Consider the emission of a photon when an atom decays from an excited state to its ground state. In most cases, this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction. However, quantum optics tell us that photons are essentially associated with the amplitude of plane wave modes forming an electromagnetic field. Therefore, the small wave-packet representation seems rather suspicious...
What does the emitted photon really "look" like then ?
Assuming that the emitted photon has a corresponding wavelength much larger than the size of the atom, the atom should essentially behave as a small electric dipole. Therefore, considering lifetime broadening, it seems to me that the photon should propagate away from the atom as a pulsed dipolar emission. Is this correct ?
Answer
The photon comes out with a direction and a polarization that have a particular probability distribution. The rate of spontaneous emission per solid angle as the atom transitions from initial state $i$ to final state $f$ is
$$\frac{dw_{if}}{d\Omega}=\frac{\alpha\omega_{if}^3}{2\pi c^2}\sum_{j=1,2}|\vec{\epsilon}_j\cdot\vec{d}_{if}|^2$$
where $\alpha$ is the fine structure constant, $\omega_{if}$ is the frequency of the transition, $c$ is the speed of light, $\vec{\epsilon}_j$ are two orthogonal polarization unit vectors perpendicular to the direction of emission, and $\vec{d}_{if}$ is the matrix element of the dipole moment between the initial state $i$ and the final state $f$.
The dependence on direction and polarization is probably consistent with the flux of energy and momentum from a classical oscillating electric dipole, but I haven’t checked that.
I think of the emitted photon as a point particle (but with a polarization vector) traveling in a straight line from the atom to the measuring device. It is simply more likely to come out in some directions than others, and to prefer certain polarizations. But everybody has their own picture of what is "really" happening in quantum mechanics.
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