The relation between Hamiltonian and Energy

I know Hamiltonian can be energy and be a constant of motion if and only if:

1. Lagrangian be time-independent,
2. potential be independent of velocity,
3. coordinate be time independent.

Otherwise $$H\neq E\neq {\rm const},$$ or $$H=E\neq {\rm const},$$ or $$H\neq E={\rm const}.$$

I am looking for examples of these three situation.

• "Hamiltonian" expressed via unknown variables $q$ and $p$ is a Hamiltonian and serves for writing down equations of motion. "Hamiltonian" expressed via solutions $q(t)$ and $p(t)$ is energy $E$. Energy is not obliged to be conserved. For example, a ball bouncing elastically from a still wall has a constant energy, but the same ball in a moving reference frame (where the wall hits the still ball) acquires energy due to collision. In the latter case the moving wall is described as a time-dependent potential $U(q,t)=V(q-vt)$ Mar 25 '13 at 15:37
• @VladimirKalitvianski Strange answer I believe. When energy is not conserved, would you still continue to talk about energy ? For instance, in QM, time dependent problems usually have no defined energy, am I wrong ? Mar 25 '13 at 15:41
• Related: physics.stackexchange.com/q/11905/2451 and links therein. Mar 25 '13 at 16:47
• @Oaoa: Yes, you are wrong. In QM any measured energy is an eigenvalue $E_n$ and a state without certain energy has these eigenstates in a superposition or mixture. Mar 25 '13 at 17:53
• @VladimirKalitvianski Ok that's just a terminology problem. I would not call this state having definite energy. But you're essentially right of course :-) Thanks. Mar 25 '13 at 20:52

Example. Time-dependent gravitational acceleration ($H=E$ but $\dot E \neq 0$)
Consider a particle falling under the influence of gravity near the surface of a large, spherically symmetric planet. Suppose that the mass of the planet changes with time, so that the acceleration due to gravity near the surface is some function $g(t)$ of time. Then the Lagrangian is $$L(t, z, \dot z) = \frac{1}{2}m\dot z^2 - mg(t)z$$ then the canonical momentum conjugate to $z$ is $$p_z = \frac{\partial L}{\partial\dot z} = m\dot z$$ and the Hamiltonian is $$H = p_z\dot z - L = \frac{p_z^2}{2m} +mg z$$ Notice that in this case $H(t) = E(t)$; the Hamiltonian is equal to the total energy. Now, in this case, the equations of motion are $$\dot p_z(t) = -mg(t)$$ So for any solution $z(t)$ to the equations of motion, we have $$\dot E(t) = p_z\dot p_z + m(\dot gz + g\dot z) = p_z(\dot p_z + mg) + m\dot g z = m\dot g z\neq 0$$ Total energy is not conserved, it changes as a function of time due to the fact that the gravitational acceleration depends on time.
• Joshphysics example is quite realistic if the gravitational force changes not because of varying mass $m(t)$, but because of moving planet: $g(t)=-G\cdot M/R^2 (t)$. Mar 25 '13 at 18:40
• maybe you missed $m$ so that $H=\frac{p_z^2}{2m} + mgz$? Mar 14 '18 at 1:48