Hamiltonian in Quantum vs Classical

[Electric to be]

Only if the system is in an eigenstate of the Hamiltonian. If the system is not in an eigenstate of the Hamiltonian, then it doesn't have a definite energy; all you have is probabilities of different energies being measured if you measure the energy.

Well sure it can be in a superposition of possible energies as well. But the energy possibilities are defined as the eigenvalues, and when measured is one of these possibilities.

What makes you think that? The field is part of the total system, and the Hamiltonian operator acts on the total system.

I feel like this is certainty true in QED where the electron field and em field are both part of an interacting system together. But here I am speaking specifically about Schrodinger equation non-relativistic style QM. Where the fields are not quantized, and the energy is all assigned to the particle. In classical Hamiltonian mechanics, if there is a time dependency of the field it is acknowledged that the Hamiltonian can't really be defined as the total energy in any meaningful way, but nonetheless it is still used to solve the equations of the motion with the Lagrangian. However, in QM the quantized Hamiltonian is still used to find the total energy in these situations.

 

[PeterDonis]

here I am speaking specifically about Schrodinger equation non-relativistic style QM. Where the fields are not quantized

Ah, yes, in that case the "system" is the particle. But the "system" in terms of QM in this case is not "everything that is there", as it would be in the QFT case where we are modeling the field as well as the particle.

in QM the quantized Hamiltonian is still used to find the total energy in these situations

It's the operator corresponding to the "energy" observable, yes. But this observable only measures the energy in the "system", i.e., the part that we are modeling using QM. It can't tell you anything about the part that we are not modeling using QM, i.e., the field. So if you consider the field as part of "the total system", then you would have to say that "the total system" is not completely modeled by QM in this case, so the "energy" you get from the QM operator is not the "total energy".

 

[Electric to be]

It's the operator corresponding to the "energy" observable, yes. But this observable only measures the energy in the "system", i.e., the part that we are modeling using QM. It can't tell you anything about the part that we are not modeling using QM, i.e., the field. So if you consider the field as part of "the total system", then you would have to say that "the total system" is not completely modeled by QM in this case, so the "energy" you get from the QM operator is not the "total energy".

Yes agreed just the system. And in this case the system is the particle. However, as I mentioned - even in classical Lagrangian mechanics the energy of a single particle isn't really defined for such changing fields since there is no concept of a scalar potential. As a result - do we simply define energy as the values at which differences can be taken to determine possible photon absorptions?

 

[PeterDonis]

do we simply define energy as the values at which differences can be taken to determine possible photon absorptions?

I don't understand what this means.

 

[Electric to be]

I don't understand what this means.

I guess what I mean is, in situations where it isn't possible to describe energy E as K+U in shrodingers equation, does the energy simply become some quantity that can be increased by the absorption of photons and their energy? And not simply K+U.

 

[vanhees71]

If you want to describe absorption and emission processes of photons you use the QED Hamiltonian, where such processes are automatically built in. In QFT everything is expressed in terms of field operators. In the most simple version of QED you have a Dirac field describing electrons and positrons and a massless U(1) gauge field, describing the electrons. Total energy is conserved since the Hamiltonian is not explicitly time dependent, and thus in this case it's clearly representing the total energy of the system.