Geometric Phase vs Dynamic phase On the Wikipedia page for the Adiabatic Theorem (https://en.wikipedia.org/wiki/Adiabatic_theorem), formulas for both the dynamic and geometric phases are given. The dynamic phase is expressed as:
$$\theta_m(t)=-\frac{1}{\hbar}\int_0^tE_m(t')\,\mathrm{d}t'$$
And the geometric phase is expressed as:
$$\gamma_m(t)=i\int_0^t\langle m(t')|\dot{m}(t')\rangle \,\mathrm{d}t'$$
where $|m\rangle$ is previously declared to be a stationary state of the Hamiltonian. However, by the time-dependent Schrodinger equation, I would think that $\langle m(t)|\dot{m}(t)\rangle\equiv E(t)$.
In fact, a review article by Franscisco De Zela (https://cdn.intechopen.com/pdfs/29588.pdf) defines this exact quantity to be the dynamic phase, even without requiring $|m\rangle$ be a stationary state. Under what conditions is this integral considered to be the geometric phase?
 A: The equation $$\langle m(t)|\frac{\mathrm{d}}{\mathrm{d}t}|m(t)\rangle\overset{!}=E_m(t)$$ mentioned in the question is
not true in general. I suspect you evaluated that by considering $\mathrm{i}\hbar\partial_t|m(t)\rangle\overset{!}=H(t)|m\rangle$ and $E_m=\langle m|H|m\rangle$. This is incorrect because $|m(t)\rangle$ are not defined to be solutions to the time-dependent Schrödinger equation. They are defined to be "instantaneous eigenstates": they satisfy the TISE for a Hamiltonian $H(\mathbf{R})$ where $\mathbf{R}$ is a set of parameters (like strengths of external applied fields or distances between atomic nucleii in the system) at a fixed time. This is stated by equation (2) in the reference linked in the question: $$H(\mathbf{R})|m(\mathbf{R})\rangle =E_m(\mathbf{R})|m(\mathbf{R})\rangle.$$ The time dependence emerges when you allow the parameters that are the components of $\mathbf{R}$ to vary slowly.
Moreover, it's possible to see why we should not expect $|m(\mathbf{R}(t))\rangle$ to satisfy the time-dependent Schrödinger equation: the time evolution that is implicit when we simplify the TDSE to the TISE (tacking on a phase like $\exp(-iE_mt/\hbar)$) is derived from the assumption that the wave function is separable.
The derivations of the equations for the dynamical phase and the geometric phase are well derived in numerous sources (particularly Girvin and Yang's Modern Condensed Matter Physics, Chapter 13); they're actually found by guessing that the solution to the TDSE might be expressed neatly in terms of the instantanous eigenstates. Specifically, for $\theta$ and $\gamma$ defined in the question, as long as the system is slowly varying and transitions between levels can be neglected,$$|\Psi_m\rangle=\mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{i\gamma}|m\rangle\tag{*}$$ is a solution to the TDSE.
You can think of the solution expressed in ($*$) intuitively: at a time $t$, when the system's parameters are $\mathbf{R}(t)$, the wave function "looks" like the stationary state corresponding to a static (i.e. not varying in time) Hamiltonian with that set of parameters. However, it picks up important phases: the dynamical phase is the equivalent of the familiar $\exp(i\omega t)$, but it's now an integral because the energy of the $n$th state changes as a function of time. Furthermore, there is a "geometrical phase" that is picked up; you can manipulate the expression for $\gamma$ to find that it does not depend upon the time taken by the system to reach a parameter configuration $\mathbf{R}'$ from an initial configuration $\mathbf{R}_0$, but it does depend upon the path in parameter space that is taken, i.e. it does depend upon the configurations $\mathbf{R}''$ that the system takes before it reaches the configuration $\mathbf{R}'$. Clearly, you cannot simply solve for $|m(t)\rangle$ and call it your solution to the TDSE: the phases it picks up can matter.
