= Solution
Insert the expansion in the time-dependent <Schrödinger equation>, cancel the unperturbed evolution, and project onto $\langle\psi_n|$. With $V_{nm}(t)=\langle\psi_n|V(t)|\psi_m\rangle$,
$$
i\hbar\dot c_n(t)=\sum_m e^{i(\omega_n-\omega_m)t}V_{nm}(t)c_m(t).
$$
In first-order <time-dependent perturbation theory>, put $c_m(t)=\delta_{m0}$ on the right and use $c_n(-\infty)=0$ for $n\ne0$. Integration gives
$$
\boxed{c_n(t)=\frac1{i\hbar}\int_{-\infty}^t e^{i(\omega_n-\omega_0)t'}V_{n0}(t')\,dt'+O(V^2).}
$$
The initial state must be the specified unperturbed eigenstate, the integral must exist under the switching prescription, and total transition probability must stay small. The diagonal term produces a first-order phase; setting $c_0=1$ in the off-diagonal integral changes those amplitudes only at second order. A weak peak perturbation alone is insufficient if its accumulated action invalidates this expansion.
For the <infinite square well>, retain the source's zero-based labels:
$$
\psi_n(x)=\sqrt{\frac2w}\sin\frac{(n+1)\pi x}{w},\qquad E_n=\frac{\pi^2\hbar^2(n+1)^2}{2m_ew^2},\qquad n=0,1,\ldots.
$$
An electron in a field along $+x$ has perturbation $V(t)=+eE_0xe^{-t/\tau}$ for $t\geq0$. About the well centre, $x-w/2$ is odd, whereas $\psi_0$ and $\psi_2$ are both even. The constant $w/2$ has zero off-diagonal matrix element, so the <first-order parity selection in an infinite well> gives \b[$\langle\psi_2|x|\psi_0\rangle=0$]. Excitation to that state is forbidden at first order, not necessarily at higher orders.
For the first excited state, direct integration or the supplied integral gives
$$
x_{10}=\frac2w\int_0^w x\sin\frac{2\pi x}w\sin\frac{\pi x}w\,dx=-\frac{16w}{9\pi^2},\qquad
\omega_{10}=\frac{3\pi^2\hbar}{2m_ew^2}.
$$
The Fourier amplitude of the <exponentially decaying electric-field pulse> is $\int_0^\infty e^{-t/\tau+i\omega t}dt=\tau/(1-i\omega\tau)$. Thus
$$
c_1(\infty)=\frac{eE_0x_{10}\tau}{i\hbar(1-i\omega_{10}\tau)},\qquad
\boxed{P_{0\to1}=\frac{256e^2E_0^2w^2\tau^2}{81\pi^4\hbar^2\left[1+\left(3\pi^2\hbar\tau/(2m_ew^2)\right)^2\right]}.}
$$
This is the leading, quadratic-in-field transition probability; the total probability of all excitations must remain small.
Starting instead in $\psi_1$, which is odd about the centre, the pulse can transfer amplitude to the ground state and to even-labelled excited states $\psi_2,\psi_4,\ldots$. Other odd-labelled states have zero first-order dipole amplitude. Both downward and upward transitions are possible; their relative probabilities depend on dipole matrix elements and the pulse's spectral weight at the corresponding energy gaps. Most population remains in $\psi_1$, with a small depletion equal to the summed leading transition probabilities. The <common diagonal dipole phase in an infinite well> is $-eE_0w\tau/(2\hbar)$ and has no population effect. After the pulse, the remaining coherent superposition evolves under the unperturbed <Hamiltonian>; no spontaneous relaxation is implied by this closed-system model.
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