Classical angular momentum components are numbers. Can they be generators of some symmetry group?












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In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=ihbarepsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?






quantum-mechanics classical-mechanics angular-momentum symmetry rotation







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    $begingroup$


    In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=ihbarepsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?






    quantum-mechanics classical-mechanics angular-momentum symmetry rotation







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=ihbarepsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?






      quantum-mechanics classical-mechanics angular-momentum symmetry rotation







      share|cite|improve this question











      $endgroup$




      In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=ihbarepsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?






      quantum-mechanics classical-mechanics angular-momentum symmetry rotation




      quantum-mechanics classical-mechanics angular-momentum symmetry rotation






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      edited 12 hours ago









      Qmechanic♦

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      asked 12 hours ago









      SRSSRS

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          $begingroup$

          As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.



          Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are
          $$
          {L_i,L_j}=epsilon_{ijk}L_k,
          $$
          where the Poisson bracket is defined as
          $$
          {f,g} = sum_i left[ frac{partial f}{partial x_i} frac{partial g}{partial p_i} - frac{partial f}{partial p_i} frac{partial g}{partial x_i} right],
          $$
          and the group action generated by the angular momentum on a function $f$ works as
          $$
          f mapsto f+{L_i,f}
          $$
          in its infinitesimal version, and as
          $$
          f mapstoexpleft(theta{L_i,cdot}right) f
          $$
          for a finite angle $theta$.



          And, of course, the group that they generate is simply $rm SO(3)$, acting on the space of real functions on phase space.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
            $endgroup$
            – SRS
            12 hours ago










          • $begingroup$
            @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
            $endgroup$
            – Emilio Pisanty
            11 hours ago











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          1






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          1 Answer
          1






          active

          oldest

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          active

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          active

          oldest

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          4












          $begingroup$

          As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.



          Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are
          $$
          {L_i,L_j}=epsilon_{ijk}L_k,
          $$
          where the Poisson bracket is defined as
          $$
          {f,g} = sum_i left[ frac{partial f}{partial x_i} frac{partial g}{partial p_i} - frac{partial f}{partial p_i} frac{partial g}{partial x_i} right],
          $$
          and the group action generated by the angular momentum on a function $f$ works as
          $$
          f mapsto f+{L_i,f}
          $$
          in its infinitesimal version, and as
          $$
          f mapstoexpleft(theta{L_i,cdot}right) f
          $$
          for a finite angle $theta$.



          And, of course, the group that they generate is simply $rm SO(3)$, acting on the space of real functions on phase space.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
            $endgroup$
            – SRS
            12 hours ago










          • $begingroup$
            @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
            $endgroup$
            – Emilio Pisanty
            11 hours ago
















          4












          $begingroup$

          As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.



          Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are
          $$
          {L_i,L_j}=epsilon_{ijk}L_k,
          $$
          where the Poisson bracket is defined as
          $$
          {f,g} = sum_i left[ frac{partial f}{partial x_i} frac{partial g}{partial p_i} - frac{partial f}{partial p_i} frac{partial g}{partial x_i} right],
          $$
          and the group action generated by the angular momentum on a function $f$ works as
          $$
          f mapsto f+{L_i,f}
          $$
          in its infinitesimal version, and as
          $$
          f mapstoexpleft(theta{L_i,cdot}right) f
          $$
          for a finite angle $theta$.



          And, of course, the group that they generate is simply $rm SO(3)$, acting on the space of real functions on phase space.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
            $endgroup$
            – SRS
            12 hours ago










          • $begingroup$
            @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
            $endgroup$
            – Emilio Pisanty
            11 hours ago














          4












          4








          4





          $begingroup$

          As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.



          Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are
          $$
          {L_i,L_j}=epsilon_{ijk}L_k,
          $$
          where the Poisson bracket is defined as
          $$
          {f,g} = sum_i left[ frac{partial f}{partial x_i} frac{partial g}{partial p_i} - frac{partial f}{partial p_i} frac{partial g}{partial x_i} right],
          $$
          and the group action generated by the angular momentum on a function $f$ works as
          $$
          f mapsto f+{L_i,f}
          $$
          in its infinitesimal version, and as
          $$
          f mapstoexpleft(theta{L_i,cdot}right) f
          $$
          for a finite angle $theta$.



          And, of course, the group that they generate is simply $rm SO(3)$, acting on the space of real functions on phase space.






          share|cite|improve this answer









          $endgroup$



          As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.



          Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are
          $$
          {L_i,L_j}=epsilon_{ijk}L_k,
          $$
          where the Poisson bracket is defined as
          $$
          {f,g} = sum_i left[ frac{partial f}{partial x_i} frac{partial g}{partial p_i} - frac{partial f}{partial p_i} frac{partial g}{partial x_i} right],
          $$
          and the group action generated by the angular momentum on a function $f$ works as
          $$
          f mapsto f+{L_i,f}
          $$
          in its infinitesimal version, and as
          $$
          f mapstoexpleft(theta{L_i,cdot}right) f
          $$
          for a finite angle $theta$.



          And, of course, the group that they generate is simply $rm SO(3)$, acting on the space of real functions on phase space.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 12 hours ago









          Emilio PisantyEmilio Pisanty

          83.2k22203417




          83.2k22203417












          • $begingroup$
            Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
            $endgroup$
            – SRS
            12 hours ago










          • $begingroup$
            @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
            $endgroup$
            – Emilio Pisanty
            11 hours ago


















          • $begingroup$
            Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
            $endgroup$
            – SRS
            12 hours ago










          • $begingroup$
            @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
            $endgroup$
            – Emilio Pisanty
            11 hours ago
















          $begingroup$
          Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
          $endgroup$
          – SRS
          12 hours ago




          $begingroup$
          Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation ${L_i,cdot}$ stand for? @EmilioPisanty
          $endgroup$
          – SRS
          12 hours ago












          $begingroup$
          @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
          $endgroup$
          – Emilio Pisanty
          11 hours ago




          $begingroup$
          @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $${L_i,cdot}: fmapsto {L_i, f}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function ${L_i, f}$, but of the operator ${L_i,cdot}$, so begin{align}exp(theta{L_i,cdot}) & = sum_{n=0^infty} frac{1}{n!} {L_i,cdot}^n \ & = 1+ theta {L_i,cdot} + frac{1}{2!}theta^2{L_i,{L_i,cdot}} + frac{1}{3!} theta^3 {L_i,{L_i,{L_i,cdot}}} + cdots .end{align}
          $endgroup$
          – Emilio Pisanty
          11 hours ago


















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