Let's start by looking at the behavior of an atom irradiated by a resonant electromagnetic wave of constant amplitude and frequency $\omega$. For simplicity, let's use hydrogen.
Classical electric field
The electric field polarizes the atom by moving the electron along the axis of polarization. To appropriately account for the discrete energy eigenstates of the electron (
$|1s\rangle$,
$|2s\rangle$,
$|2p\rangle$, etc.) we need to treat it quantum mechanically. So instead of a point particle we need to describe it by its wavefunction. In the ground state it occupies a
$1s$ orbital.
If you, for example, excite it at the frequency resonant with the
$1s \to 2p_x$ transistion, the wavefunction describing the electron is displaced by the electic field along the
$x$ axis. It can then be described by a superposition of the wavefunctions associated with the
$1s$ and
$2p_x$ orbitals.
$$\left| \Psi(t) \right\rangle = \alpha(t) \left| 1s \right\rangle + \beta(t) e^{-i \omega t} \left| 2p_x \right\rangle$$
After some time the electron will be fully in the
$\left| 2p_x \right\rangle$ state. Driving it further at the same frequency
$\omega$ will not inrease its energy (as expected from a harmonic oscillator), but transition it back to the ground state. This is in line with how Einstein described stimulated emission in his famous 1917 paper
Zur Quantentheorie der Strahlung, Phys. Z. 18 121-128 (1917). In §2b) he wrote:
If a Planck resonator [the atom] is in a radiation field, the energy of the resonator can be changed by the transfer of energy from the electromagnetic field to the resonator; this energy can be positive or negative depending on the phases of the resonator and of the oscillating field.
Better than any wordy description is a visualization of the wavefunctions, like
this applet. I got a pretty good intuition for this process by just watching the electron undergoing many
Rabi cycles.
The movement of the electron creates an electric field by itself, which interferes with the incident beam. On the transmission side this interference is destructive, so that a weak, strongly focused beam can be completely blocked.
Single photon
Since a single photon holds only a finite amount of energy you need to care about its spatial extend. Typically, single photons are created by other atoms, which in free space emit in all directions. For simplicity, let's not care about this and assume that we managed to transform its emission pattern into a
Gaussian beam focused on our observed atom.
The temporal envelope of a photon created by spontaneous emission is a decaying exponential, but the excitation probability is highest, when it's a rising exponential. In fact,
if the photon temporally and spatially matches a time-reversed spontaneously emitted photon it would be absorbed with a probability of $100\,\%$. In that case the photon is completely gone at the moment the absorption is finished.
$$\left| 1s \right\rangle_{\text{electron}} \left| 1 \right\rangle_{\text{photon}} \rightarrow \left| 2p_x \right\rangle_{\text{electron}} \left| 0 \right\rangle_{\text{photon}}$$
The more realistic case is that the photon does not fulfill the previous conditions and therefore is only partially absorbed, so that the final state is a superposition of the photon being absorbed and having passed the atom:
$$\alpha \, \left| 1s \right\rangle_{\text{electron}} \left| 1 \right\rangle_{\text{photon}} + \beta \, \left| 2p_x \right\rangle_{\text{electron}} \left| 0 \right\rangle_{\text{photon}}$$
One can increase the interaction efficiency
$\beta$ for non-ideal photon shapes by embedding the atom in a suitable environment,
for example in a cavity.
The latter superposition can be collapsed by a measurement. For example if you place a detector after the atom and you detect a photon you know that it must have passed the atom. Since atom and photon are
entangled it also affects the state of the atom. It will be completely in the
$| 1s \rangle$ state then. Else, if you detected no photon (and unrealistically made sure that you would inevitably detect one if it was there) the atom would end up completely in the excited state
$| 2p_x \rangle$.
