Momentum Change from Spontaneous Emission In page 188 of Christopher Foot's book on atomic physics, he talks about the Doppler cooling limit, and how it comes from a random walk caused by spontaneous emission. He claims that each spontaneous photon gives a recoil kick in the z-direction of $\hbar k$, where $k$ is the wave number of the laser light.
I have trouble proving that is true. I know that the momentum from a photon spontaneously decaying should be $\hbar \omega_0 /c$, where $\omega_0$ is the energy level difference. I also know that because the atom is moving we might have to make some adjustment to the momentum in the lab frame. Can someone show me how I can make this adjustment to get to the result of $\hbar k$? Or am I missing something else?
 A: In order to achieve the Doppler cooling limit, the laser is typically tuned half a natural linewidth red of the transition frequency. This makes the difference between $k$ and $\omega_0/c$ negligibly small for most purposes: $\frac{\omega_0-\omega}{\omega} \approx \frac{\textrm{a few MHz}}{\textrm{a few hundred THz}} \approx 1\times 10^{-8}$. Foot might just be ignoring the distinction between these quantities in his treatment.
However, even in a more careful treatment, I believe the spontaneous emission recoil momentum will usually be closer to $\hbar k$ than $\hbar \omega_0/c$. Conservation of energy implies that the change in the atom's kinetic energy is given by the energy of the absorbed photon minus the energy of the emitted photon. Computing the kinetic energy change from the recoil momentum (i.e. $\Delta KE = \frac{(\hbar k)^2}{2 m}$) for the 780 nm cooling transition in rubidium-87, for example, gives a kinetic energy change of $\approx 1.5 \times 10^{-11}\,\mathrm{eV}$ per photon. This is much smaller than the energy difference between the laser photons and the atomic transition at a few MHz detuning, $\approx 1 \times 10^{-8}\,\mathrm{eV}$, so the emitted photon energy/wavenumber in the lab frame will be very similar to the absorbed photon energy/wavenumber.
This result is consistent with the empirical observation that it usually takes at least several hundred photon scatters to bring the velocity of an atom from the maximum "capture" velocity of the cooling laser down to zero. 
A: This is simpler than you think. Consider an isolated atom initially at rest i.e. we choose a reference frame in which the momentum of the atom is zero. Our atom emits a photon with momentum $p$, so to conserve momentum the atom must recoil with a momentum $-p$. So to find the impulse imparted to the atom we just need to find the photon momentum.
It is a standard result that the momentum of any particle is given by:
$$ p = \frac{h}{\lambda} $$
where $\lambda$ is the de Broglie wavelength. If this needs to be proved a convenient way of doing it is to use relativistic expression for energy:
$$ E^2 = p^2c^2 + m^2c^4 $$
For a photon the energy is $E=h\nu = hc/\lambda$ and the mass is zero, and substituting this we get:
$$ \frac{hc}{\lambda} = pc $$
from which the result follows. Your book gives this as $\hbar k$, but since $k=2\pi/\lambda$ and of course $\hbar=h/2\pi$ this is the same expression.
