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Like in the Hamiltonian for a particle in an electromagnetic field. This is not a conservative field so the Hamiltonian doesn't represent the energy of the system. And yet the time independent Schrodinger equation still reads $H \psi = E \psi$ (for example that's what Griffiths did in page 183 of his book). Which suggests that the Hamiltonian does represent the energy of the system. What am I missing?

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  • $\begingroup$ I couldn't find that example on page 183 of Griffiths' (maybe I have a different edition?), but Schrödinger's equation is indeed only valid for conservative systems (particularly, where you can write $H=T+V$). If you can't find such a Hamiltonian, I think you need to do other quantization techniques. $\endgroup$ – peguerosdc Mar 22 at 23:23
  • $\begingroup$ I have the 3rd edition. It's in the electromagnetic interactions section. And i'm not sure about that last statement of yours. As far as I have seen Schrodinger's equation is very much applicable to non-conservative fields. You write the classical hamiltonian and then do the canonical substitution (i.e. convert the hamiltonian to operator form) then plug it in Schrodinger's equation. $\endgroup$ – Leonid Mar 23 at 0:19
  • $\begingroup$ Depends on what the system is. If it is atom plus em field, then the Hamiltonian is incomplete. If it is just the atom then the Hamiltonian is non conservative. But the energy levels of atoms are given by it because it represents the total energy of the atom. $\endgroup$ – Superfast Jellyfish Mar 23 at 4:49
  • $\begingroup$ Particle in a static electromagnetic field is a conservative system. Its Hamiltonian is constant. Only this makes it possible to write down the time-independent Schrödinger's equation. And this then gives you energy eigenstates. See also this discussion of conservativity, in particular the replies by Ulrich Mutze. $\endgroup$ – Ruslan Mar 23 at 5:59
  • $\begingroup$ Related: When is the Hamiltonian of a system not equal to its total energy? $\endgroup$ – Ruslan Mar 23 at 6:06
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You are quite right to say that the Hamiltonian is not always the energy in the sense of being kinetic plus potential energy. This statement is often repeated in courses on analytical mechanics, to drill the idea that the Hamiltonian is the energy function in specific variables, namely position and momentum. But it is always a conserved quantity, and the actual values the Hamiltonian takes are energy values. Since eigenvalues are the values of the Hamiltonian that can be measured, and since these values do not depend on the variables in which they were expressed, they can be called energies.

On the other hand, your saying that the electromagnetic field is not conservative is possibly a misunderstanding: For velocity-independent forces we call the force field conservative if it does no work on a particle moving on a closed circuit. This definition fails for the Lorentz-force, which I assume you have in mind, because it never performs any work on a moving particle.

As an example, for a particle in a magnetic field, the energy is always $$ \frac m2(v_x^2+v_y^2) $$

But the Hamiltonian is expressed in terms of a nonstandard momentum, and is given by $$ \frac1{2m}\left[(p_x - \frac{e B}{m c}y)^2 + p_y^2\right] $$

but it has the same values as the energy. Its expression is different because we deal with momenta.

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A non-conservative field just means that energy can be added or removed from the system. It does not change the fact that the Hamiltonian represents energy.

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  • $\begingroup$ You want to be careful here as there are situations where the Hamiltonian is conserved but is not the total energy of the system, i.e. does not boil down to $T+U$. This happens (classically) with beads on a rotating wire for instance. $\endgroup$ – ZeroTheHero Mar 22 at 21:45
  • $\begingroup$ On the other hand, if you can write a Hamiltonian then the field must be conservative. $\endgroup$ – lcv Mar 22 at 23:20
  • $\begingroup$ No. You can still write down Lagrangians/Hamiltonians for non-conservative systems. Like the electromagnetic field to name one. In fact Lagrangians are still applicable in curved spacetime (where conservation laws need not hold). $\endgroup$ – Leonid Mar 23 at 0:23

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