Why can one equate the the zeroth order coefficient with the initial state in time-dependent perturbation theory in quantum mechanics?

Question

Setup

In the typical treatment of time-dependent perturbation theory in quantum mechanics, one arrives at the set of equations

$$ i \dot{a}^{(r + 1)}_m(t) = \sum_n \langle m |H_1(t)|n \rangle e^{i \omega_{mn} t} a^{(r)}_m(t) $$ for integers $r > 1$ and $\dot{a}^{(0)}_m(t) = 0$. Here, the Hamilton is $H(t) = H_0 + \lambda H_1(t)$, the states $\{|n\rangle\}$ satisfy $H_0 |n\rangle = \varepsilon_n |n\rangle$ and $\omega_{mn} = \varepsilon_m - \varepsilon_n$. The expansion coefficients are defined by $$ a_m(t) = \sum_{r = 0}^{\infty} \lambda^r a^{(r)}_m(t) $$ where $a_m(t)$ are the coefficients in the expansion $$ |\psi(t)\rangle = \sum_n a_n(t) e^{-i \varepsilon_n t} |n\rangle $$ where $$ i \frac{d |\psi(t)\rangle}{dt} = H(t) |\psi(t)\rangle $$ Note then that $a^{(0)}_m(t) = a^{(0)}_m(0) = a^{(0)}_m$, say.

Question

The literate then claims that if $|\psi(0)\rangle$ is known, then $a^{(0)}_m = \langle m |\psi(0)\rangle$.

Does this really follow from the above? I feel an additional assumption (perhaps a boundary condition on $t$) is required to conclude the association of the initial state with the zeroth order coefficient. (Though obviously we have $a_m(0) = \langle m |\psi(0)\rangle).$

Answer 1

The zeroth order terms in the expansion simply correspond to the solutions for the value $\lambda=0$, the unperturbed system. That's clearly the case because all the other corrections are proportional to positive powers of $\lambda$ (or other, nonperturbative functions of it) which require $\lambda\neq 0$ for them to contribute a nonzero amount.

Now, for $\lambda=0$, you may easily solve the system, and because the basis $|n\rangle$ used in all the formulae above is obviously a basis of eigenstates of $H_0$ – as seen, for example, from the variable $\epsilon_n$ used in one of the formulae (it only makes sense to attribute a single energy to a state if it is an energy eigenstate) – it follows that their evolution in time is very simple and in the notation above, it is simply $a_n(t)=a_n(0)$.

The coefficients are constant as functions of time. The evolution by Schrödinger's equation simply adds the phase $\exp(-i\epsilon_n t)$ to each amplitude but this is already incorporated in the prescription for $|\psi(t)\rangle$ which implies that $a_n(t)$ are just constant (independent of time).