Kleinert - Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets - ed.4 - 2006 (523104), страница 21
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First, thecorrect Hamiltonian operator (1.416) does not agree with the canonically quantizedone which would be obtained by inserting Eqs. (1.407) into (1.406). The correctresult would, however, arise by distributing dummy factorsg −1/4 = r −1 sin−1/2 θ,g 1/4 = r sin1/2 θ(1.420)between the canonical momentum operators as observed earlier in Eq. (1.395). Second, just as in the case of polar coordinates, the correct Hamiltonian operator isequal toh̄2Ĥ = −∆,(1.421)2Mwhere ∆ is the Laplace-Beltrami operator associated with the metricgµν = ri.e.,∆=1.15100 sin2 θ2!,(1.422)111∂θ (sin θ∂θ ) +∂2 .2r sin θsin2 θ ϕ(1.423)Spinning TopFor a spinning top, the optimal starting point is again not the classical Lagrangianbut the Hamiltonian expressed in terms of the classical angular momenta. In thesymmetric case in which two moments of inertia coincide, it is written asH=11(Lξ 2 + Lη 2 ) +L 2,2Iξ2Iζ ζ(1.424)where Lξ , Lη , Lζ are the components of the orbital angular momentum in the directions of the principal body axes with Iξ , Iη ≡ Iξ , Iζ being the corresponding momentsof inertia.
The classical angular momentum of an aggregate of mass points is givenbyXL=xν × pν ,(1.425)νwhere the sum over ν runs over all mass points. The angular momentum possessesa unique operatorXL̂ =x̂ν × p̂ν ,(1.426)νwith the commutation rules (1.411) between the components L̂i . Since rotationsdo not change the distances between the mass points, they commute with the constraints of the rigid body. If the center of mass of the rigid body is placed at theorigin, the only dynamical degrees of freedom are the orientations in space. Theycan uniquely be specified by the rotation matrix which brings the body from somestandard orientation to the actual one.
We may choose the standard orientationH. Kleinert, PATH INTEGRALS611.15 Spinning Topto have the principal body axes aligned with the x, y, z-directions, respectively. Anarbitrary orientation is obtained by applying all finite rotations to each point of thebody. They are specified by the 3 × 3 orthonormal matrices Rij . The space of thesematrices has three degrees of freedom. It can be decomposed, omitting the matrixindices asR(α, β, γ) = R3 (α)R2 (β)R3 (γ),(1.427)where R3 (α), R3 (γ) are rotations around the z-axis by angles α, γ, respectively,and R2 (β) is a rotation around the y-axis by β. These rotation matrices can beexpressed as exponentialsRi (δ) ≡ e−iδLi /h̄ ,(1.428)where δ is the rotation angle and Li are the 3 × 3 matrix generators of the rotationswith the elements(Li )jk = −ih̄ijk .(1.429)It is easy to check that these generators satisfy the commutation rules (1.411) ofangular momentum operators.
The angles α, β, γ are referred to as Euler angles.The 3 × 3 rotation matrices make it possible to express the infinitesimal rotations around the three coordinate axes as differential operators of the three Eulerangles. Let ψ(R) be the wave function of the spinning top describing the probabilityamplitude of the different orientations which arise from a standard orientation bythe rotation matrix R = R(α, β, γ). Under a further rotation by R(α0 , β 0 , γ 0 ), thewave function goes over into ψ 0 (R) = ψ(R−1 (α0 , β 0, γ 0 )R).
The transformation maybe described by a unitary differential operator000Û (α0 , β 0, γ 0 ) ≡ e−iα L̂3 e−iβ L̂2 e−iγ L̂3 ,(1.430)where L̂i is the representation of the generators in terms of differential operators.To calculate these we note that the 3 × 3 -matrix R−1 (α, β, γ) has the followingderivatives−ih̄∂α R−1 = R−1 L3 ,−ih̄∂β R−1 = R−1 (cos α L2 − sin α L1 ),−ih̄∂γ R−1= R−1(1.431)[cos β L3 + sin β(cos α L1 + sin α L2 )] .The first relation is trivial, the second follows from the rotation of the generatore−iαL3 /h̄ L2 eiαL3 /h̄ = cos α L2 − sin α L1 ,(1.432)which is a consequence of Lie’s expansion formulae−iA BeiA = 1 − i[A, B] +i2[A, [A, B]] + .
. . ,2!(1.433)together with the commutation rules (1.429) of the 3 × 3 matrices Li . The thirdrequires, in addition, the rotatione−iβL2 /h̄ L3 eiβL2 /h̄ = cos βL3 + sin βL1 .(1.434)621 FundamentalsInverting the relations (1.431), we find the differential operators generating therotations [7]:L̂1L̂2!cos α= ih̄ cos α cot β ∂α + sin α ∂β −∂ ,sin β γ!sin α∂ ,= ih̄ sin α cot β ∂α − cos α ∂β −sin β γ(1.435)L̂3 = −ih̄∂α .After exponentiating these differential operators we deriveÛ (α0 , β 0, γ 0 )R−1 Û −1 (α0 , β 0, γ 0 )(α, β, γ) = R−1 (α, β, γ)R(α0, β 0 , γ 0 ),Û (α0 , β 0 , γ 0 )R(α, β, γ)Û −1(α0 , β 0 , γ 0 ) = R−1 (α0 , β 0, γ 0 )R(α, β, γ), (1.436)so that Û (α0 , β 0 , γ 0 )ψ(R) = ψ 0 (R), as desired.In the Hamiltonian (1.424), we need the components of L̂ along the body axes.They are obtained by rotating the 3 × 3 matrices Li by R(α, β, γ) intoLξ = RL1 R−1 = cos γ cos β(cos α L1 + sin α L2 )+ sin γ(cos α L2 − sin α L1 ) − cos γ sin β L3 ,Lη = RL2 R−1 = − sin γ cos β(cos α L1 + sin α L2 )+ cos γ(cos α L2 − sin α L1 ) + sin γ sin β L3 ,(1.437)Lζ = RL3 R−1 = cos β L3 + sin β(cos α L1 + sin α L2 ),and replacing Li → L̂i in the final expressions.
Inserting (1.435), we find the operatorsL̂ξL̂η!cos γ∂ ,= ih̄ − cos γ cot β ∂γ − sin γ ∂β +sin β α)sin γ= ih̄ sin γ cot β ∂γ − cos γ ∂β −∂ ,sin β α(1.438)L̂ζ = −ih̄∂γ .Note that these commutation rules have an opposite sign with respect to those inEqs.
(1.411) of the operators L̂i :18[L̂ξ , L̂η ] = −ih̄L̂ζ ,ξ, η, ζ = cyclic.(1.439)The sign is most simply understood by writingL̂ξ = aiξ L̂i ,L̂η = aiη L̂i ,L̂ζ = aiζ L̂i ,(1.440)18When applied to functions not depending on α, then, after replacing β → θ and γ → ϕ, theoperators agree with those in (1.413), up to the sign of L̂1 .H. Kleinert, PATH INTEGRALS631.15 Spinning Topwhere aiξ , aiη , aiζ , are the components of the body axes. Under rotations these behavelike [L̂i , ajξ ] = ih̄ijk akξ , i.e., they are vector operators.
It is easy to check that thisproperty produces the sign reversal in (1.439) with respect to (1.411).The correspondence principle is now applied to the Hamiltonian in Eq. (1.424)by placing operator hats on the La ’s. The energy spectrum and the wave functionscan then be obtained by using only the group commutators between L̂ξ , L̂η , L̂ζ . Thespectrum is! #"1112L(L + 1) +−Λ2 ,(1.441)ELΛ = h̄2Iξ2Iζ2Iξwhere L(L + 1) with L = 0, 1, 2, . . .
are the eigenvalues of L̂2 , and Λ = −L, . . . , Lare the eigenvalues of L̂ζ . The wave functions are the representation functions ofthe rotation group. If the Euler angles α, β, γ are used to specify the body axes, thewave functions areLψLΛm (α, β, γ) = DmΛ(−α, −β, −γ).(1.442)LHere m0 are the eigenvalues of L̂3 , the magnetic quantum numbers, and DmΛ(α, β, γ)are the representation matrices of angular momentum L. In accordance with (1.430),one may decompose0−i(mα+m γ) LLdmm0 (β),Dmm0 (α, β, γ) = e(1.443)with the matricesdLmm0 (β)(L + m0 )!(L − m0 )!=(L + m)!(L − m)!"×βcos2!m+m0#1/2β− sin2!m−m0(m0 −m,m0 +m)PL−m0(cos β).(1.444)For j = 1/2, these form the spinor representation of the rotations around the y-axis1/2dm0 m (β)=cos β/2 − sin β/2sin β/2cos β/2!.(1.445)The indices have the order +1/2, −1/2.
The full spinor representation functionD 1/2 (α, β, γ) in (1.443) is most easily obtained by inserting into the general expression (1.430) the representation matrices of spin 1/2 for the generators L̂i with thecommutation rules (1.411) the famous Pauli spin matrices:σ1 =0 11 0!, σ2 =0 −ii 0!, σ3 =100 −1!.(1.446)Thus we can writeD 1/2 (α, β, γ) = e−iασ3 /2 e−iβσ2 /2 e−iγσ3 /2 .(1.447)641 FundamentalsThe first and the third factor yield the pure phase factors in (1.443). The function21/2dm0 m (β) is obtained by a simple power series expansion of e−iβσ /2 , using the factthat (σ 2 )2n = 1 and (σ 2 )2n+1 = σ 2 :e−iβσ2 /2= cos β/2 − i sin β/2 σ 2,(1.448)which is equal to (1.445).For j = 1, the representation functions (1.444) form the vector representation1(1 + cos β)2√1 sin β21(1 − cos β)2d1m0 m (β) = − √12 sin β 12 (1 − cos β).cos β− √12 sin β11√ sin β(1+cosβ)22(1.449)where the indices have the order +1/2, −1/2.
The vector representation goes overinto the ordinary rotation matrices Rij (β) by mapping the states |1mi onto thespherical unit vectors (0) = ẑ, (±1) = ∓(x̂ ± iŷ)/2 using the matrix elementsPhi|1mi = i (m). Hence R(β) (m) = 1m0 =−1 (m0 )d1m0 m (β).The representation functions D 1 (α, β, γ) can also be obtained by inserting intothe general exponential (1.430) the representation matrices of spin 1 for the generators L̂i with the commutation rules (1.411). In Cartesian coordinates, these aresimply (L̂i )jk = −iijk , where ijk is the completely antisymmetric tensor with123 = 1. In the spherical basis, these become (L̂i )mm0 = hm|ii(L̂i )ij hj|m0 i =∗i (m)(L̂i )ij j (m0 ). The exponential (e−iβ L̂2 )mm0 is equal to (1.449).(α,β)The functions Pl(z) are the Jacobi polynomials [8], which can be expressedin terms of hypergeometric functions as(α,β)Pl≡(−1)l Γ(l + β + 1)F (−l, l + 1 + α + β; 1 + β; (1 + z)/2),l!Γ(β + 1)(1.450)a(a + 1) b(b + 1) z 2abF (a, b; c; z) ≡ 1 + z ++ ...
.cc(c + 1)2!(1.451)whereThe rotation functions dLmm0 (β) satisfy the differential equation2m2 + m0 − 2mm0 cos β Ld2d−+dmm0 (β) = L(L + 1)dLmm0 (β). (1.452)−cotβdβdβ 2sin2 βThe scalar products of two wave functions have to be calculated with a measure ofintegration which is invariant under rotations:hψ2 |ψ1 i ≡Z2π0Z0πZ02πdαdβ sin βdγ ψ2∗ (α, β, γ)ψ1 (α, β, γ).(1.453)The above eigenstates (1.443) satisfy the orthogonality relationZ02πZ0πZ02πL1 ∗L2dαdβ sin βdγ Dm0 m (α, β, γ)Dm0 m (α, β, γ)211= δm01 m02 δm1 m2 δL1 L228π 2.2L1 + 1(1.454)H. Kleinert, PATH INTEGRALS651.15 Spinning TopLet us also contrast in this example the correct quantization via the commutationrules between group generators with the canonical approach which would start outwith the classical Lagrangian.
In terms of Euler angles, the Lagrangian reads1L = [Iξ (ωξ 2 + ωη 2 ) + Iζ ωζ 2 ],2(1.455)where ωξ , ωη , ωζ are the angular velocities measured along the principal axes of thetop. To find these we note that the components in the rest system ω1 , ω2 , ω3 areobtained from the relationωk Lk = iṘR−1(1.456)asω1 = −β̇ sin α + γ̇ sin β cos α,ω2 = β̇ cos α + γ̇ sin β sin α,ω3 = γ̇ cos β + α̇.(1.457)After the rotation (1.437) into the body-fixed system, these becomeωξ = β̇ sin γ − α̇ sin β cos γ,ωη = β̇ cos γ + α̇ sin β sin γ,ωζ = α̇ cos β + γ̇.(1.458)Explicitly, the Lagrangian is1L = [Iξ (β̇ 2 + α̇2 sin2 β) + Iζ (α̇ cos β + γ̇)2 ].2(1.459)Considering α, β, γ as Lagrange coordinates q µ with µ = 1, 2, 3, this can be writtenin the form (1.384) with the Hessian metric [recall (1.12) and (1.385)]:gµνwhose determinant isIξ sin2 β + Iζ cos2 β 0 Iζ cos β0Iξ0=,Iζ cos β0Iζ√g = Iξ2 Iζ sin2 β.(1.460)(1.461)Hence the measure d3 q g in the scalar product (1.381) agrees with the rotationinvariant measure (1.453) up to a trivial constant factor.
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