A Jaynes-Cummings Hamiltonian: Where did the time dependence go?

[yucheng]

TL;DR Summary

Semiclassically, the electric field varies harmonically in time, but sometimes, in the JC Hamiltonian, the time dependence disappears. What???

Consider the interaction of a two level atom and an electric field (semiclassically, we treat the field as 'external' i.e. not influenced by the atom; the full quantum treats the change in the field as well)

Electric field in semiclassical Hamiltonian: plane wave

$H_{int,~semiclassical}=-\mu \cdot E=-\mu \cdot E_{0}\cos \nu t$

Electric field in Jaynes-Cummings Hamiltonian, single mode i.e. plane wave
(Schrodinger picture)

$H_{int}=\hbar g(\sigma _{+}a+\sigma _{-}a^{\dagger })$

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We realize $H_{int}$ is time independent! So where did the time-harmonic
dependence go? How does this compare to the classical case?

Also, how are we supposed to go to the interaction picture, with a constant
hamiltonian?

Furthermore, $H_{int,~semiclassical}$ is time-dependent, but isn't this the
Schrodinger picture Hamiltonian? Shouldn't it be time-independent?

Possibly related: Sakurai and Napolitano, Modern Quantum Mechanics: constant
perturbation turned on at t=0!?

Also see:
https://www.physicsforums.com/threa...hen-the-hamiltonian-is-time-dependent.971007/

 

[Demystifier]

In the JC Hamiltonian, the field is quantized. It's not a semiclassical approximation.
https://en.wikipedia.org/wiki/Jaynes–Cummings_model#Jaynes–Cummings_Hamiltonian_1

 

[f95toli]

It is also possible to add a time-dependent drive term to the J-C Hamiltonian.
However, in the simples case you have a situation where the energy is continuously moving between the cavity and the two-level systems; i.e. it is a closed system and there is no real time-dependence.

Note that the J-C Hamiltonian in the "strong driving" regime gets really complicated.

 

[yucheng]

In the JC Hamiltonian, the field is quantized. It's not a semiclassical approximation.

Yes I understand that it's quantized, but... why must find dependence disappear if it's quantized?

it is a closed system and there is no real time-dependence.

Perhaps this is a good hint. Closed system=energy conservation=no time dependence, but:
How do we know if the system is closed?
Why no time dependence=energy conservation?....

 

[Demystifier]

Yes I understand that it's quantized, but... why must find dependence disappear if it's quantized?

Because the time dependence of operators usually disappears in the Schrodinger picture.