An Exactly Solvable 3 Body Problem The most famous n-body problem is one where particles interact by an inverse square-law force. However, there is a class of exactly solvable n-body problems in which interactions are harmonic. In this section a Lie-algebraic method is illustrated by treating a solvable 3-body problem. Lagrange s equations of motion are so, defining canonical momentum d L L = 0 () dt ẋ i x i p i = L ẋ i, the Hamiltonian H(x, p, t) = p i ẋ i L i is a function of phase-space coordinates x and p. In general, one has 2n coordinates, x = {x,..., x n } and p = {p,..., p n }. This allows one to obtain Hamilton s Canonical Equations: ẋ i = H p i, ṗ i = H x i. (2) The time derivative of any function A( x, p ) of phase-space coordinates is da dt = n i= ( A ẋ i + A ) ṗ i x i p i So, following the sign convention of Perelomov [], one can define the Poisson bracket between any two functions of phase-space variables A and B [ A, B ] = ( A B B ) A. (3) p i i x i p i x i. So using Eqs. (2) and (3), da dt is the bracket of the Hamiltonian H with A da dt = [ H, A ].
If [ H, A ] = 0, then A is a constant of the motion, or a conserved quantity. The Poisson bracket is another example of a Lie bracket [ x, y ] which, as the reader will recall, has three main properties:. [x, y] = [y, x] -antisymmetry 2. [x + αy, z] = [x, z] + α[y, z] -linearity 3. [[x, y], z] + [[y, z], x] + [[z, x], y] = 0 -Jacobi identity As noted before, a Lie algebra is an algebra with a Lie bracket product. Mechanics concerns the Lie algebra A over C generated by functions of the phase space variables including the Hamiltonian function, with the Poisson bracket as the product. (So it is infinite dimensional.) The Poisson bracket has the following additional property: [x, yz] = y[x, z] + [x, y]z -Leibniz rule. The Three Body Problem: The three-body problem involves three particles, interacting with each other via harmonic forces. I choose these particles to have equal masses in order not to obscure the important points of the following discussion with inessential details. Thus, m = m 2 = m 3 = m, which have respective positions r, r 2, and r 3. (Here harmonic forces are forces associated with harmonic motion, where the Hamiltonian is of quadratic form.) There are three particles in this problem each with three degrees of freedom, corresponding to six phase space coordinates. Also, there is an external force similar to the force caused by a magnetic field acting on moving charges. (The actual magnetic force is not considered, since if the origin of the force were actually through electric charge, there would also be mutual electric repulsion of inverse square type, rendering the problem unsolvable.) So, the potential energy is V ( r) = k ( r α r β ) 2, 2 α β where k is like a spring constant. This potential leads to a homogeneous quadratic Hamiltonian, so that the motion is governed by a set of linear 2
homogenous differential equations with constant coefficients. This is what makes the problem exactly solvable. The Hamiltonian is H = 2m 3 p α q A c 2 + 2 α = α β k ( r α r β 2 ), (4) where again the symmetric vector potential A = Bo (xŷ yˆx) is chosen. The 2 total angular momentum L is ( ) L = ( r 3 + r 2 + r 3 ) ( p + p 2 + p 3 ). (5) In order to solve the problem, it is helpful, as usual, to identify the conserved quantities. Due to the magnetic field, the vector L is not conserved. However, since B is parallel to the z-axis, the problem is invariant with respect to rotation about this axis. Recall Nöther s theorem states that to each symmetry of the the Lagrangian corresponds a conserved quantity of the system. This guarantees that L z is conserved. Also, one finds that the z-components L αβz of the relative angular momentum vector of two particles α and β with respect to one another is conserved L αβ = ( r β r α ) ( p β p α ). (6) Finally, the z-component of the total linear momentum vector of the system is also conserved p z = (p z + p 2z + p 3z ). (7) Thus, in Poisson bracket notation [ H, L z ] = 0, [ H, L αβz ] = 0, [ H, p z ] = 0. (8) Consider first A, the finite dimensional sub-algebra of A generated by H and the 2n phase space variables. All of the elements such that the bracket between any two gives zero, form a sub Lie algebra of A which is the Cartan sub-algebra. In the three-body problem, the Cartan sub-algebra contains five elements L z, L 2z, p z,, and H. As usual, a linear combination of all the basis elements of the algebra A can be constructed such that it is an eigenvector of the Poisson bracket operation with each generator of the Cartan sub-algebra {K}, including the Hamiltonian. One defines the set of 8-dimensional phase-space coordinates as ξ = {ξ, ξ 2,..., ξ 8 } = {x, y, z, x 2,..., p 3y, p 3z }. 3
Thus, an eigenvector of adk (where k K) constructed from combinations of ξ is of the form 8 u = a k ξ k, k = where {a k } are complex coefficients to be determined. Thus, as before, treat the set of generators ξ = {ξ, ξ 2,..., ξ 8 } as a basis of independent vectors and compare coefficients. In this way, one determines what the coefficients must be in order to form, for example, an eigenvector of the eigenvalue equation [ H, u α ] = λ α u α. (9) In other words, the motivation here is to solve the differential equation u = λu, with elementary solutions that have exponential time dependence u(t) = u(0)e λt (since, again, this is a set of linear homogeneous DE s with constant coefficients). This differential equation describes the motion of u throughout time. Since u is a linear combination of the phase-space coordinates of the system, the solutions will allow one to find the position and momentum of each particle at any given time. Using the oscillator frequency ω o and cyclotron frequency ω b ω o = k m and making the substitution ω b = qb o 2mc, Ω = ω b2 + 3ω o2, there are two roots that appear in the solution to the eigenvalue equation, Eq.(9), namely r = ω b + Ω and r + = ω b + Ω. Therefore, solving the set of DE s, for the Hamiltonian alone, one obtains the following table of eigenvalues and eigenvectors: 4
λ eigenvector. 0 u = p z + p z2 + p z3 2. 0 u 2 = p y + p y2 + p y3 + mx ω b + mx 2 ω b + mx 3 ω b 3. 0 u 3 = p x + p x2 + p x3 my ω b my 2 ω b my 3 ω b 4. 0 u 4 = z + z 2 + z 3 (generalized) 5. 2iω b u 5 = p y + p y2 + p y3 i(p x + p x2 + p x3 ) mω b [(x + x 2 + x 3 ) + i(y + y 2 + y 3 )] 6. 2iω b u 6 = p y + p y2 + p y3 + i(p x + p x2 + p x3 ) mω b [(x + x 2 + x 3 ) i(y + y 2 + y 3 )] 7. i 3ω o u 7 = p z3 p z + i 3mω o (z z 3 ) 8. i 3ω o u 8 = p z2 p z + i 3mω o (z z 2 ) 9. i 3ω o u 9 = p z3 p z i 3mω o (z z 3 ) 0. i 3ω o u 0 = p z2 p z i 3mω o (z z 2 ). ir + u = p y3 p y i(p x3 p x ) + mω(x x 3 ) + imω(y y 3 ) 2. ir + u 2 = p y2 p y i(p x2 p x ) + mω(x x 2 ) + imω(y y 2 ) 3. ir + u 3 = p y3 p y + i(p x3 p x ) + mω(x x 3 ) imω(y y 3 ) 4. ir + u 4 = p y2 p y + i(p x2 p x ) + mω(x x 2 ) imω(y y 2 ) 5. ir u 5 = p y3 p y + i(p x3 p x ) mω(x x 3 ) + imω(y y 3 ) 6. ir u 6 = p y2 p y + i(p x2 p x ) mω(x x 2 ) + imω(y y 2 ) 7. ir u 7 = p y3 p y i(p x3 p x ) mω(x x 3 ) imω(y y 3 ) 8. ir u 8 = p y2 p y i(p x2 p x ) mω(x x 2 ) imω(y y 2 ) 5
Physically, the eigenvectors {u α } correspond to the normal mode coordinates of the three-mass system, providing those solutions to the motion that have exponential time dependence. Each eigenvector defines a mode of the system, the frequency of which is given by the corresponding eigenvalue λ = iω o. From the table, it is clear that in the 3-body problem, the eigenvectors appear in complex conjugate pairs. These pairs each correspond to two possible motions, both of which have the same frequency. Complete solutions to the equations of motion for this system involve combinations of the normal modes weighted with appropriate amplitude and phase factors [2]. A solution to the physical motion of the system is the real or imaginary part of the complex conjugate combination. From Table I (of solutions to Eq.(9)), it is clear that the problem decouples into xy and z parts. Thus, the Hamiltonian can be rewritten as H = ( (p 2 xα + p 2 2m yα) 2q ) c L zα + q2 c 2 (A2 xα + A 2 yα) + 2 α β α k ( x αˆx y α ŷ x β ˆx y β ŷ ) 2 + 2m p 2 zα + k( z α ẑ z β ẑ ) 2. 2 The motion along the z-axis is completely determined by eigenvectors u, u 4, u 7, u 8, u 9, and u 0. The eigenvector u corresponds to a translational mode along the z-axis, unaffected by the field. The eigenvectors u 7, u 8, u 9, and u 0 describe an oscillatory motion of two particles with respect to a third, confined to the z-axis moving with oscillator frequency ±ω o. The motion corresponding to the vector u 4 does not have an exponential solution. This vector is a generalized eigenvector of order greater than one. The occurrence of this generalized eigenvector means the entire Lie algebra fails to be semisimple. This adds unnecessary complication to the analysis of the problem. Thus, for simplicity, I consider only the xy portion in the following discussion. Motivated by the desire to construct raising and lowering operators for this system in the corresponding quantum mechanical problem (see section I), one looks for a set of eigenvectors that satisfy simultaneously the eigenvalue α α β 6
equations corresponding to all of the elements in the Cartan subalgebra [ H, u α ] = λ α u α, [ L z, u α ] = γ α u α, [ L 2z, u α ] = µ α u α. (0) In order to construct such a set of eigenvectors, start by obtaining solutions to the eigenvalue equation corresponding to L z by the same method as was employed for H. Then, collect the eigenvectors {u α } degenerate in both λ and γ. Finally, combine these degenerate eigenvectors into linear combinations v β = α a β α u α to form the desired set {v β } of vectors that satisfy the entire system. (The coefficient a may be zero.) In the process, one finds twelve ladder operators for the spectrum listed in Table II. 7
Table II λ γ µ operators. 0 i 0 v 2 = p y + p y2 + p y3 + mω b (x + x 2 + x 3 ) i(p x + p x2 + p x3 mω b (y + y 2 + y 3 )) 2. 0 i 0 v 3 = p y + p y2 + p y3 + mω b (x + x 2 + x 3 ) +i(p x + p x2 + p x3 mω b (y + y 2 + y 3 )) 3. 2iω b i 0 v 5 = i(p x + p x2 + p x3 ) + p y + p y2 + p y3 mω b ((x + x 2 + x 3 ) + i(y + y 2 + y 3 )) 4. 2iω b i 0 v 6 = i(p x + p x2 + p x3 ) + p y + p y2 + p y3 mω b ((x + x 2 + x 3 ) i(y + y 2 + y 3 )) 5. i(ω b + Ω) 0 0 v = i(p x + p x2 2p x3 ) + p y + p y2 2p y3 mω((x + x 2 2x 3 ) + i(y + y 2 2y 3 )) 6. i(ω b + Ω) 0 2i v 2 = i(p x p x2 ) p y + p y2 +mω((x x 2 ) + i(y y 2 )) 7. i(ω b + Ω) 0 0 v 3 = i(p x + p x2 2p x3 ) + p y + p y2 2p y3 mω((x + x 2 2x 3 ) i(y + y 2 2y 3 )) 8. i(ω b + Ω) 0 2i v 4 = i(p x p x2 ) p y + p y2 +mω((x x 2 ) i(y y 2 )) 9. i(ω ω b ) 0 0 v 5 = i(p x + p x2 2p x3 ) + p y + p y2 2p y3 +mω((x + x 2 2x 3 ) i(y + y 2 2y 3 )) 0. i(ω ω b ) 0 2i v 6 = i(p x p x2 ) p y + p y2 +mω((x 2 x ) + i(y y 2 )). i(ω ω b ) 0 0 v 7 = i(p x + p x2 2p x3 ) + p y + p y2 2p y3 +mω((x + x 2 2x 3 ) + i(y + y 2 2y 3 )) 2. i(ω ω b ) 0 2i v 8 = i(p x p x2 ) p y + p y2 +mω((x 2 x 2 ) i(y y 2 )) 8
Once one has the list of eigenvectors which satisfy by all of the bracket relations simultaneously, Eq.(0), the Hamiltonian can be rewritten in terms of these vectors. Classically, these eigenvectors are the elementary solutions of the system. The new eigenvectors of the total system appear in complex conjugate pairs v 2 = v v 4 = v3 v 7 = v5 v 8 = v6 v = v9 v 2 = v0 With this in mind, the Hamiltonian of the system becomes: H xy = 24mΩ 2 [ ω 2 b v 9 v 9 + 3ω 2 b v 0 v 0 + 4ω 2 b v 3 v 3 ω b Ωv 9 v 9 3ω b Ωv 0 v 0+ ω b (Ω + ω b )v 5 v 5 + 3ω b (Ω + ω b )v 6 v 6 + 3ω 2 o(v 5 v 5 + 3v 6 v 6 + v 9 v 9 + 3v 0 v 0 + 4v 3 v 3) ]. Converting each of the phase-space coordinates into combinations of the eigenvectors {v β }, one obtains Table III. As was the case with the raising and lowering operators in section I, these elementary solutions with exponential time dependence can be used to determine the motion of the system corresponding to each mode. (To make this a quantum mechanical problem, the Hamiltonian must be rewritten in terms of symmetrized combinations of the eigenvectors 2 α v αv α + v αv α.) In order to trace out this motion, begin by collecting the appropriate elementary solutions. For example when analyzing the mode corresponding to frequency λ = (ω b + Ω), one must form linear combinations of the elementary solutions v, v 2, v 3, and v 4. When combined properly, these elementary solutions lead to real functions, which constitute actual motion. So, including the proper time dependence, v α = c α e λαt (α =, 2,..., 8), one can graph the motion of each particle as a function of time. Therefore, the coefficients c α are the initial conditions of the system v(t) = v(0) e λ αt. In this way, the entire path of each particle is determined over all time. A 2-D parametric plot of the motion corresponding to each mode of the system is shown in Figure. From fig., it is clear that the solutions v 5 and v 6 represent a mode of the system in which all three particles move in a circle 9
about the center of mass. Analysis of initial conditions for this mode reveals that all three particles begin at the same position on the x or y axis, and move in unison around the circle. The mode corresponding to solutions v, v 2, v 3, and v 4 is slightly more complicated. There are a number of possible situations for motion in this mode, depending on the initial conditions given. The final oscillatory mode in the xy plane is determined by elementary solutions v 5, v 6, v 7, and v 8. From the figure, it is clear that this mode is characterized by the same motion as was displayed for the previous mode. The only difference is in the angular frequency of rotation and the direction of motion. The final mode corresponding to xy motion of the particles (not included in the figure) is the zero frequency mode given by solutions v 2 and v 3. Here, the particles are stationary relative to each other throughout time, only translating in the xy plane effected by the magnetic field. All possible motion of this system can be described by appropriate linear combinations of these normal mode solutions. Thus, by analyzing the elementary solutions that satisfy the entire Cartan subalgebra (Eq.(0)) simultaneously, one can solve this harmonic 3-body problem completely. 0
Table III x 24mω b Ω (ω b(v + 3v 2 + v 3 + 3v 4 v 5 3v 6 v 7 3v 8 ) +2Ω(v 2 + v 3 v 5 v 6 )) x 2 24mω b Ω (ω b(v 3v 2 + v 3 3v 4 v 5 + 3v 6 v 7 + 3v 8 ) +2Ω(v 2 + v 3 v 5 v 6 )) x 3 2mω b Ω (ω b( v v 3 + v 5 + v 7 ) +Ω(v 2 + v 3 v 5 v 6 )) y y 2 y 3 p x p x2 p x3 i (ω 24mω b Ω b(v + 3v 2 v 3 3v 4 + v 5 + 3v 6 v 7 3v 8 ) +2Ω(v 2 v 3 v 5 + v 6 )) i (ω 24mω b Ω b(v 3v 2 v 3 + 3v 4 + v 5 3v 6 v 7 + 3v 8 ) +2Ω(v 2 v 3 v 5 + v 6 )) i (ω 2mω b Ω b( v + v 3 v 5 + v 7 ) +Ω(v 2 v 3 v 5 + v 6 )) i 24 (v + 3v 2 v 3 3v 4 v 5 3v 6 + v 7 + 3v 8 2v 2 + 2v 3 2v 5 + 2v 6 ) i 24 (v 3v 2 v 3 + 3v 4 v 5 + 3v 6 + v 7 3v 8 2v 2 + 2v 3 2v 5 + 2v 6 ) i (v 2 v 3 v 5 + v 7 + v 2 v 3 + v 5 v 6 ) p y 24 ( v 3v 2 v 3 3v 4 v 5 3v 6 v 7 3v 8 + 2v 2 + 2v 3 + 2v 5 + 2v 6 ) p y2 24 ( v + 3v 2 v 3 + 3v 4 v 5 + 3v 6 v 7 + 3v 8 + 2v 2 + 2v 3 + 2v 5 + 2v 6 ) p y3 2 (v + v 3 + v 5 + v 7 + v 2 + v 3 + v 5 + v 6 )
y y x x (a) (b) y y x x (c) (d) Figure : 2D parametric plots of possible xy motion generated by combinations of oscillatory elementary solutions, which determine three separate modes of the system. Paths (a) - (c) correspond to orbits generated by solutions v - v 8 (double arrows indicate direction of motion), while path (d) represents the mode generated by v 5 and v 6. 2
References [] A. M. Perelomov. Integrable Systems Of Classical Mechanics and Lie Algebras, volume I. Birkhäuser, Boston, MA, 990. [2] Herbert Goldstein. Classical Mechanics. Addison Wesley, San Francisco, CA, 3rd edition, 2002. 3