How to simplify second-order derivative of Ket in Dirac notation?

Question

I am currently playing around with Dirac notation in the context of interband transitions and came across a second derivative of a Ket. Under what conditions will this second derivative be zero: $langle m | ddot{n} rangle = 0$? Or, what other simplifications can I make?
I am used to seeing people drop second-order terms in expressions, but I am not sure about that here. Maybe I can simplify by considering only linear $H$, and inserting the identity $langle m | dot{n} rangle =frac{langle m | dot{H} | n rangle}{(E_n-E_m)}$? But it's not clear to me.
For a 2-band model, $|mrangle$ and $|nrangle$ are eigenstates from $H|mrangle=E_m|mrangle$, and the second derivative is with respect to some other parameter (such as momentum) that is not time.

JoshuaTS · Accepted Answer

From the Schroedinger equation
$$frac{partial}{partial t}|psirangle=-frac{i}{hbar}H|psirangle.$$
Therefore,
$$frac{partial^2}{partial t^2}|psirangle=-frac{i}{hbar}frac{partial}{partial t}(H|psirangle)=-frac{i}{hbar}dot{H}|psirangle-frac{i}{hbar}Hleft(frac{partial}{partial t}|psirangleright)=-frac{i}{hbar}dot{H}|psirangle-frac{1}{hbar^2}H^2|psirangle.$$
I'm not entirely sure what your $|mrangle$ and $|nrangle$ are (are they eigenstates of some time-independent part of the Hamiltonian?). Hopefully, this is enough to get you on the right track.

How to simplify second-order derivative of Ket in Dirac notation?

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