The question: whenever you are dealing with quantum optics you will encounter a lot of unitary transformations very soon. A question that is in my mind for quite a long time is: Why does the discussion in a certain reference frame gives a global result that can be observed in the lab?

What I mean by that: Normaly, before starting discussing a physical system quantum physicists often happily transform the Hamiltonian with all kind of transformations until it looks simple enough to discuss it with the tools of linear algebra. Normally one uses simply the interaction picture: $ \tilde{H} = U H U^{\dagger}$ or a more sophisticated approach that conserves the Schrödinger Equation: $ \tilde{H} = U^{\dagger} H U + i \hbar \dot{U} U^{\dagger} $ for time-dependent transformations. I am aware that a unitary transformation is simply a change of reference frame which are equitable. But looking at it from a general relativity view, physics is not the same in accelerated frames. However a transformation, like the one in a rotating frame with $U = e^{i\omega |e\rangle \langle e| t}$ can be associated with transforming into an accelerated frame, can't it? So why does the physics that pops up at the end can be treated as a general result?

An example Consider the rabi oscillations happining to a two-level atom under the impact of a single mode "classical" light. We start with the dipole Hamiltonian, do some unitary transformation here and there and at the end we get a easy 2x2 matrix that describes the system and as a conclusion the rabi oscillations. But isn't that only the physics in the reference frame we transformed in?

Thanks for your hints ;) Philipp

  • $\begingroup$ I'm not sure that you can really interpret the unitarty transformations as "changing frames" in its most general sense. I need to think, but my guess is that all you're doing is switching from inertial frames to inertial frames, so the physics is "morally" the same. Indeed, the preservation of probability has no reason to hold in accelerated frames, does it? $\endgroup$ – sure Oct 3 '15 at 10:16
  • $\begingroup$ "Every unitary transformation transforms one basis into another" Source If you take a look at the proof I don't see any reason why the basis vectors shouldn't be accelerated in respect to the initial frame and therefore non-inertial. $\endgroup$ – P. Aumann Oct 3 '15 at 11:58
  • $\begingroup$ The point is, you're staying in the same Hilbert space and are just doing a change of basis. There's no reason that the physics of an accelerated observer is described by the same Hilbert space, nor that the representation of the Hamiltonian to be equivalent to the one of an inertial frame. $\endgroup$ – sure Oct 3 '15 at 12:10
  • $\begingroup$ Thanks for your answer! Let me again draw the parallel to relativity: In special relativity if we switch between observers we always stay at the same "Minkowski space" and simply do basis transformations, don't we? Also very similar in general relativity, only that we don't have a minkowski but a schwarzschild-metric or sth. similar. (If I remember correctly) So isn't that the same as staying in the same Hilbert-space but only changing the basis? $\endgroup$ – P. Aumann Oct 3 '15 at 12:33

Unless they are representations of the Lorentz group, unitary transformations are not a "change of reference frame". They are, however, changes of the basis we use for the Hilbert space.

All physically relevant things, like expectation values and transition probabilities, are invariant under unitary transformations.

So what physics you derive in one basis of the Hilbert space is valid in all of them.

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  • $\begingroup$ Thanks for your answer! It seems I was somewhat misled by the phrase of a "rotating frame" (when doing the Rotating Wave Approximation) since it seems that it's not a different frame. $\endgroup$ – P. Aumann Oct 3 '15 at 13:33

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