Free particle on SU(2) group manifold

The dynamics of a free particle on $SU\left(2\right)$ group manifold is described by the Lagrangian $$L=\text{〈}{g}^{-1}ġg^{-1}ġ\text{〉}$$ where $g\in SU\left(2\right)$ and $\text{〈}\text{〉}$ denotes the normalized trace $$\text{〈}·\text{〉}=-\mathrm{½}Tr(·)$$ which defines a scalar product in $su\left(2\right)$ algebra. This Lagrangian gives rise to equations of motion $$\frac{d}{dt}{g}^{-1}ġ=0$$ that describe dynamics of particle on group manifold. Also, one can notice that it has $SU\left(2\right)$ "right" and $SU\left(2\right)$ "left" symmetry. It means that it is invariant under the following transformations $$\begin{gathered} g\to h_{1}g \\ g\to g{h}_2 \end{gathered}$$ where $h_1,h_2\in SU\left(2\right)$

According to the Noether's theorem these symmetries lead to the matrix valued conserved quantities $$C=g^{-1}ġ\frac{d}{dt}C=0$$ and $$S=ġg^{-1}\frac{d}{dt}S=0$$ To construct integrals of motion out of $C$ and $S$ let us introduce the basis of $su\left(2\right)$ algebra — three matrices: $$T_1=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)T_2=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)T_3=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$$ The elements of $su\left(2\right)$ are traceless anti-hermitian matrices, and any $A\in su\left(2\right)$ can be parameterized in the following way $$A=A^n{T}_{n}n=1,2,3$$ Scalar product $$AB=\text{〈}AB\text{〉}=-\mathrm{½}Tr\left(AB\right)$$ ensures that $$A^n=\text{〈}A{T}_n\text{〉}(\text{〈}{T}_n{T}_m\text{〉}={\delta }_{nm})$$ Now we can introduce six functions $$\begin{gathered} C_n=\text{〈}{T}_{n}C\text{〉}n=1,2,3C=C^n{T}_n \\ {S}_n=\text{〈}{T}_{n}S\text{〉}n=1,2,3S=S^n{T}_n \end{gathered}$$ which are integrals of motion.

Conservation of $C$ and $S$ leads to general solution of Euler-Lagrange equations $$\begin{gathered} \frac{d}{dt}{g}^{-1}ġ=0\Rightarrow g^{-1}ġ=const \\ g=e^{Ct}g\left(0\right) \end{gathered}$$ These are well known geodesics on Lie group.

Working in a first order Hamiltonian formalism we can construct new Lagrangian which is equivalent to the initial one $$\Lambda =\text{〈}C(g^{-1}ġ-v)\text{〉}+\mathrm{½}\text{〈}{v}^2\text{〉}$$ in sense that variation of C provides $$g^{-1}ġ=v$$ and $\Lambda$ reduces to $L$. Variation of $v$ gives $C=v$ and therefore we can rewrite equivalent Lagrangian $\Lambda$ in terms of C and g variables $$\Lambda =\text{〈}C{g}^{-1}ġ\text{〉}-\mathrm{½}\text{〈}{C}^2\text{〉}$$ where function $$H=\mathrm{½}\text{〈}{C}^2\text{〉}$$ plays the role of Hamiltonian and one-form $\text{〈}C{g}^{-1}dg\text{〉}$ is a symplectic potential $\theta$. External differential of $\theta$ is the symplectic form $$\omega =d\theta =-\text{〈}{g}^{-1}dg\wedge dC\text{〉}-\text{〈}C{g}^{-1}dg\wedge g^{-1}dg\text{〉}$$ that determines Poisson brackets, the form of Hamilton's equation and provides isomorphism between vector fields and one-forms $$X\to i_X\omega$$ For any smooth $SU\left(2\right)$ valued smooth function $f\in SU\left(2\right)$ one can define Hamiltonian vector field $X_f$ by $$i_{X_f}\omega =-df$$ where $i_X\omega$ denotes the contraction of $X$ with $\omega$. According to its definition Poisson bracket of two functions is $$\{f,g\}=L_{X_f}g=i_{X_f}dg=\omega (X_f,X_g)$$ where $L_{X_f}g$ denotes Lie derivative of $g$ with respect to vector filed $X_f$. The skew symmetry of $\omega$ provides skew symmetry of Poisson bracket.

Hamiltonian vector fields that correspond to $C_n,S_m$ and $g$ functions are $$\begin{gathered} X_n=X_{C_n}=\left(\right[C,T_n],g{T}_n) \\ {Y}_m=X_{S_m}=\left(\right[C,g{T}_m{g}^{-1}],T_{m}g) \end{gathered}$$ and give rise to the following commutation relations $$\begin{gathered} \{S_n,S_m\}=-2{\epsilon }_{nm}^k{S}_k \\ \{C_n,C_m\}=2{\epsilon }_{nm}^k{C}_k \\ \{C_n,S_m\}=0 \\ \{C_n,g\}=g{T}_n \\ \{S_m,g\}=T_{m}g \end{gathered}$$ The results are natural. $C$ and $S$ that correspond respectively to the "right" and "left" symmetry commute with each other and independently form $su\left(2\right)$ algebras. Now knowing Poisson bracket structure one can write down Hamilton's equations $$ġ=\{H,g\}=gR$$ $$Ċ=\{H,C\}=0$$

Let's introduce operators $${Ĉ}_n=\frac{i}{2}{L}_{X_n}$$ $${Ŝ}_m=-\frac{i}{2}{L}_{Y_m}$$ They act on the square integrable functions (see Appendix A) on $SU\left(2\right)$ and satisfy quantum commutation relations $$[{Ŝ}_n,{Ŝ}_m]=i{\epsilon }_{nm}^k{Ŝ}_k$$ $$[{Ĉ}_n,{Ĉ}_m]=i{\epsilon }_{nm}^k{Ĉ}_k$$ $$[{Ĉ}_n,{Ŝ}_m]=0$$ The Hamiltonian is defined as $$Ĥ={Ĉ}^2={Ŝ}^2$$ and the complete set of observables that commute with each other is $$Ĥ,{Ĉ}_a,{Ŝ}_b$$ with 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 form $$Ĥ{\psi }_{jsc}=j(j+1){\psi }_{jsc}$$ where $j$ takes positive integer and half integer values $$j=0,\frac{1}{2},1,\frac{3}{2},2...$$ $${Ĉ}_a{\psi }_{jsc}=c{\psi }_{jsc}$$ $${Ŝ}_b{\psi }_{jsc}=s{\psi }_{jsc}$$ with $c$ and $s$ taking values in the following range $$-j,-j+1,...,j-1,j$$ Further we construct the corresponding eigenfunctions ${\psi }_{jsc}$. The first step of this construction is to note that the function $\text{〈}Tg\text{〉}$ where $T=(1+i{T}_a)(1+i{T}_b)$ is an eigenfunction of $Ĥ,{Ĉ}_a$ and ${Ŝ}_b$ with eigenvalues $\mathrm{¾},\mathrm{½},\mathrm{½}$ respectively. Proof of this proposition is straightforward. Using $\text{〈}Tg\text{〉}$ one can construct the complete set of eigenfunctions of $Ĥ,{Ĉ}_a$ and ${Ŝ}_b$ operators $${\psi }_{jsc}={Ŝ}_{-}^{j-s}{Ĉ}_{-}^{j-c}\text{〈}Tg{\text{〉}}^{2j}$$ in the manner described in Appendix B.

Free particle on $2D$ sphere can be obtained from our model by gauging $U\left(1\right)$ symmetry. In other words let's consider the following local gauge transformations $$g\to h\left(t\right)g$$ Where $h\left(t\right)\in U\left(1\right)\subset SU\left(2\right)$ is an element of $U\left(1\right)$. Without loss of generality we can take $$h=e^{\beta \left(t\right)T_3}$$ Since $T_3$ is antihermitian $h\left(t\right)\in U\left(1\right)$ and since $h\left(t\right)$ depends on $t$ Lagrangian $$L=\text{〈}{g}^{-1}ġg^{-1}ġ\text{〉}$$ is not invariant under (38) local gauge transformations.

To make (40) gauge invariant we should replace time derivative with covariant derivative $$\frac{d}{dt}g\to \nabla g=(\frac{d}{dt}+B)g$$ where $B$ can be represented as follows $$B=b{T}_3\in su\left(2\right)$$ with transformation rule $$B\to hB{h}^{-1}-\frac{dh}{dt}{h}^{-1}$$ or in terms of $b$ variable $$b\to b-\frac{d\beta }{dt}$$ The new Lagrangian $$L_G=\text{〈}{g}^{-1}\nabla g{g}^{-1}\nabla g\text{〉}$$ is invariant under (38) local gauge transformations. But this Lagrangian as well as every gauge invariant Lagrangian is singular. It contains additional non-physical degrees of freedom. To eliminate them we should eliminate $B$ using Lagrange equations $$\frac{\partial L_G}{\partial B}\to b=-\text{〈}ġg^{-1}{T}_3\text{〉}$$ put it back in (45) and rewrite last obtained Lagrangian in terms of gauge invariant variables. $$L_G=\text{〈}(g^{-1}ġ-S_3{T}_3{)}^2\text{〉}$$ It's obvious that the following $$Z=g^{-1}{T}_{3}g\in su\left(2\right)$$ element of $su\left(2\right)$ algebra is gauge invariant. Since $Z\in su\left(2\right)$ it can be parameterized as follows $$Z=z^a{T}_a$$ where $z^a$ are real functions on $SU\left(2\right)$ $$z_a=\text{〈}Z{T}_a\text{〉}$$

So we have three gauge invariant variables $z^a(a=1,2,3)$ but it's easy to check that only two of them are independent. Indeed $$\text{〈}{Z}^2\text{〉}=\text{〈}{g}^{-1}{T}_{3}g{g}^{-1}{T}_{3}g\text{〉}=\text{〈}{T}_3^2\text{〉}=1$$ otherwise $$\text{〈}{Z}^2\text{〉}=\text{〈}{z}^a{T}_a{z}^b{T}_b\text{〉}=z^a{z}_a$$

So configuration space of $SU\left(2\right)/U\left(1\right)$ coset model is sphere. By direct calculations one can check that after being rewritten in terms of gauge invariant variables $L_G$ takes the form $$L_G=\mathrm{¼}\text{〈}{Z}^{-1}ŻZ^{-1}Ż\text{〉}$$ This Lagrangian describes free particle on the sphere. Indeed, since $Z=z^a{T}_a$ it's easy to show that $$L_G=\mathrm{¼}\text{〈}{Z}^{-1}ŻZ^{-1}Ż\text{〉}=\mathrm{¼}\text{〈}ZŻZŻ\text{〉}=\mathrm{½}{ż}^a{ż}_a$$ So $SU\left(2\right)/U\left(1\right)$ coset model describes free particle on $S^2$ manifold.

Working in a first order Hamiltonian formalism one can introduce equivalent Lagrangian $${\Lambda }_G=\text{〈}C(g^{-1}ġ-u)\text{〉}+\mathrm{½}\text{〈}(u+g^{-1}Bg{)}^2\text{〉}$$ variation of $u$ provides $$\begin{gathered} C=u+g^{-1}Bg \\ u=C-g^{-1}Bg \end{gathered}$$ Rewriting ${\Lambda }_G$ in terms of $C$ and $g$ leads to $$\begin{gathered} {\Lambda }_G=\text{〈}C{g}^{-1}ġ\text{〉}-\mathrm{½}\text{〈}{C}^2\text{〉}-\text{〈}BgC{g}^{-1}\text{〉}=\text{〈}C{g}^{-1}ġ\text{〉} \\ -\mathrm{½}\text{〈}{C}^2\text{〉}-b\text{〈}gC{g}^{-1}{T}_3\text{〉}=\text{〈}C{g}^{-1}ġ\text{〉}-\mathrm{½}\text{〈}{C}^2\text{〉}-b{S}_3 \end{gathered}$$ Due to the gauge invariance of ${\Lambda }_G$ we obtain constrained Hamiltonian system, where $\text{〈}C{g}^{-1}dg\text{〉}$ is symplectic potential, function $$H=\mathrm{½}\text{〈}{C}^2\text{〉}$$ plays the role of Hamiltonian and $b$ is a Lagrange multiple leading to the first class constrain $$\phi =\text{〈}gC{g}^{-1}{T}_3\text{〉}=\text{〈}S{T}_3\text{〉}=S_3=0$$ So coset model is equivalent to the initial one with (59) constrain. Using technique of the constrained quantization, instead of quantizing coset model we can subject quantum model that corresponds to the free particle on $SU\left(2\right)$, to the following operator constrain $${Ŝ}_3|\psi \text{〉}=0$$ Hilbert space of the initial system, that is linear span of $${\psi }_{jcs}j=0,\frac{1}{2},1,\frac{3}{2},2,...$$ wave functions, reduces to the linear span of $${\psi }_{jc0}j=0,1,2,3,...$$ wave functions. Indeed, ${Ŝ}_3{\psi }_{jcs}=0$ 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 ${\psi }_{jcs}$ rewriten in terms of gauge invariant variables up to a constant multiple should coincide with well known spherical harmonics $${\psi }_{jc0}\sim J_{jc}$$ One can check the following $$\begin{gathered} {\psi }_{jc0}\sim {Ŝ}_{-}^j{Ĉ}_{-}^{j-c}\text{〈}Tg{\text{〉}}^{2j}\sim {Ĉ}_{-}^{j-c}\text{〈}{T}_{+}{g}^{-1}{T}_{3}g{\text{〉}}^j \\ \sim {Ĉ}_{-}^{j-c}si{n}^j\theta e^{ij\theta }\sim {Ĉ}_{-}^{j-c}{J}_{jj}\sim J_{jc} \end{gathered}$$ This is an example of using large initial model in quantization of coset model.

Scalar product in Hilbert space is defined as follows $$\text{〈}{\psi }_1|{\psi }_2\text{〉}=\underset{SU\left(2\right)}{\int }\prod _{a=1}^3\text{〈}{g}^{-1}dg{T}_a\text{〉}({\psi }_1{)}^{\text{†}}{\psi }_2$$ It's easy to prove that under this scalar product operators ${Ĉ}_n$ and ${Ŝ}_m$ are hermitian. Indeed $$\begin{gathered} \text{〈}{\psi }_1|{Ĉ}_n{\psi }_2\text{〉}=\underset{SU\left(2\right)}{\int }\prod _{a=1}^3\text{〈}{g}^{-1}dg{T}_a\text{〉}({\psi }_1{)}^{\text{†}}\left(\frac{i}{2}{L}_{X_n}{\psi }_2\right) \\ =\underset{SU\left(2\right)}{\int }\prod _{a=1}^3\text{〈}{g}^{-1}dg{T}_a\text{〉}(\frac{i}{2}{L}_{X_n}{\psi }_1{)}^{\text{†}}{\psi }_2 \end{gathered}$$ Where integration by part has been used and the additional term coming from measure $$\prod _{a=1}^3\text{〈}{g}^{-1}dg{T}_a\text{〉}$$ vanished since $$L_{X_n}\text{〈}{g}^{-1}dg{T}_a\text{〉}=0$$ For more transparency one can introduce the following parameterization of $SU\left(2\right)$. For any $g\in SU\left(2\right)$. $$g=e^{q^a{T}_a}$$ Then the symplectic potential takes the form $$\text{〈}C{g}^{-1}dg\text{〉}=C_{a}d{q}^a$$ and scalar product becomes $$\text{〈}{\psi }_1|{\psi }_2\text{〉}=\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}{d}^{3}q({\psi }_1{)}^{\text{†}}{\psi }_2$$ that coincides with (65) because of $$d{q}_a=\text{〈}{g}^{-1}dg{T}_a\text{〉}$$

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 ${Ĉ}_3$ have at least one common eigenfunction $$\begin{gathered} Ĥ\psi =E\psi \\ {Ĉ}_3\psi =c\psi \\ {Ŝ}_3\psi =s\psi \end{gathered}$$ it is easy to show that eigenvalues of $Ĥ$ are non-negative $E\ge 0$ and conditions $$\begin{gathered} E-c^2\ge 0 \\ E-s^2\ge 0 \end{gathered}$$ are satisfied. Indeed, operators $Ĉ$ and $Ŝ$ are selfadjoint so $$\begin{gathered} \text{〈}\psi \left|Ĥ\right|\psi \text{〉}=\text{〈}\psi \left|{Ĉ}^2\right|\psi \text{〉}=\text{〈}\psi \left|{Ĉ}_a{Ĉ}^a\right|\psi \text{〉}=\text{〈}\psi \left|\right({Ĉ}_a{)}^{\text{†}}{Ĉ}^a|\psi \text{〉}= \\ \text{〈}{Ĉ}_a\psi |{Ĉ}^a\psi \text{〉}=\parallel {Ĉ}_a\psi \parallel \ge 0 \end{gathered}$$ To prove (74) we shall consider ${Ĉ}_1^2+{Ĉ}_2^2$ and ${Ŝ}_1^2+{Ŝ}_2^2$ operators $$\text{〈}\psi |{Ĉ}_1^2+{Ĉ}_2^2|\psi \text{〉}=\parallel {Ĉ}_1\psi \parallel +\parallel {Ĉ}_2\psi \parallel \ge 0$$ and $$\text{〈}\psi |{Ĉ}_1^2+{Ĉ}_2^2|\psi \text{〉}=\text{〈}\psi |Ĥ-{Ĉ}_3^2|\psi \text{〉}=(E-c^2)\text{〈}\psi |\psi \text{〉}$$ thus $E-c^2\ge 0$.

Now let's introduce new operators $${Ĉ}_{+}=i{Ĉ}_1+{Ĉ}_2{Ĉ}_{-}=i{Ĉ}_1-{Ĉ}_2$$ $${Ŝ}_{+}=i{Ŝ}_1+{Ŝ}_2{Ŝ}_{-}=i{Ŝ}_1-{Ŝ}_2$$ These operators are not selfadjoint, but $({Ĉ}_{-}{)}^{\text{†}}={Ĉ}_{+}$ and $({Ŝ}_{-}{)}^{\text{†}}={Ŝ}_{+}$ and they fulfill the following commutation relations $$[{Ĉ}_{\pm },{Ĉ}_3]=\pm {Ĉ}_{\pm }[{Ŝ}_{\pm },{Ŝ}_3]=\pm {Ŝ}_{\pm }$$ $$[{Ĉ}_{+},{Ĉ}_{-}]=2{Ĉ}_3[{Ŝ}_{+},{Ŝ}_{-}]=2{Ŝ}_3$$ $$[{Ĉ}_{\text{•}},{Ŝ}_{\text{•}}]=0$$ where $\text{•}$ takes values $+,-,3$ using these commutation relations it is easy to show that if ${\psi }_{\lambda cs}$ is eigenfunction of $Ĥ,{Ŝ}_3$ and ${Ĉ}_3$ with corresponding eigenvalues : $$\begin{gathered} Ĥ{\psi }_{\lambda cs}=\lambda {\psi }_{\lambda cs} \\ {Ŝ}_3{\psi }_{\lambda cs}=s{\psi }_{\lambda cs} \\ {Ĉ}_3{\psi }_{\lambda cs}=c{\psi }_{\lambda cs} \end{gathered}$$ then ${Ĉ}_{\pm }{\psi }_{\lambda cs}$ and ${Ŝ}_{\pm }{\psi }_{\lambda cs}$ are the eigenfunctions with corresponding eigenvalues $\lambda ,s\pm 1,c$ and $\lambda ,s,c\pm 1$. Consequently using ${Ĉ}_{\pm },{Ŝ}_{\pm }$ operators one can construct a family of eigenfunctions with eigenvalues $$\begin{gathered} c,c\pm 1,c\pm 2,c\pm 3,... \\ s,s\pm 1,s\pm 2,s\pm 3,... \end{gathered}$$ but conditions (74) give restrictions on a possible range of eigenvalues. Namely we must have $$\begin{gathered} \lambda -c^2\ge 0 \\ \lambda -s^2\ge 0 \end{gathered}$$ In other words, in order to interrupt (84) sequences we must assume $$\begin{gathered} {Ŝ}_{+}{\psi }_{\lambda cj}=0{Ŝ}_{-}{\psi }_{\lambda c,-j}=0 \\ {Ĉ}_{+}{\psi }_{\lambda ks}=0{Ĉ}_{-}{\psi }_{\lambda ,-ks}=0 \end{gathered}$$ for some $j$ and $k$, therefore $s$ and $c$ could take only the following values $$\begin{gathered} -j,-j+1,...,j-1,j \\ -k,-k+1,...,k-1,k \end{gathered}$$ The number of values is $2j+1$ and $2k+1$ respectively. Since number of values should be integer, $j$ and $k$ should take integer or half integer values $$\begin{gathered} j=0,\frac{1}{2},1,\frac{3}{2},2,... \\ k=0,\frac{1}{2},1,\frac{3}{2},2,... \end{gathered}$$ Now using commutation relations we can rewrite $Ĥ$ in terms of ${Ĉ}_{\pm },{Ĉ}_3$ operators $$Ĥ={Ĉ}_{+}{Ĉ}_{-}+{Ĉ}_3^2+{Ĉ}_3$$ and it is clear that $j=k$ and $\lambda =j(j+1)=k(k+1)$