Free particle on SU(2) group manifold

The dynamics of a free particle on SU(2) group manifold is described by the LagrangianL = 〈g^{− 1}ġg^{− 1}ġ〉where g ∈ SU(2) and 〈 〉 denotes the normalized trace〈 · 〉 = − ½Tr( · )which defines a scalar product in su(2) algebra. This Lagrangian gives rise to equations of motionddtg^{− 1}ġ = 0that describe dynamics of particle on group manifold.Also, one can notice that it has SU(2) "right" and SU(2) "left" symmetry.It means that it is invariant under the following transformationsg       →      h₁gg       →    gh₂where h₁, h₂ ∈ SU(2)

According to the Noether's theorem these symmetries lead to the matrix valued conserved quantitiesC = g^{− 1}ġ           ddtC = 0andS = ġg^{− 1}           ddtS = 0To construct integrals of motion out of C and S let us introduce the basis ofsu(2) algebra — three matrices:T₁ =i00− i     T₂ =0− 110     T₃ =0ii0The elements of su(2) are traceless anti-hermitian matrices, and anyA ∈ su(2) can be parameterized in the following wayA = AⁿT_n            n = 1, 2, 3Scalar product AB = 〈AB〉 = − ½Tr(AB) ensures thatAⁿ = 〈AT_n〉            (〈T_nT_m〉 = δ_nm)Now we can introduce six functionsC_n = 〈T_nC〉          n = 1, 2, 3           C = CⁿT_nS_n = 〈T_nS〉           n = 1, 2, 3            S = SⁿT_nwhich are integrals of motion.

Conservation of C and S leads to general solution of Euler-Lagrange equationsddtg^{− 1}ġ = 0           ⇒           g^{− 1}ġ = constg = e^Ctg(0)These are well known geodesics on Lie group.

Working in a first order Hamiltonian formalism we can construct new Lagrangianwhich is equivalent to the initial oneΛ = 〈C(g^{− 1}ġ − v)〉 + ½〈v²〉in sense that variation of C providesg^{− 1}ġ = vand Λ reduces to L.Variation of v gives C = v and therefore we can rewriteequivalent Lagrangian Λ in terms of C and g variablesΛ = 〈Cg^{− 1}ġ〉 − ½ 〈C²〉where functionH = ½〈C²〉plays the role of Hamiltonian andone-form 〈Cg− 1dg〉 is a symplectic potential θ.External differential of θ is the symplectic formω = dθ = − 〈g^{− 1}dg ∧ dC〉 − 〈Cg^{− 1} dg ∧ g^{− 1}dg〉that determines Poisson brackets, the form of Hamilton's equationand provides isomorphism between vector fields and one-formsX      →      i_XωFor any smooth SU(2) valued smooth functionf ∈ SU(2) one can define Hamiltonian vector field X_f byi_Xfω = − dfwhere i_Xω denotes the contraction of X with ω.According to its definition Poisson bracket of two functions is{f , g} = L_Xfg = i_Xfdg = ω(X_f , X_g)where L_Xfg denotes Lie derivative of g with respect to vector filed X_f.The skew symmetry of ω provides skew symmetry of Poisson bracket.

Hamiltonian vector fields that correspond to C_n, S_m and g functions areX_n = X_Cn = ([C ,T_n] , gT_n)Y_m = X_Sm = ([C , gT_mg^{− 1}] , T_mg )and give rise to the following commutation relations{S_n , S_m} = − 2ε_nm^k S_k{C_n , C_m} = 2ε_nm^k C_k{C_n , S_m} = 0{C_n , g} = gT_n{S_m , g} = T_mgThe results are natural. C and S that correspond respectively to the "right"and "left" symmetry commute with each other and independently form su(2)algebras. Now knowing Poisson bracket structure one can write down Hamilton's equationsġ = {H , g} = gRĊ = {H , C} = 0

Let's introduce operatorsĈ_n = i2L_XnŜ_m = − i2L_YmThey act on the square integrable functions (see Appendix A) on SU(2) and satisfy quantumcommutation relations[Ŝ_n , Ŝ_m] = iε_nm^k Ŝ_k[Ĉ_n , Ĉ_m] = iε_nm^k Ĉ_k[Ĉ_n , Ŝ_m] = 0The Hamiltonian is defined asĤ = Ĉ² = Ŝ²and the complete set of observables that commute with each other isĤ,           Ĉ_a,            Ŝ_bwith some fixed a and b. Using a simple generalization of a well known algebraic construction (see Appendix B)one can check that the eigenvalues of the quantum observablesĤ, Ĉ_a and Ŝ_b have the formĤψ_jsc = j(j + 1)ψ_jscwhere j takes positive integer and half integer valuesj = 0, 12, 1, 32, 2 ...Ĉ_aψ_jsc = cψ_jscŜ_bψ_jsc = sψ_jscwith c and s taking values in the following range− j, − j + 1, ... , j − 1, jFurther we construct the corresponding eigenfunctionsψ_jsc. The first step of this construction is to note thatthe function 〈Tg〉 where T = (1 + iT_a)(1 + iT_b)is an eigenfunction of Ĥ, Ĉ_a and Ŝ_bwith eigenvalues ¾, ½, ½ respectively.Proof of this proposition is straightforward.Using 〈Tg〉 one can construct the complete set of eigenfunctions ofĤ, Ĉ_a and Ŝ_b operatorsψ_jsc =Ŝ₋^{j − s}Ĉ₋^{j − c}〈Tg〉^2j in the manner described in Appendix B.

Free particle on 2D sphere can be obtained from our model by gauging U(1) symmetry.In other words let's consider the following local gauge transformationsg      →      h(t)gWhere h(t) ∈ U(1) ⊂ SU(2) is an element of U(1). Without loss of generality we can takeh = e^β(t)T₃Since T₃ is antihermitian h(t) ∈ U(1) and since h(t) depends on t LagrangianL = 〈g^{− 1}ġg^{− 1}ġ〉is not invariant under (38) local gauge transformations.

To make (40) gauge invariant we should replace time derivativewith covariant derivativeddtg     →    ∇g = (ddt + B)gwhere B can be represented as followsB = bT₃ ∈ su(2)with transformation ruleB      →     hBh^{− 1} − dhdth^{− 1}or in terms of b variableb    →     b − dβdtThe new LagrangianL_G = 〈g^{− 1}∇gg^{− 1}∇g〉is invariant under (38) local gauge transformations. But thisLagrangian as well as every gauge invariant Lagrangian is singular.It contains additional non-physical degrees of freedom. Toeliminate them we should eliminate B using Lagrange equations∂L_G∂B      →      b = − 〈ġg^{− 1}T₃〉put it back in (45) and rewrite last obtained Lagrangian in terms of gauge invariant variables.L_G = 〈(g^{− 1}ġ − S₃T₃)²〉It's obvious that the followingZ = g^{− 1}T₃g ∈ su(2)element of su(2) algebra is gauge invariant. Since Z ∈ su(2) it can be parameterized as followsZ = z^aT_awhere za are real functions on SU(2)z_a = 〈ZT_a〉

So we have three gauge invariant variables za (a = 1, 2, 3) but it's easy tocheck that only two of them are independent. Indeed〈Z²〉 = 〈g^{− 1}T₃gg^{− 1}T₃g〉 = 〈T₃²〉 = 1otherwise〈Z²〉 = 〈z^aT_az^bT_b〉 = z^az_a

So configuration space of SU(2)/U(1) coset model is sphere.By direct calculations one can check that after being rewritten in terms of gauge invariant variables L_Gtakes the formL_G = ¼〈Z^{− 1}ŻZ^{− 1}Ż〉This Lagrangian describes free particle on the sphere. Indeed,since Z = zaT_a it's easy to show thatL_G = ¼〈Z^{− 1}ŻZ^{− 1}Ż〉 =¼〈ZŻZŻ〉 = ½ż^aż_aSo SU(2)/U(1) coset model describes free particle on S2 manifold.

Working in a first order Hamiltonian formalism one can introduce equivalent LagrangianΛ_G = 〈C(g^{− 1}ġ − u)〉 + ½ 〈(u + g^{− 1}Bg)²〉variation of u providesC = u + g^{− 1}Bg u = C − g^{− 1}BgRewriting Λ_G in terms of C and g leads toΛ_G = 〈Cg^{− 1}ġ〉 − ½ 〈C²〉 − 〈BgCg^{− 1}〉 = 〈Cg^{− 1}ġ〉 − ½ 〈C²〉 − b〈gCg^{− 1}T₃〉 = 〈Cg^{− 1}ġ〉 − ½ 〈C²〉 − bS₃Due to the gauge invariance of Λ_G we obtain constrained Hamiltonian system,where 〈Cg− 1dg〉 is symplectic potential, functionH =½〈C²〉 plays the role of Hamiltonian andb is a Lagrange multiple leading to the first class constrainφ = 〈gCg^{− 1}T₃〉 = 〈ST₃〉 = S₃ = 0So coset model is equivalent to the initial one with (59) constrain.Using technique of the constrained quantization, instead ofquantizing coset model we can subject quantum model that corresponds to the free particle on SU(2),to the following operator constrainŜ₃|ψ〉 = 0Hilbert space of the initial system, that is linear span ofψ_jcs           j = 0, 12, 1, 32, 2, ...wave functions, reduces tothe linear span ofψ_jc0            j = 0, 1, 2, 3, ...wave functions. Indeed,Ŝ₃ψ_jcs = 0 implies s = 0, and if s = 0 then j is integer.Thus c takes − j, − j + 1, ..., j − 1, j integer values only.Wave functions ψ_jcs rewriten in terms of gauge invariantvariables up to a constant multiple should coincide with well knownspherical harmonicsψ_jc0 ∼ J_jcOne can check the followingψ_jc0 ∼ Ŝ₋^jĈ₋^{j − c} 〈Tg〉^2j ∼ Ĉ₋^{j − c} 〈T₊g^{− 1}T₃g〉^j ∼ Ĉ₋^{j − c}sin^jθe^ijθ ∼ Ĉ₋^{j − c}J_jj ∼J_jcThis is an example of using large initial model in quantization ofcoset model.

Scalar product in Hilbert space is defined as follows〈ψ₁|ψ₂〉 =∫SU(2)3∏a = 1〈g^{− 1}dgT_a〉(ψ₁)^†ψ₂It's easy to prove that under this scalar product operatorsĈ_n and Ŝ_m are hermitian.Indeed〈ψ₁|Ĉ_nψ₂〉 =∫SU(2)3∏a = 1〈g^{− 1}dgT_a〉(ψ₁)^†(i2L_Xnψ₂) = ∫SU(2)3∏a = 1〈g^{− 1}dgT_a〉(i2L_Xnψ₁)^†ψ₂Where integration by part has been used and the additional term coming from measure3∏a = 1 〈g^{− 1}dgT_a〉vanished sinceL_Xn〈g^{− 1}dgT_a〉 = 0For more transparency one can introduce the following parameterization ofSU(2). For any g ∈ SU(2).g = e^qaT_aThen the symplectic potential takes the form〈Cg^{− 1}dg〉 = C_adq^aand scalar product becomes〈ψ₁|ψ₂〉 =2π∫02π∫02π∫0d³q(ψ₁)^†ψ₂that coincides with (65) because ofdq_a = 〈g^{− 1}dg T_a〉

Without loss of generality we can takeĤ, Ŝ₃ andĈ₃ as a complete set of observables.Assuming that operators Ĥ, Ŝ₃ and Ĉ₃have at least one common eigenfunctionĤψ = EψĈ₃ψ = cψŜ₃ψ = sψit is easy to show that eigenvalues of Ĥ are non-negative E ≥ 0and conditionsE − c² ≥ 0E − s² ≥ 0are satisfied. Indeed, operators Ĉ and Ŝ are selfadjoint so〈ψ|Ĥ|ψ〉 = 〈ψ|Ĉ²|ψ〉 = 〈ψ|Ĉ_aĈ^a|ψ〉 =〈ψ|(Ĉ_a)^†Ĉ^a|ψ〉 =〈Ĉ_aψ|Ĉ^aψ〉 = ∥Ĉ_aψ∥ ≥ 0To prove (74) we shall considerĈ₁² + Ĉ₂² andŜ₁² + Ŝ₂² operators〈ψ|Ĉ₁² + Ĉ₂²|ψ〉 =∥Ĉ₁ ψ∥ + ∥Ĉ₂ ψ∥ ≥ 0and〈ψ|Ĉ₁² + Ĉ₂²|ψ〉 =〈ψ|Ĥ − Ĉ₃²|ψ〉 = (E − c²)〈ψ|ψ〉thus E − c2 ≥ 0.

Now let's introduce new operatorsĈ₊ = iĈ₁ + Ĉ₂            Ĉ₋ =iĈ₁ − Ĉ₂Ŝ₊ = iŜ₁ + Ŝ₂            Ŝ₋ =iŜ₁ − Ŝ₂These operators are not selfadjoint, but (Ĉ₋)^† = Ĉ₊ and(Ŝ₋)^† = Ŝ₊and they fulfill the following commutation relations[Ĉ_± , Ĉ₃] = ± Ĉ_±           [Ŝ_± , Ŝ₃] = ± Ŝ_±[Ĉ₊ , Ĉ₋] = 2Ĉ₃           [Ŝ₊ , Ŝ₋] = 2Ŝ₃[Ĉ_• , Ŝ_•] = 0where • takes values +, −, 3 using these commutation relations it is easy to showthat if ψ_λcs is eigenfunction ofĤ, Ŝ₃ andĈ₃ with corresponding eigenvalues :Ĥψ_λcs = λψ_λcsŜ₃ψ_λcs = sψ_λcsĈ₃ψ_λcs = cψ_λcsthen Ĉ_±ψ_λcs andŜ_±ψ_λcsare the eigenfunctions with corresponding eigenvaluesλ, s ± 1, c and λ , s, c ± 1.Consequently using Ĉ_±, Ŝ_± operators one can constructa family of eigenfunctions with eigenvaluesc, c ± 1, c ± 2, c ± 3, ...s, s ± 1, s ± 2, s ± 3, ...but conditions (74) give restrictions on a possible range of eigenvalues.Namely we must haveλ − c² ≥ 0λ − s² ≥ 0In other words, in order to interrupt (84) sequences we must assumeŜ₊ ψ_λcj = 0            Ŝ₋ψ_{λc, − j} = 0Ĉ₊ψ_λks = 0           Ĉ₋ψ_{λ, − ks} = 0for some j and k, therefore s and c could take only the following values− j, − j + 1, ... , j − 1, j− k, − k + 1, ... , k − 1, kThe number of values is 2j + 1 and 2k + 1 respectively. Since number of valuesshould be integer, j and k should take integer or half integer valuesj = 0, 12, 1, 32, 2, ...k = 0, 12, 1, 32, 2, ...Now using commutation relations we can rewrite Ĥ in terms ofĈ_±, Ĉ₃ operatorsĤ = Ĉ₊ Ĉ₋ + Ĉ₃² + Ĉ₃and it is clear that j = k and λ = j(j + 1) = k(k + 1)