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I do not completely understand, why do we consider Hamilton–Jacobi equation $H(q,p)=E$ as equation of motion, whereas it is looks like energy-conservation law?

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  • $\begingroup$ This question is too old. Since, I read many books and removed my misunderstanding. $\endgroup$
    – Sergio
    Commented Oct 18, 2019 at 10:06

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That is not the hamilton-jacobi equation. What you have above is an equality between the time-ind. Hamiltonian and the energy of a particle with generalized postion q and momentum p. Energy is conserved in some systems, but it depends on the exact mathematical form of the hamiltonian and has nothing to do with this equation as far as I know.

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    $\begingroup$ By the way, the hamilton jacobi equation is just a partial differential equation for the transformation between the hamiltonian and a kamiltonian where the generalized coordinates are identically zero (the intial state of the system is described). It also uses an F2 type generating function. cheers! $\endgroup$
    – Timtam
    Commented Jan 31, 2012 at 9:20
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As Timtam pointed out, this is not the Hamilton-Jacobi Equation. If you say, quantum mechanically, this does not even make any sense because, there the Hamiltonian is an operator which has to act on some eigenstate to give the energy of that state. Effectively, energy of the different states becomes the eigenvalue of the Hamiltonian. As far as conservation laws are concerned, it comes from the symmetry of the problem. A symmetry is defined as a transformation which keeps the action invariant \begin{equation} A = \int L dt \end{equation} (If, identity belongs to the group of transformation, then it becomes a continuous symmetry. The symmetries gives rise to Noether currents which are conserved. You get your conservation laws because the system has symmetries, not because the eigenvalues of the Hamiltonian is the energy which is anyways true. Classically, if we consider a non-dispersive system, then the Hamiltonian evaluated at a definite $p$ and $q$ gives you the energy of the system at that point. Go to another point and evaluate $H$, you should get the same number (since, it is non-dissipative). However, to know if energy will be conserved at all or not, you need the explicit form of the Hamiltonian.

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  • $\begingroup$ this is a good explanation of how energy conservation arises $\endgroup$
    – Timtam
    Commented Jan 31, 2012 at 9:32
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Hamilton Jacobi equation corresponds to a transformation to a phase space where canonically conjugate variables remain constant over time. {Q,P}=1 The new hamiltonian $K= H+\frac{ds}{dt}$ with the condition $K=0$ In quantum mechanics Schrodinger equation gives rise to HJ equation at the limit $\hbar$ tends to 0.

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