How can I think about Poisson commutation geometrically?












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In the Hamiltonian formulation of classical mechanics, we have the result that, in a system with Hamiltonian $H(q_i,p_i), i=1,dots,N$, a quantity $f(q_i,p_i)$ time-evolves according to
$$ frac{mathrm{d} f}{mathrm{d} t} = {f,H} = frac{partial f}{partial q_i}frac{partial H}{partial p_i} - frac{partial f}{partial p_i}frac{partial H}{partial q_i} $$
and so $f$ is conserved if and only if it Poisson commutes with $H$, i.e. ${f,H} = 0$.



In the case $N=1$ so that the phase space is 2-dimensional, and contour surfaces of functions are curves. Contours of $H$ are trajectories, and $f$ is a conserved quantity if and only if the contours of $f$ align with those of $H$. But for $Ngeq2$, the contour surfaces of $H$ and $f$ are $(2N-1)$-dimensional manifolds in the $2N$-dimensional phase space. What can you say about these surfaces when ${f,H} = 0$?



In quantum mechanics, when the functions $f$ and $H$ are replaced by operators $hat{f}$ and $hat{H}$ and the Poisson bracket is replaced by the commutator, then $[hat{f},hat{H}] = 0$ means that the two operators have coinciding eigenspaces. Is there such a nicely geometric way of visualising the classical equivalent?










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    In the Hamiltonian formulation of classical mechanics, we have the result that, in a system with Hamiltonian $H(q_i,p_i), i=1,dots,N$, a quantity $f(q_i,p_i)$ time-evolves according to
    $$ frac{mathrm{d} f}{mathrm{d} t} = {f,H} = frac{partial f}{partial q_i}frac{partial H}{partial p_i} - frac{partial f}{partial p_i}frac{partial H}{partial q_i} $$
    and so $f$ is conserved if and only if it Poisson commutes with $H$, i.e. ${f,H} = 0$.



    In the case $N=1$ so that the phase space is 2-dimensional, and contour surfaces of functions are curves. Contours of $H$ are trajectories, and $f$ is a conserved quantity if and only if the contours of $f$ align with those of $H$. But for $Ngeq2$, the contour surfaces of $H$ and $f$ are $(2N-1)$-dimensional manifolds in the $2N$-dimensional phase space. What can you say about these surfaces when ${f,H} = 0$?



    In quantum mechanics, when the functions $f$ and $H$ are replaced by operators $hat{f}$ and $hat{H}$ and the Poisson bracket is replaced by the commutator, then $[hat{f},hat{H}] = 0$ means that the two operators have coinciding eigenspaces. Is there such a nicely geometric way of visualising the classical equivalent?










    share|cite|improve this question

























      2












      2








      2







      In the Hamiltonian formulation of classical mechanics, we have the result that, in a system with Hamiltonian $H(q_i,p_i), i=1,dots,N$, a quantity $f(q_i,p_i)$ time-evolves according to
      $$ frac{mathrm{d} f}{mathrm{d} t} = {f,H} = frac{partial f}{partial q_i}frac{partial H}{partial p_i} - frac{partial f}{partial p_i}frac{partial H}{partial q_i} $$
      and so $f$ is conserved if and only if it Poisson commutes with $H$, i.e. ${f,H} = 0$.



      In the case $N=1$ so that the phase space is 2-dimensional, and contour surfaces of functions are curves. Contours of $H$ are trajectories, and $f$ is a conserved quantity if and only if the contours of $f$ align with those of $H$. But for $Ngeq2$, the contour surfaces of $H$ and $f$ are $(2N-1)$-dimensional manifolds in the $2N$-dimensional phase space. What can you say about these surfaces when ${f,H} = 0$?



      In quantum mechanics, when the functions $f$ and $H$ are replaced by operators $hat{f}$ and $hat{H}$ and the Poisson bracket is replaced by the commutator, then $[hat{f},hat{H}] = 0$ means that the two operators have coinciding eigenspaces. Is there such a nicely geometric way of visualising the classical equivalent?










      share|cite|improve this question













      In the Hamiltonian formulation of classical mechanics, we have the result that, in a system with Hamiltonian $H(q_i,p_i), i=1,dots,N$, a quantity $f(q_i,p_i)$ time-evolves according to
      $$ frac{mathrm{d} f}{mathrm{d} t} = {f,H} = frac{partial f}{partial q_i}frac{partial H}{partial p_i} - frac{partial f}{partial p_i}frac{partial H}{partial q_i} $$
      and so $f$ is conserved if and only if it Poisson commutes with $H$, i.e. ${f,H} = 0$.



      In the case $N=1$ so that the phase space is 2-dimensional, and contour surfaces of functions are curves. Contours of $H$ are trajectories, and $f$ is a conserved quantity if and only if the contours of $f$ align with those of $H$. But for $Ngeq2$, the contour surfaces of $H$ and $f$ are $(2N-1)$-dimensional manifolds in the $2N$-dimensional phase space. What can you say about these surfaces when ${f,H} = 0$?



      In quantum mechanics, when the functions $f$ and $H$ are replaced by operators $hat{f}$ and $hat{H}$ and the Poisson bracket is replaced by the commutator, then $[hat{f},hat{H}] = 0$ means that the two operators have coinciding eigenspaces. Is there such a nicely geometric way of visualising the classical equivalent?







      classical-mechanics quantum-mechanics hamilton-equations






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      asked Jun 10 '17 at 0:01









      Jonathan Michael Foonlan Tsang

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          So your velocity in phase-space $(dot q_1, dot q_2, ..., dot q_N,dot p_1,...,dot p_N )$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.



          You may see these ones intersect pairwise; so the common intersection of all of them on which your velocity (trajectory) must lie is at least 1-dimensional. In N =1 you cannot have another independent quantity beyond H itself, the energy. One more would "freeze" the trajectory to an immovable point.



          A system with k=N such quantities, is called completely integrable , and the leaves of the invariant foliation are invariant tori, displayed visibly by action-angle variables. Think of N constant momenta and uniform motion of the N coordinates.



          Systems with more invariants, k>N , are superintegrable, and are most elegantly described by Nambu mechanics.



          Typically, the most extreme ("maximally") superintegrable systems have their 1-dim trajectory completely specified by the 2N -1 constants of the motion, being the intersection of their hypersurfaces. For example, in 3 dimensions, so 6-dim phase space, the Kepler problem (central potential) has 3 conserved angular momentum components, the hamiltonian, and an extra independent conserved hypersurface associated with the Laplace-Runge-Lenz vector; so the closed Kepler planetary phase-space trajectories are explicitly described by their collective Nambu 6-bracket.



          You might also illustrate this with the 3-d harmonic oscillator, also maximally superintegrable.






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            1 Answer
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            So your velocity in phase-space $(dot q_1, dot q_2, ..., dot q_N,dot p_1,...,dot p_N )$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.



            You may see these ones intersect pairwise; so the common intersection of all of them on which your velocity (trajectory) must lie is at least 1-dimensional. In N =1 you cannot have another independent quantity beyond H itself, the energy. One more would "freeze" the trajectory to an immovable point.



            A system with k=N such quantities, is called completely integrable , and the leaves of the invariant foliation are invariant tori, displayed visibly by action-angle variables. Think of N constant momenta and uniform motion of the N coordinates.



            Systems with more invariants, k>N , are superintegrable, and are most elegantly described by Nambu mechanics.



            Typically, the most extreme ("maximally") superintegrable systems have their 1-dim trajectory completely specified by the 2N -1 constants of the motion, being the intersection of their hypersurfaces. For example, in 3 dimensions, so 6-dim phase space, the Kepler problem (central potential) has 3 conserved angular momentum components, the hamiltonian, and an extra independent conserved hypersurface associated with the Laplace-Runge-Lenz vector; so the closed Kepler planetary phase-space trajectories are explicitly described by their collective Nambu 6-bracket.



            You might also illustrate this with the 3-d harmonic oscillator, also maximally superintegrable.






            share|cite|improve this answer




























              0














              So your velocity in phase-space $(dot q_1, dot q_2, ..., dot q_N,dot p_1,...,dot p_N )$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.



              You may see these ones intersect pairwise; so the common intersection of all of them on which your velocity (trajectory) must lie is at least 1-dimensional. In N =1 you cannot have another independent quantity beyond H itself, the energy. One more would "freeze" the trajectory to an immovable point.



              A system with k=N such quantities, is called completely integrable , and the leaves of the invariant foliation are invariant tori, displayed visibly by action-angle variables. Think of N constant momenta and uniform motion of the N coordinates.



              Systems with more invariants, k>N , are superintegrable, and are most elegantly described by Nambu mechanics.



              Typically, the most extreme ("maximally") superintegrable systems have their 1-dim trajectory completely specified by the 2N -1 constants of the motion, being the intersection of their hypersurfaces. For example, in 3 dimensions, so 6-dim phase space, the Kepler problem (central potential) has 3 conserved angular momentum components, the hamiltonian, and an extra independent conserved hypersurface associated with the Laplace-Runge-Lenz vector; so the closed Kepler planetary phase-space trajectories are explicitly described by their collective Nambu 6-bracket.



              You might also illustrate this with the 3-d harmonic oscillator, also maximally superintegrable.






              share|cite|improve this answer


























                0












                0








                0






                So your velocity in phase-space $(dot q_1, dot q_2, ..., dot q_N,dot p_1,...,dot p_N )$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.



                You may see these ones intersect pairwise; so the common intersection of all of them on which your velocity (trajectory) must lie is at least 1-dimensional. In N =1 you cannot have another independent quantity beyond H itself, the energy. One more would "freeze" the trajectory to an immovable point.



                A system with k=N such quantities, is called completely integrable , and the leaves of the invariant foliation are invariant tori, displayed visibly by action-angle variables. Think of N constant momenta and uniform motion of the N coordinates.



                Systems with more invariants, k>N , are superintegrable, and are most elegantly described by Nambu mechanics.



                Typically, the most extreme ("maximally") superintegrable systems have their 1-dim trajectory completely specified by the 2N -1 constants of the motion, being the intersection of their hypersurfaces. For example, in 3 dimensions, so 6-dim phase space, the Kepler problem (central potential) has 3 conserved angular momentum components, the hamiltonian, and an extra independent conserved hypersurface associated with the Laplace-Runge-Lenz vector; so the closed Kepler planetary phase-space trajectories are explicitly described by their collective Nambu 6-bracket.



                You might also illustrate this with the 3-d harmonic oscillator, also maximally superintegrable.






                share|cite|improve this answer














                So your velocity in phase-space $(dot q_1, dot q_2, ..., dot q_N,dot p_1,...,dot p_N )$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.



                You may see these ones intersect pairwise; so the common intersection of all of them on which your velocity (trajectory) must lie is at least 1-dimensional. In N =1 you cannot have another independent quantity beyond H itself, the energy. One more would "freeze" the trajectory to an immovable point.



                A system with k=N such quantities, is called completely integrable , and the leaves of the invariant foliation are invariant tori, displayed visibly by action-angle variables. Think of N constant momenta and uniform motion of the N coordinates.



                Systems with more invariants, k>N , are superintegrable, and are most elegantly described by Nambu mechanics.



                Typically, the most extreme ("maximally") superintegrable systems have their 1-dim trajectory completely specified by the 2N -1 constants of the motion, being the intersection of their hypersurfaces. For example, in 3 dimensions, so 6-dim phase space, the Kepler problem (central potential) has 3 conserved angular momentum components, the hamiltonian, and an extra independent conserved hypersurface associated with the Laplace-Runge-Lenz vector; so the closed Kepler planetary phase-space trajectories are explicitly described by their collective Nambu 6-bracket.



                You might also illustrate this with the 3-d harmonic oscillator, also maximally superintegrable.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 25 at 23:15

























                answered Nov 24 at 22:14









                Cosmas Zachos

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                1,447520






























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