Shortest possible damping time in Optical Molasses

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SUMMARY

The discussion focuses on calculating the shortest possible damping time for potassium atoms in optical molasses after being slowed by a Zeeman slower. The relevant transition is $$4s^2S_{1/2} --4p^2P_{3/2}$$ with a wavelength of $$\lambda = 767nm$$ and a laser intensity of $$I=0.75W/m^2$$. The damping constant $$\alpha$$ is derived from the scattering force equations, and the damping time is expressed as $$\tau = \frac{M}{2\alpha}$$. Key challenges include determining the saturation intensity and understanding the implications of detuning in the context of the damping coefficient.

PREREQUISITES
  • Understanding of Zeeman slowing and its application in atomic physics.
  • Familiarity with optical molasses and laser cooling principles.
  • Knowledge of scattering force equations and their derivations.
  • Basic concepts of atomic transitions and resonance conditions.
NEXT STEPS
  • Calculate the saturation intensity for the $$4s^2S_{1/2} --4p^2P_{3/2}$$ transition in potassium.
  • Explore the effects of detuning on the damping coefficient $$\alpha$$ in optical molasses.
  • Investigate the relationship between laser intensity and the friction coefficient $$\alpha$$ in laser cooling scenarios.
  • Review experimental setups for measuring damping times in optical molasses using potassium atoms.
USEFUL FOR

Physicists, particularly those specializing in atomic and laser physics, as well as researchers working on laser cooling techniques and optical trapping of atoms.

Poirot
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Homework Statement


A beam of $$^{39}K$$ atoms is slowed with a Zeeman slower with laser light tuned to the $$4s^2S_{1/2} --4p^2P_{3/2}$$ transition with $$F=2, m_F=2 -> F'=3,m_{F'}=3 (\lambda =767nm)$$ Nuclear spin $$I=3/2$$. The most probably velocity the atoms escape from the oven with is $$v_0 =450m/s$$ and the lifetime of the excited state is $$26nm$$.

The first few parts of the question ask for the stopping distance, maximum possible scattering force, and the number of scattering events taking place in the time it takes for the atom to slow down.

The bit I'm stuck on:

After Zeeman slowing, atoms are now placed in an optical molasses. What is the shortest possible damping time in the absence of heat due to fluctuations if the laser intensity is $$I=0.75W/m^2?$$ (Note $$I/I_{sat} << 1).$$

Homework Equations


$$F_{scatt} = \frac{\hbar k\Gamma}{2} \frac{I/I_{Sat}}{1+ I/I_{Sat} +(\frac{2\delta}{\Gamma})^2}$$
$$F_{molasses} = -2kv \frac{\partial F_{scat}}{\partial \omega}=-\alpha v$$

where
$$\alpha = 4\hbar k^2 \frac{I}{I_{sat}} \frac{-2\delta/\Gamma}{(1+(2\delta/\Gamma)^2)^2}$$
is the damping constant.

the damping time is:
$$\tau = \frac{M}{2\alpha}$$

The Attempt at a Solution


I've completed the first few parts of the question with relative ease. The issues I have are arising from calculating the damping coefficient. I also think that due to the values given in the question (which I haven't had to use most of yet) .

There's a few things I have deduced from my lecture notes such as

$$\delta = \omega -\omega_0 +kv = \frac{\mu}{\hbar} B(z)$$
where ω is laser frequency, ω0 is atomic resonance, k is wavenumber, v is the doppler shift, and the magnetic field B is there to counteract the doppler shift to stay on resonance.
$$B(z)= B_{bias} + B_0(1-\frac{z}{L_0})^{1/2}$$ where
$$ B_{bias} = \frac{\hbar}{\mu}(\omega -\omega_0)$$
which I think can be taken as 0 since it's on resonance so ω = ω0? And with the 1-z/L0, does z=L0 since the molasses comes after we've already Zeeman slowed? But with these assumptions I think this makes the B field zero and so all the preamble of the question is unnecessary so something's probably wrong.

I also need to calculate the initial magnetic field $$B_0 = v_0 \hbar k/\mu$$ where $$\mu=(g_F'm_F' - g_Fm_F)\mu_B$$ can be calculated hyperfine lande factor.

I think all of this takes care of the detuning δ in the equation of the damping coefficient but I'm not sure if I'm barking up the wrong tree entirely.

I also have that $$\tau = \frac{1}{\Gamma}$$ where tau is the lifetime of the state and is given, so this takes care of the gamma in the equation.

I'm now confused as the how I find the saturation intensity? I think this is the only value unknown now (if my assumptions are correct). I haven't calculated anything yet because my knowledge isn't there yet so I fear it would be a waste of time.

Thanks in advance for any help!
 
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This article helped me with gaining a somewhat practical feel for what these variables actually represent in real life (albeit "lab life" haha).
https://cuax.mit.edu/apwiki/Optical_Molasses

Usually if there are gaps in what you've been given vs what you need for your calcs, assumptions can be based on the properties/ nature /definition of the "stuff" mentioned in a problem statement, e.i. optical molasses and potassium. Other assumptions could also be preferences of your lecturer or the textbook for a course, or simply an "industry habit" :-) It doesn't hurt to ask ourselves "what would I have done if I had been the first person on Earth to work this out?" :-)

For example they write...
α describes the viscosity imparted by the light force to the atom, reflecting the restoring force applied when the atom is not at zero velocity. This configuration is known as an optical molasses, because of this restoring force, which makes the light behave like a thick, viscous medium for the atoms in it.

and

Cooling at high intensities
Let us consider an example, of laser cooling at high laser intensities. Keep in mind that laser cooling works because F=−αv and α>0. Assume we have a detuning of about one linewith, δ=−Γ. Now plot the friction coefficient α as a function of intensity:

408px-Chapter1-intro-to-cooling-Lec1-high-inten.png

Initially, at small intensities, α increases as a function of intensity. Don't be confused by the fact that the Doppler limit is achieved at low intensities. The diffusion coefficient is also linear in intensity at low intensity. α increases with I/I0 at first, and peaks around 0.5, but above the saturation limit α actually changes sign and starts heating. When α<0 then, counter-intuitively, blue detuned light can be used to cool atoms.


I'm not sure if your question specified one-dimensional or three-dimensional optical molasses but it covers both.

I also found the graphs and tables in these two papers quite helpful. They are specific to potassium.
https://arxiv.org/pdf/1310.4014.pdf
http://www.tobiastiecke.nl/archive/PotassiumProperties.pdf
 

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