Schrodinger or Interaction picture hamiltonian?

[Pu ZHANG]

Dear all, 

These days I'm learning QuTiP by following some examples in the tutorials or documentation. It seems to me in some examples Schrodinger picture hamiltonian is input (e.g., the JC model), yet sometimes it's the interaction picture hamiltonian that is used (the three-level atom single photon source example). So I feel a bit confused about this. 

Which equation is solved in the behind and which hamiltonian should be considered? Thanks! 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

[nonher...@gmail.com]

Schrodinger or interaction doesnt matter. The time evolution of the state vector or density matrix has the same form in both representations. 

-P

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[Pu ZHANG]

Thanks! But is there any difference in postprocessing? 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

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[nonher...@gmail.com]

Since it is a unitary operator that moves you to the interaction frame there is little difference in general. In terms of qutip, the functions are the same. However, the Hamiltonian in the interaction picture may be time independent providing some advantages. 

-P

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[Pu ZHANG]

Thanks a lot for your explanation! 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

[Pu ZHANG]

I tried to play around with the Hamiltonians, and find the results do change when I use different pictures. Here's one example from the tutorial (but of an old version). The script can be seen from 

http://qutip.org/docs/2.2.0/examples/td/ex-41.html

The Hamiltonian used is 

H = -g * (sigma_ge.dag() * a + a.dag() * sigma_ge) - w * (sigma_ue.dag() + sigma_ue)

which is in interaction picture (if I understand correctly). 

I then add terms to include -w_u * |u><u| - w_g * |g><g|, which bring us to Schrondinger picture. 

The results from the two calculations turn out not identical. How do we understand the difference? 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

On Tue, Aug 23, 2016 at 8:30 AM, <nonher...@gmail.com> wrote:

[nonher...@gmail.com]

The interaction picture states is related to the schrodinger picture states by a known unitary transform. Therefore, if you calculate the evolution of the state vector in the interaction frame you need only to undo the unitary to get back to the schrodinger.This Is   why I said the analysis remains relatively unchanged. You still have a schrodinger equation but under the action of a modified Hamiltonian. It is best to think of the interaction frame as intermediate to the Schrodinger and Heisenberg frames. In this sense you can understand how the evolution varies. 

-P 

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[Pu ZHANG]

I understand that the states are related via a unitary transform between the two pictures, but the outputs (like population, photon number, etc.) are expectation values, which should be the same regardless of the transform. 

In the interaction picture, the energies of the atomic levels are not fed into the code, so the results won't depend on them. However in the Schrodinger picture, the energies are present, and have influence on the results. So to me there seems to be contradictions. Please point me out if I'm wrong about anything. 

Thanks! 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

[Pu ZHANG]

Maybe I should make it clear that I'm actually trying to solve the Lindblad master equation, not pure Schrodinger equation. In this case, what kind of Hamiltonian should one use, with or without terms like \omega_2 * |2><2| - \omega_1 * |1><1|? 

Thanks! 

Best regards, Pu Zhang

--

Faculty at School of Physics, Huazhong University of Science and Technology

Room 819 (N.), Yifu Science and Technology Building

1037 Luoyu Road, Wuhan, China

E-mail: puzha...@hust.edu.cn

Homepage: www.researchgate.net/profile/Pu_Zhang4

Phone: +86 18871860394

On Tue, Aug 23, 2016 at 10:17 PM, <nonher...@gmail.com> wrote: