# What is better than time-dependent perturbation theory if the pointer states aren't energy eigenstates?

Time-dependent perturbation theory works excellently if the interaction is weak and the pointer states are approximately energy eigenstates. However, what if the pointer states are not remotely energy eigenstates? Then, the pointer states themselves evolve in time nontrivially by more than just a phase rotation.

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If the pointer states form a complete basis of the Hilbert space, any function can be expressed in the form $\Psi(\vec{x},t)=\Sum\,a(t)\psi(\vec{x})$. In the case you describe, it will just be a grotesque decomposition, rather than a simple one where the largest terms are the first few couple of them, and there will be no simple relationship of the basis states to the Hamiltonian. – Jerry Schirmer Aug 11 '12 at 15:10
@JerrySchirmer: Your formula should read $\Psi(t)=\sum_y a(y,t) \psi(y)$. – Arnold Neumaier Aug 11 '12 at 15:51
Yes, of course. I don't know why the LateX broke there. – Jerry Schirmer Aug 11 '12 at 16:09
Perturbation theory has nothing to do with pointer states, that's for measurement. The perturbation theory is on the level of the system only, not the pointer, and I don't understand the level-mixing. What's the question? Is the system not in an energy eigenstate? – Ron Maimon Aug 12 '12 at 21:25
It's the Redfield equation. See cmmp.ucl.ac.uk/~ajf/course_notes/node42.html – user11433 Aug 16 '12 at 13:21