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Symmetries and solutions of the three-dimensional Paul trap

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Abstract

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

©2001 Optical Society of America

1 Introduction

We will discuss the Paul trap from a different point of view: using the Paul trap’s symmetries 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: 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. Taking a different look at a pronlem is exemplified in Ref. [1].

2 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,z,t)=Vx(x,t)+Vy(y,t)+Vz(z,t),

where

Vx(x,t)=+e2r02V(t)x2g(t)x2,
Vy(y,t)=+e2r02V(t)y2g(t)y2,
Vz(z,t)=er02V(t)z2g3(t)z2.

These potentials can be used to set up the classical motion problem

ẍcl=Fx(x,t)=dVx(x,t)dx=2g(t)xcl,
ÿcl=Fy(y,t)=dVy(y,t)dy=2g(t)ycl,
z̈cl=Fz(z,t)=dVz(z,t)dz=2g3(t)zcl.

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).

3 The quantum-mechanical Paul trap

In the quantum mechanical treatment of the Paul trap, one must solve the Schrödinger equation, which in Cartesian coordinates has the form (ħ=m=1)

SΨ(x,y,z,t)={xx+yy+zz+2it2g(t)(x2+y2)2g3(t)z2}Ψ(x,y,z,t)=0.

The time-dependent functions g and g 3 in Eq. (8) are

g(t)=+e2r02V(t),g3(t)=er02V(t).

where

V(t)=VdcVaccosω(tt0)

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,

x=rcosθ,y=rsinθ,z=z,

the Schrödinger equation becomes

ScylΦ(r,θ,z,t)={rr+1rr+1r2θθ+zz+2it2g(t)r22g3(t)z2}Φ(r,θ,z,t)
=0.

In addition, the forms of Eqs. (8) and (12) suggest that we can factorize the solutions and equations into the forms

Ψ(x,y,z,t)=X(x,t)Y(y,t)Z(z,t),
SxX(x,t)={xx+2it2g(t)x2}X(x,t)=0,
SyY(y,t)={yy+2it2g(t)y2}Y(y,t)=0,
SzZ(z,t)={zz+2it2g3(t)z2}Z(z,t)=0,

and

Φ(x,y,z,t)=Ω(r,θ,t)Z(z,t),
SrθΩ(r,θ,t)={rr+1rr+1r2θθ2g(t)r2}Ω(r,θ,t)=0.

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. But in the end this procedure [9] does turn out to be justified [10]. So, now we can just go blindly on.

4 Lie symmetries and separable coordinates

Lie symmetries for the Schrödinger equation (8) can be obtained by solving the operator equation [9, 11]

[S,L]=(x,y,z,t)S.

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

L=C0t+C1x+C2y+C3z+C,

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:

x=xϕ1/2(t),y=yϕ1/2(t),z=zϕ31/2(t),t=t,

and

ρ=x2+y2ϕ1/2(t),θ=sin1(yx2+y2),z=zϕ31/2(t),t=t.

The t-dependent functions ϕ(t) and ϕ3(t) are given by

ϕ=2ξξ¯ϕ3=2ξ3ξ¯3,

where {ξ(t), ξ̄(t)} and {ξ 3(t), ξ̄3(t)} are the complex solutions of the second-order, linear, differential equations in time

γ̈+2g(t)γ=0,γ̈3+2g3(t)γ3=0,

respectively, and satisfy the Wronskians

W(ξ,ξ¯)=W(ξ3,ξ¯3)=i.

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.

5 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 ξ3’s and ξg3’s, etc., lose the subscripts 3. [Elsewhere, we will discuss the symmetries of the Paul trap in much greater detail [10].]

One can find, or simply verify, that the Schrödinger equation has the symmetry operators

Jz=ξ3ziξ˙3z=+12(ξ¯3ξ3)1/2[z+z(1i2ϕ˙3)],
Jz+=ξ¯3z+iξ¯˙3z=12(ξ¯3ξ3)1/2[zz(1+i2ϕ˙3)].

These operators satisfy the nonzero commutation relation

[Jz,Jz+]=I,

and so form a complex Heisenberg Weyl algebra ωc . This means that the operators generate a set of “number states” given by

JzZnz(z,t)=nzZnz1(z,t),
Jz+Znz(z,t)=nz+1Znz+1.

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 Znz(z,t) are not to be construed as energy eigenstates. Znz is a solution to the time-dependent Schrödinger equation (16). It is generally not an eigenfunction of the Hamiltonian. That is,

itZnz=HZnz=i[itZnzZnz]Znz[Const]Znz.

In line with the exortation of Klauder [12], we use the terminology “extremal state” for Z 0 and “higher-order” states for Znz.. 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:

JzZ0(z,t)=0.

Solving it yields

Z0(z,t)=f(t)exp{z22[1iϕ˙32]},

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,” [π ϕ3(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

Z0(z,t)=(πϕ3)1/4(ξ¯3ξ3)1/4exp{z22[1iϕ˙32]}.

So why is the phase factor (ξ̄3/ξ3)1/4 there? It is there because, in this time-dependent Schrödinger equation, it is the necessary generalization of the simple-harmonic oscillator ground-state energy exponential, exp[-iħω/2]. (This follows since ξ3(t)→(2ω)-1/2 exp[iωt].) This phase factor is necessary [13, 14, 15] for Eq. (34) to be a solution of Eq. (16).

Now repeated application of Eq. (30) gives

Znz(z,t)=[nz!]1/2[Jz+]nzZ0(z,t).

But the right hand side of Eq. (35) is proportional to [10] the Rodrigues formula for the Hermite polynomials [16]. Using this yields the result

Znz(z,t)=(2nznz!)1/2(πϕ3)1/4(ξ¯3ξ3)1/2(nz+12)
Hnz(z)exp{z22[1iϕ˙32]}.

The forms of Xnx(x,t) and Yny(y,t) follow immediately by just changing notation,

Xnx(x,t)=Znznx(zx,ξ3ξ,ϕ3ϕ,t),
Yny(y,t)=Znzny(zy,ξ3ξ,ϕ3ϕ,t),

so that

Ψnx,ny,nz(x,y,z,t)=Xnx(x,t)Yny(y,t)Znz(z,t).

6 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]:

a=a+=12(Jx+iJy)=12[ξ(x+iy)iξ˙(x+iy)]
=12(ξξ¯)1/2eiθ[ρ+iρθ+ρ(1i2ϕ˙)],
c=c+=12(JxiJy)=12[ξ(xiy)iξ˙(xiy)]
=12(ξξ¯)1/2eiθ[ρiρθ+ρ(1i2ϕ˙)].

If we add to the above four operators I, the operator

κ=Jx+Jx+Jy+Jy+1=a+a+c+c+1,

and the angular momentum operator

z=i(yxxy)=iθ,

we have that the set {I, a ±, c ±,K,Lz} forms a closed algebra. Further, if we make the transformations

f12(κz)=a+a+12,
d12(κ+z)=c+c+12,

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 Gx,y =osa (1)+osc (1), with the two Casimir operators C1=a+af=12 and C2=c+cd=12.

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)∈Z0+, we have

fΩn,m=(n+12)Ωn,m,dΩn,m=(m+12)Ωn,m,
aΩn,m=nΩn1,m,cΩn,m=mΩn,m1,
a+Ωn,m=n+1Ωn+1,m,c+Ωn,m=m+1Ωn,m+1,
zΩn,m=(fd)Ωn,m=(mn)Ωn,m,
κΩn,m=(d+f)Ωn,m=(n+m+1)Ωn,m,

In order that the spectrum of d and f be bounded below, we also have

aΩ0,0=0,cΩ0,0=0.

Eqs. (43) and (49) immediately tell us that the (normalized) θ dependence of the Ω n,m (r, θ, t) can be given by

Ωn,m,(r,θ,t)=𝓡n,m(r,t)Θn,m(θ)=𝓡n,m(r,t)exp[i(mn)θ]2π.

Therefore, Ω0,0(r, θ, t) is a function only of r or ρ and t. In this (n,m)=(0, 0) case, both the equations in (51) have the same first-order differential form. So, we can solve for Ω0,0 similarly to as was done for the Z0 solution. The normalized solution to the Schrödinger equation (18) is found to be

Ω0,0(r,θ,t)=12π[2ϕ]1/2(ξ¯ξ)12exp{ρ22[1iϕ˙2]}=X0(x,t)Y0(y,t).

Repeated application of Eqs. (48) gives us the general result:

Ωn,m(ρ,θ,t)=(n!m!)1/2a+nc+mΩ0,0(ρ,θ,t).

This turns out also to be a Rodrigues-type formula, although more complicated than before [10]. This time it is for the generalized Laguerre polynomials [16]. The end result turns out to be

Ωn,m(r,θ,t)=exp[i(mn)θ]2π()kk!(n!m!)1/2(2ϕ(t))1/2(ξ¯ξ)12(n+m+1)
ρnmLk(nm)(ρ2)exp{ρ22[1iϕ˙2]},
k12(n+mmn).

Now let us change to the more standard cylindrical quantum numbers:

=z,z=mn,nr=m+n0.

We find that

Ωn,m(r,t)Rnr,(r,t)Θz(θ),
Θz(θ)=exp[izθ]2π,
Rnr,(r,t)=(1)(nr)/2[2[nr2]!ϕ(t)[nr+2]!]1/2(ξ¯ξ)12(n+m+1)
ρL12(nr)()(ρ2)exp{ρ22[1iϕ˙2]}.

Except for our minus-sign phase convention, Eq. (60) resembles the standard result for the ordinary two-dimensional harmonic oscillator [17].

This standard solution has the known property that nr and z cannot differ by an odd integer to allow a normalizable solution. [12(nr) must be an integer.] If one solves the Schrödinger equation directly, this falls out, but the physical reason is not transparent. However, from the symmetry point of view, the reason is clear. The n and m quantum numbers, which reflect the fundamental symmetries, can be all non-negative integers. The nr and z quantum numbers reflect a rotation of the axes by 45°. (See figure 1.) So, reaching the allowed positions of quantum numbers along these diagonals, scaled by 1=√2, means all integer values of (nr ,z ) are not allowed.

 figure: Fig. 1.

Fig. 1. A plot of the allowed quantum numbers for the polar coordinates of the Paul trap. Shown are the (n,m) quantum numbers of Eqs. (54) and (55) as well as the (nr,ℓz ) quantum numbers of Eqs. (57) and (60).

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This is similar in spirit to Pauli’s being able to show, by symmetry methods, that there is an extra conserved quantity, the Runge-Lenz vector, om the hydrogen atom. This allowed an understanding of the extra degenerate quantum number.

Finally [18], we can write out the complete solution in spherical coordinates as

Φnr,z,nz(x,y,z,t)=Rnr,(r,t)Θz(θ)Znz(z,t).

Acknowledgements

MMN acknowledges the support of the United States Department of Energy. DRT acknowledges a grant from the Natural Sciences and Engineering Research Council of Canada. It is a pleasure to honor Joe Eberly.

References and links

1. J. H. Eberly and L. P. S. Singh, “Time operators, partial stationarity, and the energy-time uncertainty relation,” Phys. Rev. D 7, 359–362 (1973). [CrossRef]  

2. P. H. Dawson, Quadrupole Mass Spectrometry and its Applications (Elsevier, Amsterdam, 1976), Chaps. I–IV. Reprinted by (AIP, Woodbury, NY, 1995).

3. D. J. Wineland, W. M. Itano, and R. S. Van Dyck Jr., “High-resolution spectroscopy of stored ions,” Adv. Atomic Mol. Phys. 19, 135–186 (1983). [CrossRef]  

4. W. Paul, “Electromagnetic traps for charged and neutral particles,” Rev. Mod. Phys. 62, 531–540 (1998). [CrossRef]  

5. M. Combescure, “A quantum particle in a quadrupole radio-frequency trap,” Ann. Inst. Henri Poincare 44, 293–314 (1986).

6. M. Feng, J. H. Wu, and K. L. Wang, “A Study of the characteristics of the wave packets of a Paul-trapped ion,” Commun. Theoret. Phys. 29, 497–502 (1998).

7. M. M. Nieto and D. R. Truax, “Coherent states sometimes look like squeezed states, and visa versa: The Paul trap,” New J. Phys. 2, 18.1–18.9 (2000). Eprint quant-ph/0002050. [CrossRef]  

8. G. Schrade, V. I. Man’ko, W. P. Schleich, and R. J. Glauber, “Wigner functions in the Paul trap,” Quantum Semiclass. Opt. 7, 307–325 (1995). [CrossRef]  

9. W. Miller Jr., Symmetry and Separation of Variables (Addison-Wesley, Reading, MA, 1977).

10. M. M. Nieto and D. R. Truax (in preparation).

11. V. A. Kosteleck, V. I. Man’ko, M. M. Nieto, and D. R. Truax, “Supersymmetry and a time-dependent Landau system,” Phys. Rev. A 48, 951–963 (1993). [CrossRef]   [PubMed]  

12. J. R. Klauder, private communication.

13. The phase factor can also be obtained [14, 15], by solving the eigenvalue equation M3Znz=(nz+12)Znz,, where M3=i{ϕ3t+12(zz+12)i4ϕ¨3z2}.. Then, solving the equation Jz -Z0=0 will yield the extremal state function up to a factor of (π)-1/4.

14. D. R. Truax, “Symmetry of time-dependent Schrödinger equations. II. Exact solutions for the equation {xx +2i∂t -2g2(t)x2-2g1(t)x-2g0(t)}Ψ(x, t]=0.” J. Math. Phys. 23, 43–54 (1982). [CrossRef]  

15. M. M. Nieto and D. R. Truax, “Displacement operator squeezed states. I. Time-dependent systems having isomorphic symmetry algebras,” J. Math. Phys. 38, 84–97 (1997). [CrossRef]  

16. W. Magnus, F. Oberhettinger, and R. P. Soni, Formulas and Theorems for the Special Functions of Mathematical Physics. 3rd Edition (Springer, New York, 1966).

17. V. A. Kosteleckŷ, M. M. Nieto, and D. R. Truax, “Supersymmetry and the relationship between the Coulomb and oscillator problems in arbitrary dimensions,” Phys. Rev. D 32, 2627–2633 (1985). [CrossRef]  

18. M. M. Nieto and D. R. truax, eprint quant-ph/0011062, expands the contents of this manuscript. It contains further information on Ref. [1] and, in an appendix, on J. H. Eberly.

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Figures (1)

Fig. 1.
Fig. 1. A plot of the allowed quantum numbers for the polar coordinates of the Paul trap. Shown are the (n,m) quantum numbers of Eqs. (54) and (55) as well as the (nr,ℓz ) quantum numbers of Eqs. (57) and (60).

Equations (67)

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V ( x , y , z , t ) = V x ( x , t ) + V y ( y , t ) + V z ( z , t ) ,
V x ( x , t ) = + e 2 r 0 2 V ( t ) x 2 g ( t ) x 2 ,
V y ( y , t ) = + e 2 r 0 2 V ( t ) y 2 g ( t ) y 2 ,
V z ( z , t ) = e r 0 2 V ( t ) z 2 g 3 ( t ) z 2 .
x ̈ c l = F x ( x , t ) = d V x ( x , t ) d x = 2 g ( t ) x c l ,
y ̈ cl = F y ( y , t ) = d V y ( y , t ) dy = 2 g ( t ) y c l ,
z ̈ cl = F z ( z , t ) = d V z ( z , t ) dz = 2 g 3 ( t ) z c l .
S Ψ ( x , y , z , t ) = { x x + y y + z z + 2 i t 2 g ( t ) ( x 2 + y 2 ) 2 g 3 ( t ) z 2 } Ψ ( x , y , z , t ) = 0 .
g ( t ) = + e 2 r 0 2 V ( t ) , g 3 ( t ) = e r 0 2 V ( t ) .
V ( t ) = V d c V a c cos ω ( t t 0 )
x = r cos θ , y = r sin θ , z = z ,
S c y l Φ ( r , θ , z , t ) = { r r + 1 r r + 1 r 2 θ θ + z z + 2 i t 2 g ( t ) r 2 2 g 3 ( t ) z 2 } Φ ( r , θ , z , t )
= 0 .
Ψ ( x , y , z , t ) = X ( x , t ) Y ( y , t ) Z ( z , t ) ,
S x X ( x , t ) = { x x + 2 i t 2 g ( t ) x 2 } X ( x , t ) = 0 ,
S y Y ( y , t ) = { yy + 2 i t 2 g ( t ) y 2 } Y ( y , t ) = 0 ,
S z Z ( z , t ) = { zz + 2 i t 2 g 3 ( t ) z 2 } Z ( z , t ) = 0 ,
Φ ( x , y , z , t ) = Ω ( r , θ , t ) Z ( z , t ) ,
S r θ Ω ( r , θ , t ) = { r r + 1 r r + 1 r 2 θ θ 2 g ( t ) r 2 } Ω ( r , θ , t ) = 0 .
[ S , L ] = ( x , y , z , t ) S .
L = C 0 t + C 1 x + C 2 y + C 3 z + C ,
x = x ϕ 1 / 2 ( t ) , y = y ϕ 1 / 2 ( t ) , z = z ϕ 3 1 / 2 ( t ) , t = t ,
ρ = x 2 + y 2 ϕ 1 / 2 ( t ) , θ = sin 1 ( y x 2 + y 2 ) , z = z ϕ 3 1 / 2 ( t ) , t = t .
ϕ = 2 ξ ξ ¯ ϕ 3 = 2 ξ 3 ξ ¯ 3 ,
γ ̈ + 2 g ( t ) γ = 0 , γ ̈ 3 + 2 g 3 ( t ) γ 3 = 0 ,
W ( ξ , ξ ¯ ) = W ( ξ 3 , ξ ¯ 3 ) = i .
J z = ξ 3 z i ξ ˙ 3 z = + 1 2 ( ξ ¯ 3 ξ 3 ) 1 / 2 [ z + z ( 1 i 2 ϕ ˙ 3 ) ] ,
J z + = ξ ¯ 3 z + i ξ ¯ ˙ 3 z = 1 2 ( ξ ¯ 3 ξ 3 ) 1 / 2 [ z z ( 1 + i 2 ϕ ˙ 3 ) ] .
[ J z , J z + ] = I ,
J z Z n z ( z , t ) = n z Z n z 1 ( z , t ) ,
J z + Z n z ( z , t ) = n z + 1 Z n z + 1 .
i t Z n z = H Z n z = i [ i t Z n z Z n z ] Z n z [ Const ] Z n z .
J z Z 0 ( z , t ) = 0 .
Z 0 ( z , t ) = f ( t ) exp { z 2 2 [ 1 i ϕ ˙ 3 2 ] } ,
Z 0 ( z , t ) = ( π ϕ 3 ) 1 / 4 ( ξ ¯ 3 ξ 3 ) 1 / 4 exp { z 2 2 [ 1 i ϕ ˙ 3 2 ] } .
Z n z ( z , t ) = [ n z ! ] 1 / 2 [ J z + ] n z Z 0 ( z , t ) .
Z n z ( z , t ) = ( 2 n z n z ! ) 1 / 2 ( π ϕ 3 ) 1 / 4 ( ξ ¯ 3 ξ 3 ) 1 / 2 ( n z + 1 2 )
H n z ( z ) exp { z 2 2 [ 1 i ϕ ˙ 3 2 ] } .
X n x ( x , t ) = Z n z n x ( z x , ξ 3 ξ , ϕ 3 ϕ , t ) ,
Y n y ( y , t ) = Z n z n y ( z y , ξ 3 ξ , ϕ 3 ϕ , t ) ,
Ψ n x , n y , n z ( x , y , z , t ) = X n x ( x , t ) Y n y ( y , t ) Z n z ( z , t ) .
a = a + = 1 2 ( J x + i J y ) = 1 2 [ ξ ( x + i y ) i ξ ˙ ( x + i y ) ]
= 1 2 ( ξ ξ ¯ ) 1 / 2 e i θ [ ρ + i ρ θ + ρ ( 1 i 2 ϕ ˙ ) ] ,
c = c + = 1 2 ( J x i J y ) = 1 2 [ ξ ( x i y ) i ξ ˙ ( x i y ) ]
= 1 2 ( ξ ξ ¯ ) 1 / 2 e i θ [ ρ i ρ θ + ρ ( 1 i 2 ϕ ˙ ) ] .
κ = J x + J x + J y + J y + 1 = a + a + c + c + 1 ,
z = i ( y x x y ) = i θ ,
f 1 2 ( κ z ) = a + a + 1 2 ,
d 1 2 ( κ + z ) = c + c + 1 2 ,
f Ω n , m = ( n + 1 2 ) Ω n , m , d Ω n , m = ( m + 1 2 ) Ω n , m ,
a Ω n , m = n Ω n 1 , m , c Ω n , m = m Ω n , m 1 ,
a + Ω n , m = n + 1 Ω n + 1 , m , c + Ω n , m = m + 1 Ω n , m + 1 ,
z Ω n , m = ( f d ) Ω n , m = ( m n ) Ω n , m ,
κ Ω n , m = ( d + f ) Ω n , m = ( n + m + 1 ) Ω n , m ,
a Ω 0 , 0 = 0 , c Ω 0 , 0 = 0 .
Ω n , m , ( r , θ , t ) = 𝓡 n , m ( r , t ) Θ n , m ( θ ) = 𝓡 n , m ( r , t ) exp [ i ( m n ) θ ] 2 π .
Ω 0 , 0 ( r , θ , t ) = 1 2 π [ 2 ϕ ] 1 / 2 ( ξ ¯ ξ ) 1 2 exp { ρ 2 2 [ 1 i ϕ ˙ 2 ] } = X 0 ( x , t ) Y 0 ( y , t ) .
Ω n , m ( ρ , θ , t ) = ( n ! m ! ) 1 / 2 a + n c + m Ω 0 , 0 ( ρ , θ , t ) .
Ω n , m ( r , θ , t ) = exp [ i ( m n ) θ ] 2 π ( ) k k ! ( n ! m ! ) 1 / 2 ( 2 ϕ ( t ) ) 1 / 2 ( ξ ¯ ξ ) 1 2 ( n + m + 1 )
ρ n m L k ( n m ) ( ρ 2 ) exp { ρ 2 2 [ 1 i ϕ ˙ 2 ] } ,
k 1 2 ( n + m m n ) .
= z , z = m n , n r = m + n 0 .
Ω n , m ( r , t ) R n r , ( r , t ) Θ z ( θ ) ,
Θ z ( θ ) = exp [ i z θ ] 2 π ,
R n r , ( r , t ) = ( 1 ) ( n r ) / 2 [ 2 [ n r 2 ] ! ϕ ( t ) [ n r + 2 ] ! ] 1 / 2 ( ξ ¯ ξ ) 1 2 ( n + m + 1 )
ρ L 1 2 ( n r ) ( ) ( ρ 2 ) exp { ρ 2 2 [ 1 i ϕ ˙ 2 ] } .
Φ n r , z , n z ( x , y , z , t ) = R n r , ( r , t ) Θ z ( θ ) Z n z ( z , t ) .
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