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What is the difference between conservation of the Hamiltonian and conservation of energy?

Qmechanic
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Artemisia
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  • Related: https://physics.stackexchange.com/q/11905/2451 , https://physics.stackexchange.com/q/37725/2451 , https://physics.stackexchange.com/q/57985/2451 , and links therein. – Qmechanic Nov 28 '13 at 21:42

1 Answers1

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Consider the time derivative of the Hamiltonian

$$\frac{dH(q,p,t)}{dt}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p}+\frac{\partial H}{\partial t}=-\dot{p}\dot{q}+\dot{q}\dot{p}+\frac{\partial H}{\partial t}$$

From this you see that the Hamiltonian is conserved if it does not depend on time,$t$, explicitly. $H$ may or may not be the total energy, if it is, this means the energy is conserved. But even if it isn't, $H$ is still a constant of motion.

vnb
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  • But then my reference sheet says that the potential should be independent of velocity. How does that work? – Artemisia Nov 28 '13 at 14:52
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    The potential is velocity independent in order to have conservative forces. If $V=V(q,\dot{q})$, then the forces wouldn't be conservative and you couldn't state that $H=E$ – vnb Nov 28 '13 at 15:06
  • So if a hamiltonian is not conserved, it is time dependent and its potential is velocity dependent? – Artemisia Nov 28 '13 at 15:20
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    In general yes. But you can have cases where only the Hamiltonian is a function of time OR the potential is a function of velocity. It's not necessarily that both cases should be true at the same time. – vnb Nov 29 '13 at 07:11
  • how would it look in quantum mechanics? – MycrofD Mar 14 '18 at 01:55