I Hamiltonian in Schrödinger: necessarily total energy?

[nomadreid]

This is a basic question, so probably easy to answer. The following from Wikipedia seems pretty standard while describing the Schrödinger equation: "...and Ĥ is the Hamiltonian operator (which characterises the total energy of the system under consideration)."
On the other hand, from page 100 of Patrick Hamill's "A Student's Guide to Lagrangians and Hamiltonians", he states "... there are situations in which the Hamiltonian is not the total energy of the system."
If I accept both of these, my logical conclusion would be that the situations which Schrödinger's equation describes is not one of the situations which Hamill is referring to. The other possibility is that Wikipedia is over-simplifying. Which conclusion is correct?
The closest post I could find here on this question, https://www.physicsforums.com/threads/hamiltonian-and-total-energy.266410/, either did not answer this question, or the answer was not clear to me, which is why I am posting a new thread.

 

[bb1414]

The Hamiltonian $H$ is generally defined as the sum of the kinetic and potential energy operators or $H=T+V$. Right away, this appears to be the total energy of the system, so we could then say that for a wave function $\psi$ that

$H\psi=E\psi$

where $E$ is the energy operator. This is the generalized Schrodinger equation, and seems quite logical at first.

But yes, there are instances in which the Hamiltonian does not correspond to the total energy of the system. Rather than thinking of the Hamiltonian as the total energy of the system, I'd say it should be thought of as a way to describe the dynamics of a system. In other words, consider some closed system. Since it preserves its total energy through time, its hamiltonian is independent of time. Therefore, the value of the hamiltonian will just be the total energy of the system. For a more complex system, though, one that interacts with external systems and exchanges energy with them, its total energy is dynamic so the hamiltonian may not be exactly its total energy at a given moment.

FYI I'm taking about the Hamiltonian in the non-relativistic case. When incorporating relativity into this, I'd assume spin, relativistic mass and interactions with spacetime would be added, which would definitely make the system more dynamic. I'm not too knowledgeable in this space, but it seems logical to me that the Hamiltonian would not be equal to the total energy of the system given a relativistic system, since the energy itself would be considered relativistic, according to:

$E=\sqrt{p^{2}c^{2}+m_0^2c^{4}}$

Again, just thinking off the top of my head. Definitely worth some more research.

 

[nomadreid]

Many thanks, bb1414. That all makes sense. I would guess that the relativistic Schrödinger equation is the Klein-Gordon equation, but I still have enough problem with the non-relativistic case. For example, while I assume the operators in that equation take into account the possible uncertainty (à la Heisenberg), it looks as if the (rest) mass figures in as a constant, which seems odd to me given the uncertainty of mass given by the Energy-time uncertainty relation (and the equivalence of mass and Energy). Or another example: can one treat every Lagrangian density as a collection of applications of Schrödinger's equation (otherwise put: every path in Feynman's sum-over-histories consists of points governed by Schrödinger, or does one apply Schrödinger to the entire path somehow)? I suspect that these questions seem a bit off the wall, so I would not be offended if you stated as such.