# Free particle on SU(2) group manifold

Department of Theoretical Physics,A. Razmadze Institute of Mathematics,1 Aleksidze Street, Tbilisi 0193, Georgia
abstract. In the present paper classical and quantum dynamics of a free particle on $SU(2)$ group manifold is considered.Poisson structure of the dynamical system and commutation relations for generalized momenta arederived. Quantization is carried out and the eigenfunctions of the Hamiltonianare constructed in terms of coordinate free objects.$SU(2)/U(1)$ coset model yielding after Hamiltonian reduction free particle on $S2$ sphere is consideredand Hamiltonian reduction of coset model is carried out on both classical and quantum level.
keywords. Dynamics on group manifold; Quantization on group manifold;
msc. 70H33; 70H06; 53Z05

## Lagrangian description

The dynamics of a free particle on $SU(2)$ group manifold is described by the LagrangianL = 〈g− 1ġg− 1ġ〉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− 1ġ = 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 $h$1, h2 ∈ SU(2)
According to the Noether's theorem these symmetries lead to the matrix valued conserved quantitiesC = g− 1ġ           ddtC = 0andS = ġg− 1           ddtS = 0To construct integrals of motion out of $C$ and $S$ let us introduce the basis of$su(2)$ algebra — three matrices:T1 =i00− i     T2 =0− 110     T3 =0ii0The elements of $su(2)$ are traceless anti-hermitian matrices, and any$A ∈ su(2)$ can be parameterized in the following wayA = AnTn            n = 1, 2, 3Scalar product AB = 〈AB〉 = − ½Tr(AB) ensures thatAn = 〈ATn〉            (〈TnTm〉 = δnm)Now we can introduce six functionsCn = 〈TnC〉          n = 1, 2, 3           C = CnTnSn = 〈TnS〉           n = 1, 2, 3            S = SnTnwhich are integrals of motion.
Conservation of $C$ and $S$ leads to general solution of Euler-Lagrange equationsddtg− 1ġ = 0           ⇒           g− 1ġ = constg = eCtg(0)These are well known geodesics on Lie group.

## Hamiltonian description

Working in a first order Hamiltonian formalism we can construct new Lagrangianwhich is equivalent to the initial oneΛ = 〈C(g− 1ġ − v)〉 + ½〈v2in sense that variation of C providesg− 1ġ = 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− 1ġ〉 − ½ 〈C2where functionH = ½〈C2plays the role of Hamiltonian andone-form $〈Cg− 1dg〉$ is a symplectic potential $θ$.External differential of $θ$ is the symplectic formω = dθ = − 〈g− 1dg ∧ dC〉 − 〈Cg− 1 dg ∧ g− 1dg〉that determines Poisson brackets, the form of Hamilton's equationand provides isomorphism between vector fields and one-formsX      →      iXωFor any smooth $SU(2)$ valued smooth function$f ∈ SU(2)$ one can define Hamiltonian vector field $X$f byiXfω = − dfwhere $i$Xω denotes the contraction of $X$ with $ω$.According to its definition Poisson bracket of two functions is{f , g} = LXfg = iXfdg = ω(Xf , Xg)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, Sm and $g$ functions areXn = XCn = ([C ,Tn] , gTn)Ym = XSm = ([C , gTmg− 1] , Tmg )and give rise to the following commutation relations{Sn , Sm} = − 2εnmk Sk{Cn , Cm} = 2εnmk Ck{Cn , Sm} = 0{Cn , g} = gTn{Sm , g} = TmgThe 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

## Quantization

Let's introduce operatorsĈn = i2LXnŜm = − i2LYmThey act on the square integrable functions (see Appendix A) on $SU(2)$ and satisfy quantumcommutation relationsn , Ŝm] = iεnmk Ŝkn , Ĉm] = iεnmk Ĉkn , Ŝm] = 0The Hamiltonian is defined asĤ = Ĉ2 = Ŝ2and the complete set of observables that commute with each other isĤ,           Ĉa,            Ŝbwith 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 observables$Ĥ, Ĉ$a and $Ŝ$b have the formĤψjsc = j(j + 1)ψjscwhere $j$ takes positive integer and half integer valuesj = 0, 12, 1, 32, 2 ...Ĉaψjsc = cψjscŜbψjsc = sψjscwith $c$ and $s$ taking values in the following range− j, − j + 1, ... , j − 1, jFurther we construct the corresponding eigenfunctions$ψ$jsc. The first step of this construction is to note thatthe function $〈Tg〉$ where $T = (1 + iT$a)(1 + iTb)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 of$Ĥ, Ĉ$a and $Ŝ$b operatorsψjscj − sĈj − c〈Tg〉2j in the manner described in Appendix B.

## Free particle on S² as a SU(2)/U(1) coset model

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 = eβ(t)T3Since $T$3 is antihermitian $h(t) ∈ U(1)$ and since $h(t)$ depends on $t$ LagrangianL = 〈g− 1ġg− 1ġ〉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− 1dhdth− 1or in terms of $b$ variableb    →     b − dtThe new LagrangianLG = 〈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∂LG∂B      →      b = − 〈ġg− 1T3put it back in (45) and rewrite last obtained Lagrangian in terms of gauge invariant variables.LG = 〈(g− 1ġ − S3T3)2It's obvious that the followingZ = g− 1T3g ∈ su(2)element of $su(2)$ algebra is gauge invariant. Since $Z ∈ su(2)$ it can be parameterized as followsZ = zaTawhere $za$ are real functions on $SU(2)$za = 〈ZTa
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− 1T3gg− 1T3g〉 = 〈T32〉 = 1otherwise〈Z2〉 = 〈zaTazbTb〉 = zaza
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 formLG = ¼〈Z− 1ŻZ− 1Ż〉This Lagrangian describes free particle on the sphere. Indeed,since $Z = zaT$a it's easy to show thatLG = ¼〈Z− 1ŻZ− 1Ż〉 =¼〈ZŻZŻ〉 = ½żażaSo $SU(2)/U(1)$ coset model describes free particle on $S2$ manifold.

## Quantization of the coset model.

Working in a first order Hamiltonian formalism one can introduce equivalent LagrangianΛG = 〈C(g− 1ġ − u)〉 + ½ 〈(u + g− 1Bg)2variation of $u$ providesC = u + g− 1Bg u = C − g− 1BgRewriting $Λ$G in terms of $C$ and $g$ leads toΛG = 〈Cg− 1ġ〉 − ½ 〈C2〉 − 〈BgCg− 1〉 = 〈Cg− 1ġ〉 − ½ 〈C2〉 − b〈gCg− 1T3〉 = 〈Cg− 1ġ〉 − 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 and$b$ is a Lagrange multiple leading to the first class constrainφ = 〈gCg− 1T3〉 = 〈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 ofψjcs           j = 0, 12, 1, 32, 2, ...wave functions, reduces tothe linear span ofψjc0            j = 0, 1, 2, 3, ...wave functions. Indeed,$Ŝ$3ψjcs = 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 harmonicsψjc0 ∼ JjcOne can check the followingψjc0 ∼ ŜjĈj − c 〈Tg〉2j ∼ Ĉj − c 〈T+g− 1T3g〉j ∼ Ĉj − csinjθeijθ ∼ Ĉj − cJjj ∼JjcThis is an example of using large initial model in quantization ofcoset model.

## Appendix A

Scalar product in Hilbert space is defined as follows〈ψ12〉 =SU(2)3a = 1〈g− 1dgTa〉(ψ1)ψ2It's easy to prove that under this scalar product operators$Ĉ$n and $Ŝ$m are hermitian.Indeed〈ψ1nψ2〉 =SU(2)3a = 1〈g− 1dgTa〉(ψ1)(i2LXnψ2) = SU(2)3a = 1〈g− 1dgTa〉(i2LXnψ1)ψ2Where integration by part has been used and the additional term coming from measure3a = 1 〈g− 1dgTavanished sinceLXn〈g− 1dgTa〉 = 0For more transparency one can introduce the following parameterization of$SU(2)$. For any $g ∈ SU(2)$.g = eqaTaThen the symplectic potential takes the form〈Cg− 1dg〉 = Cadqaand scalar product becomes〈ψ12〉 =000d3q(ψ1)ψ2that coincides with (65) because ofdqa = 〈g− 1dg Ta

## Appendix B

Without loss of generality we can take$Ĥ, Ŝ$3 and$Ĉ$3 as a complete set of observables.Assuming that operators $Ĥ, Ŝ$3 and $Ĉ$3have at least one common eigenfunctionĤψ = EψĈ3ψ = cψŜ3ψ = sψit is easy to show that eigenvalues of $Ĥ$ are non-negative $E ≥ 0$and conditionsE − c2 ≥ 0E − s2 ≥ 0are satisfied. Indeed, operators $Ĉ$ and $Ŝ$ are selfadjoint so〈ψ|Ĥ|ψ〉 = 〈ψ|Ĉ2|ψ〉 = 〈ψ|ĈaĈa|ψ〉 =〈ψ|(Ĉa)Ĉa|ψ〉 =〈Ĉaψ|Ĉaψ〉 = ∥Ĉaψ∥ ≥ 0To prove (74) we shall consider$Ĉ$12 + Ĉ22 and$Ŝ$12 + 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± , Ĉ3] = ± Ĉ±           [Ŝ± , Ŝ3] = ± Ŝ±+ , Ĉ] = 2Ĉ3           [Ŝ+ , Ŝ] = 2Ŝ3 , Ŝ] = 0where $•$ takes values $+, −, 3$ using these commutation relations it is easy to showthat if $ψ$λcs is eigenfunction of$Ĥ, Ŝ$3 and$Ĉ$3 with corresponding eigenvalues :Ĥψλcs = λψλcsŜ3ψλcs = sψλcsĈ3ψλcs = cψλcsthen $Ĉ$±ψλcs and$Ŝ$±ψλcsare the eigenfunctions with corresponding eigenvalues$λ, s ± 1, c$ 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+ ψλcj = 0            Ŝψλc, − j = 0Ĉ+ψλks = 0           Ĉψλ, − 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 of$Ĉ$±, Ĉ3 operatorsĤ = Ĉ+ Ĉ + Ĉ32 + Ĉ3and it is clear that $j = k$ and $λ = j(j + 1) = k(k + 1)$

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