction.
The form of the Schrödinger equation (8) suggests that, in addition to the above Cartesian coordinate system, there is another natural coordinate system for this problem: the cylindrical coordinate system. Introducing such a change of variables, the Schrödinger equation becomes In addition, the forms of Eqs. (8) and (12) suggest that we can factorize the solutions and equations into the forms and Being physicists, we tend to blindly go ahead and accept this separation, ignoring the mathematical subtleties in separating coordinates in time-dependent partial differential equations. Fortunately, Rod insists on keeping me honest. But this procedure does turn out to be justified [9]. So, now we can just go blindly on.
The operator S is one of the the Schrödinger operators we have discussed, L is a generator of Lie symmetries, and λ is a function of the coordinates x, y, z, and t. An operator L has the general form where the coefficient in each term is a function of the coordinates and time.
First we define what are going to be useful separable coordinates: and The t-dependent functions φ(t) and φ 3 (t) are given by where {ξ(t),ξ(t)} and {ξ 3 (t),ξ 3 (t)} are the complex solutions of the second-order, linear, differential equations in timë respectively, and satisfy the Wronskians Equations (24) are the same as the classical equations of motion (5)-(7) for the Paul trap. The solutions are the same Mathieu functions. For certain values of the parameters in the potential (1), the solutions are bound, meaning the charged particle is indeed "trapped" [2]. This shows a connection between classical and quantum dynamics.

Cartesian symmetries
For this exercise we can concentrate only on the z coordinate, since formally the results are the same for the x and y solutions, with the exception that the ξ 3 's and g 3 's, etc., lose the subscripts 3. [Elsewhere, we will discuss the symmetries of the Paul trap in much greater detail [9].] One can find, or simply verify, that the Schrödinger equation has the symmetry operators These operators satisfy the nonzero commutation relation and so form a complex Heisenberg Weyl algebra w c . This means that the operators generate a set of "number states" given by There is also an su(1, 1) algebra, which we will not go into here. This is a generalization of the "squeeze algebra." A word of caution. The states Z nz (z, t) are not to be construed as energy eigenstates. Z nz is a solution to the time-dependent Schrödinger equation (16). It is generally not an eigenfunction of the Hamiltonian. That is, So, to keep John Klauder from getting mad at me, we use the terminology "extremal state" for Z 0 and "higher-order" states for Z nz . We restrict the terms "ground state" and "excited state" for problems which allow eigenstates of the Hamiltonian.
Eq. (29) is a simple first order differential equation for Z 0 : Solving it yields where the proportionality constant, f must be a function of t.
A first idea might be to take this proportionality constant as the "normalization constant," [πφ 3 (t)] −1/4 . This would conserve the probability, as one would want for the time-development of a unitary Hamiltonian. But this turns out to be incorrect. Such a solution does not satisfy the Schrödinger equation (16). Indeed, putting Eq. (33) into Eq. (16) yields a first order differential equation in t for f (t). The resulting normalized extremal-state solution is So why is the phase factor ξ 3 /ξ 3 1/4 there? It is there because, in this time- But the right hand side of Eq. (35) is proportional to [9] the Rodrigues formula for the Hermite polynomials [15]. Using this yields the result The forms of X nx (x, t) and Y ny (y, t) follow immediately by just changing notation, so that Ψ nx,ny,nz (x, y, z, t) = X nx (x, t)Y ny (y, t)Z nz (z, t). (39)

Polar symmetries
In what is intuitively interesting, finding the symmetries for the polar coordinates amounts to considering complex linear combinations of the Cartesian operators. Take the following two pairs of operators which form the basis of two Heisenberg-Weyl algebras [11]: If we add to the above four operators I, the operator and the angular momentum operator we have that the set {I, a ± , c ± , K, L z } forms a closed algebra. Further, if we make the transformations we see that the the symmetry algebra has two oscillator subalgebras, {f, a ± , I} and {d, c ± , I}, which have only the identity operator, I, in common. Therefore, the algebra is G ′ x,y = os a (1) + os c (1), with the two Casimir operators C 1 = a + a − − f = − 1 2 and C 2 = c + c − − d = − 1 2 . We restrict ourselves to the representations of os(1) that are bounded below; namely, ↑ −1/2 + ↑ −1/2 . Let Ω n,m be a member of the set of common eigenfunctions of f and d spanning the representation space. Then, for (n, m) ∈ Z + 0 , we have f Ω n,m = n + 1 2 Ω n,m , dΩ n,m = m + 1 2 Ω n,m , a − Ω n,m = √ nΩ n−1,m , c − Ω n,m = √ mΩ n,m−1 , a + Ω n,m = √ n + 1Ω n+1,m , c + Ω n,m = √ m + 1Ω n,m+1 , (48) L z Ω n,m = (f − d)Ω n,m = (m − n)Ω n,m , KΩ n,m = (d + f )Ω n,m = (n + m + 1)Ω n,m ,