Undergrad Hamiltonian in Quantum vs Classical

[PeterDonis]

The Hamiltonian is defined as the operator of time evolution.

This is not quite correct. The time evolution operator is $e^{i H t}$, where $H$ is the Hamiltonian.

the operator for the "energy" observable is defined as the Hamiltonian.

Yes, that's correct. But @vanhees71 is correct that, if the Hamiltonian is explicitly time dependent, then "energy" is not a conserved quantity--which violates many people's intuition about how "energy" should work. In other words, the issue here is not whether the operator for "energy" is the Hamiltonian or something else; the issue is that you are expecting "energy" to be conserved in situations where it isn't.

 

[vanhees71]

If $\hat{H}$ is time dependent the unitary time-evolution operator (for states in the Schrödinger picture) is
$$\hat{U}(t,t_0) = \mathcal{T} \exp \left [-\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}(t') \right].$$
So strictly speaking the Hamiltonian is the generator for time evolution.

 

[Orodruin]

This is not quite correct. The time evolution operator is $e^{i H t}$, where $H$ is the Hamiltonian.

In other words, the Hamiltonian is the generator of time translations. This is true clasically as well as in QM.

 

[Electric to be]

the issue is that you are expecting "energy" to be conserved in situations where it isn't.

Well no I'm not expecting energy to be conserved. I'm perfectly fine with the energy having some time dependence - what I am concerned with is how "energy" is defined in these situations. It seems that the basic definition of energy in QM is simply the eigenvalue of the Hamiltonian. While in situations like the hydrogen atom this makes sense classically (sum of kinetic and potential energies & when E>0 we have free particles), but in other situations it doesn't and seems to be simply defined as the Eigenvalue of the Hamiltonian without a real physical meaning like other physical observables like position and momentum.

Classically, the energy of a system that has work done on it by an external system (and therefore has the total energy increased) still has a very clear definition for it's total energy. It's the sum of kinetic energies, energy stored in fields, etc. However, here I just don't see how the energy is actually defined in the case of a changing EM field since in QM we aren't assigning any energy to the field itself, and we are treating the particle as "holding" all the energy through things like scalar potentials (when they can be defined).

One possible explanation (?) is that in certain cases the overall "energy" of the particle resembles the classical situation: kinetic + potential. This applies for static fields. Otherwise for a general EM field the "energy" simply represents possible states between which photons can be absorbed, and not the sum of it's parts (kinetic + potential).

 

[PeterDonis]

It seems that the basic definition of energy in QM is simply the eigenvalue of the Hamiltonian.

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.

in QM we aren't assigning any energy to the field itself

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

What you can't usually do in QM is split up the energy of the total system into "energy of the particle" and "energy of the field". You can only do this in particular cases when the system's state happens to be separable, i.e., it's just a single product of a particle state and a field state. But in most cases the state of the total system is entangled, so the "particle" and "field" don't have well-defined states by themselves, which means they don't have well-defined energies by themselves either.

Basically it seems to me that you are trying to apply classical thinking to the quantum case where it doesn't work.

 

[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.