This question is a more general (and shorter) version of this previous question of mine. We know that from any quantum-mechanical description of a system, we can go to an equivalent description by transforming the state vector of the system $$ | \tilde{\Psi} \rangle = U |\Psi \rangle $$ and operators (that correspond to observables) by $$ \tilde{O}_n = U O U^{\dagger}. $$ This will yield a new state vector (with a new time evolution). The system is, however, equivalent, in that for any operator, the matrix elements (and especially the expectation values) do not change: $$ \langle \tilde{\Phi} | \tilde{O}|\tilde{\Psi} \rangle = \langle \Phi | O | \Psi \rangle. $$ However, if the state is a solution of the system, then its time evolution is no longer governed by $H$ (or, in the transformed picture, by $\tilde{H}$), but instead by another operator, namely $$ \tilde{H}_N = U H U^{\dagger} + i \hbar (\partial_t U)U^{\dagger} = \tilde{H} + i \hbar (\partial_t U)U^{\dagger}. $$ Meanwhile, the time evolution of the operators also changes. If they were time independent before, they afterwards are $$ \partial_t \tilde{O} = (\partial_t U) U^{\dagger} \tilde{O} + \tilde{O} U\partial_t U^{\dagger}. $$

Which leaves me to ask: When all these frames are equivalent, then what is the right operator that measures "energy"? If I ask for the generator of time translations, then it is not clear anymore which operator that should be. In my example, is $H$ (or $\tilde{H}$) the right operator to look at when one asks, "What can the energy eigenvalues be?" Or is it $\tilde{H}_n$ (or $H_n$)?

In the Schrödinger and in the Heisenberg pictures, there is no ambiguity regarding that question, and the Hamilton operator is the same for both of them. But energy eigenvalues are also calculated in other pictures. Is this a valid procedure?

EDIT: There was a suggestion that this question is a duplicate of this question. I don't think that this is the case: The answer to the suggested question is the starting point of my question. It points out that

if one simultaneously transforms states and observables, then expectation values do not change [... but] the transformed Hamiltonian ceases to be the Hamiltonian of the system since (1) holds in place of $V(t)U_{t}V(t)^{−1}=U_{t}$. (wrong)

This result is already the starting point of my question. What I now want to is: When unitarily equivalent pictures of quantum mechanics (and as the suggested answer pointed out, we can choose any of those pictures, because expectation values do not change) have different Hamiltonians, and if the observable that we call energy is usually the observable that generates time evolution, then which of those many different inequivalent Hamiltonians is the energy of the system?

We usually don't talk about "the energy of the two-level system in the Schrödinger picture" and "the energy of the two-level system in the rotating frame," so the term "energy" seems to refer to a specific choice here. I want to know what choice that is.

At this point (guessing the one comment of Roger Vadim), I believe that the answer to my question is, "What the energy is ultimately depends on how the detector looks like, because this will determine the operator that is measured?"


1 Answer 1


...if the observable that we call energy is usually the observable that generates time evolution...

Hamiltonian operator generates time evolution, and it is not unique. Calling it "energy observable" is just lingo in some textbooks which means the author defines system energy as that Hamiltonian expectation value, or says that in general energy is not defined, but in special eigenstates it is and then it is one of the eigenvalues. Energy formula in terms of other operators such as position and momentum does not follow from any general definition, it has to be defined by us by some choice out of the infinity of possibilities that makes working with the concept reasonable.

It's like in classical mechanics, there is infinity of Hamiltonians that describe the same system and evolution, just in different coordinates. But not all Hamiltonians have value that is conventional energy: energy should obey additional conditions, such as it is sum of kinetic and potential energy (with the latter having conventional zero in infinity or elsewhere if explicitly stated). This usually selects one Hamiltonian (at least in single fixed system of canonical variables) whose value we also can call energy, and the other possible Hamiltonians that have different values will differ from energy so can't be then called energy.

Usually by energy we mean values induced by the standard Hamiltonian operator $T+V$, where $T$ is kinetic energy operator, and $V$ is potential energy of the system.

When the Hamiltonian contains a time-dependent interaction term with external field, e.g.: $$ H= H_0 - q\mathbf r\cdot \mathbf E(t) $$

then this time-dependent term is usually not part of energy of the system (at least not when we treat it as a quickly changing perturbation). Then we stick to $H_0$ when defining eigenvalues and expected average energy of the system.

However, if the time-dependence in the Hamiltonian is very slow (some parameter such as $E(t)$ changes slowly enough), then it is better to take this time-dependent term as part of the Hamiltonian that defines energy (often the potential energy part). Thus, for example, Stark effect is analyzed assuming the potential energy function in $H_0$ depends on $E$, and dependence of $E$ on $t$ is neglected when calculating eigenvalues. This choice causes the $H_0$ eigenvalues to depend on $E$, and when $E$ depends on time, also the eigenvalues depend on time.


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