I dreamed a little dream

Do I have any reader left?

After abandoning my blog for around a year and a half I felt the need to share something, but I was not sure what. In the last year and a half I discover such beautiful physics and math, but I was not sure what I wanted to share. Some months ago I finished my bachelor in Physics, so I decided to share one of the main reasons that convinced me to study this or, better said, one of the main problems: Here I present you:

Laser cooling of atoms

Some years ago I was participating in a contest to represent Mexico in the International Physics Olympiad. At the same time I was trying to decide what to do about my university; I was not sure if I wanted to study either math or physics. The latter was interesting from time to time, but I hadn’t found something that I would call amazing or beautiful at that time.

During our trainings, one of the professors brought a bunch of problems coming from different physics competitions around the world for us to try and solve. This was not an easy task and I have to say that at the moment I was not doing as well as I expected, but then I saw one of them that sounded a bit weird for me. It was the first theoretical question of the 7th Asian Physics Olympiad (APhO 2006). The problem dealt with a simple model for laser cooling of atoms (You can check the original problem in this link). A Nobel prize was awarded to Steven Chu, Claude Cohen-Tannoudji and William D. Phillips on 1997 “for development of methods to cool and trap atoms with laser light”.

The idea is fairly simple: You take two laser propagating in opposed directions and focus them into a sample of gas. The frequency of the laser is tuned to be near resonance with the first-level transition of the molecules in the gas. The photons coming from the beams “hit” the molecules of gas but, due to a relativistic effect, the “hits” have a preference to oppose to the movement of the molecules. The effect of this is some kind of damping force that reduces the speed of the molecules in the gas, thus reducing the temperature. The first experiments reached the micro-Kelvin scale, but if I remember correctly, I think now we are on the nano-Kelvin league.

Here, I write a simplified analysis of this interesting situation based on the problem I talked you about. The statements of the original problems will be written in black/gray while my solution will be in a different color. Here we go:

Theoretical Introduction

Consider a simple two-level model of the atom, with ground state energy E_g   and excited state energy E_e . The energy difference is E_e-E_g=\hbar \omega_0 ; the angular frequency of the laser used is \omega , and the laser detuning is  \delta=\omega-\omega_0<<\omega_0 . Assume that all atom velocities satisfy v << c , where с is the light speed. You can always restrict yourself to first nontrivial orders in the small parameters v/c and \delta / \omega_0 . The natural width of the excited state due to spontaneous decay is \gamma << \omega_0 : for an atom in an excited state, the probability to return to a ground state per unit time equals \gamma . When an atom returns to a ground state, it emits a photon of a frequency close to \omega_0 in a random direction.

It can be shown in quantum mechanics, that when an atom is subject to low-intensity laser radiation, the probability to excite the atom per unit time depends on the frequency of radiation  \omega_a in the reference frame of the atom, according to:

\displaystyle \gamma_p=s_0 \frac{\gamma/2}{1+4(\omega_a-\omega_0)^2/\gamma^2}<<\gamma

where s_0 is a parameter depending on the properties of the atom and the laser intensity.

2-level

In this problem properties of the gas of sodium atoms are investigated neglecting the interactions between the atoms. The laser intensity is small enough, so that the number of atoms in the excited state is always much smaller than the number of atoms in the ground state. You can also neglect the effects of the gravitation, which are compensated in real experiments by an additional magnetic field.

Numerical values

Planck constant \hbar=1.05 \times 10^{-34} J s
Boltzmann constant k_B=1.38 \times 10^{-23} J K^1
Mass of sodium atom m=3.81 \times 10^{-26} kg
Frequency of the transition used \omega_0=2 \pi (5.08 \times 10^{14} Hz)
Excited state linewidth \gamma=2\pi (9.80 \times 10^6 Hz)
Concentration of the atoms n=10^{20} m^{-3}

Questions

a) Suppose the atom is moving in the positive x direction with the velocity v_x, and the laser radiation with frequency \omega is propagating in the negative х direction.  What is the frequency of radiation in the reference frame of the atom?

Since the photon is moving towards the atom, the atom will perceive a higher frequency  due to Doppler effect which, to first order approximation, will be given by:
\displaystyle \omega_a=\omega \sqrt{\frac{1+\beta}{1-\beta}}=\omega \sqrt{(1+\beta)^2}=\omega (1+\beta)

Here I consider the coefficient \beta=\frac{v}{c} as a shorthand. Notice also that I used the Taylor approximation: (1+x)^n=1+nx for x<<1 ; I’ll keep doing this approximation every once in a while #SorryNotSorry.

b) [2.5 Points] Suppose the atom is moving in the positive x direction with the velocity v_x , and two identical laser beams shine along х direction from different sides. Laser frequencies are \omega, and intensity parameters are s_0 . Find the expression for the average force F(v_x) acting on an atom. For small υ х this force can be written as F(v_x)= - \alpha v_x . Find the expression for \alpha . What is the sign of \delta =\omega - \omega_0 , if the absolute value of the velocity of the atom decreases? Assume that momentum of an atom is much larger than the momentum of a photon.

Notice that even though the situation looks quite symmetric, the fact that the molecules perceive a higher frequency when they are approaching to the photons,  this has an influence on the probability of interaction \gamma_p . For a particle moving to the right, we can estimate that the number of photons that hit it in a time lapse \Delta t is given by \gamma_p \Delta t . Each of this punches will induce a change in momentum with a magnitude of \frac{\hbar \omega_0}{c} . Considering the opposite directions of propagation for the laser, we see that the force is given by:

\displaystyle F_x(\beta)=-\frac{\hbar \omega_0}{c} \frac{s_0 \gamma}{2}\left( \frac{1}{1+4(\omega (1+\beta)-\omega_0)^2/\gamma^2}-\frac{1}{1+4(\omega (1-\beta)-\omega_0)^2/\gamma^2} \right)

\displaystyle F_x(\beta)=-\frac{\hbar \omega_0}{c} \frac{s_0 \gamma}{2}\left( \frac{1}{1+4(\delta+\omega \beta)^2/\gamma^2}-\frac{1}{1+4(\delta-\omega \beta)^2/\gamma^2} \right)
 
Taking advantage again of the Taylor approximation and the fact that \beta <<\delta/ \omega_0 , we can rewrite this as follows:
\displaystyle F_x(\beta)=-\frac{\hbar \omega_0}{c} \frac{s_0 \gamma}{2}\left( \frac{1}{1+\frac{4\delta^2}{\gamma^2}(1+\frac{\omega \beta}{\delta})^2}-\frac{1}{1+\frac{4\delta^2}{\gamma^2}(1-\frac{\omega \beta}{\delta})^2} \right)
 \displaystyle F_x(\beta)=-\frac{\hbar \omega_0}{c} \frac{s_0 \gamma}{2}\left( \frac{1}{1+\frac{4\delta^2}{\gamma^2}(1+\frac{2\omega \beta}{\delta})}-\frac{1}{1+\frac{4\delta^2}{\gamma^2}(1-\frac{2\omega \beta}{\delta})} \right)
 \displaystyle F_x(\beta)=-\frac{\hbar \omega_0}{c} \frac{s_0 \gamma}{2}\left( \frac{-\frac{16\omega \delta}{\gamma^2}\beta}{\left( 1+\frac{4\delta^2}{\gamma^2} \right)^2- \left( \frac{8\omega \delta}{\gamma^2}\beta \right)^2} \right)
In the last expression we can neglect the quadratic term on \beta since we are dealing with a non-relativistic approximation. Also, since we obtain a quadratic term on the frequency, we can approximate \omega=\omega_0 , yielding the linear approximation:
 \displaystyle F_x(\beta)=\frac{8\hbar \omega_0^2 \delta s_0}{c\gamma\left( 1+\frac{4\delta^2}{\gamma^2} \right)^2} \beta
 Thus, the desired coefficient is given by:
\displaystyle \alpha=-\frac{8\hbar \omega_0^2 \delta s_0}{c^2\gamma\left( 1+\frac{4\delta^2}{\gamma^2} \right)^2}

This force corresponds to the damping force I was talking about before, but for it to have an actual damping effect, we need the coefficient \alpha to be positive, implying that \delta < 0  .

In what follows we will assume that the atom velocity is small enough so that one can use the linear expression for the average force.

c) If one uses 6 lasers along х, у and z axes in positive and negative directions, then for \alpha > 0 the dissipative force acts on the atoms, and their average energy decreases. This means that the temperature of the gas, which is defined through the average energy, decreases. Using the concentration of the atoms given above, estimate numerically the temperature T_Q , for which one cannot consider atoms as point-like objects because of quantum effects.

A particle can be consider to be point-like as long as the uncertainty of its position is smaller than a characteristic length of the system length. For this case, this characteristic length is given by the average separation between particles n^{-1/3} , once the uncertainty  of the position is larger than this separation I won’t be possible to distinguish classically one atom from another one. Recalling the Uncertainty principle we know that \Delta x \sim \frac{\hbar}{\Delta p_x} .

From thermodynamics we can express the average kinetic energy per atom as simple function of the temperature T; considering an isotropic situation we end up with \frac{1}{2}k_BT=\frac{1}{2m}p_x^2 . Putting together this equations we have:

\displaystyle n^{-1/3} = \frac{\hbar}{ \sqrt{m k_B T_Q} } \Rightarrow T_Q = \frac{\hbar^2 n^{2/3}}{m k_B} = 4.5 \times 10^{-7} K \sim 10^{-6} K

This is already freaking cold, but don’t worry, we don’t go that far in this treatment, stay classical… kinda. Notice also that the right-hand side of the equation I wrote for the uncertainty principle coincides with the de Broglie wavelength. Indeed, another way to have the same result is to consider that, as the momenta of the particles decrease, its associated wavelength gets larger, until it reaches the same size as the average separation, however this argument yields the same result.

In what follows we will assume that the temperature is much larger than T_Q and six lasers along х, у and z directions are used, as was explained in part с).
In part b) you calculated the average force acting on the atom. However, because of the quantum nature of photons, in each absorption or emission process the momentum of the atom changes by some discrete value and in random direction, due to the recoil processes.

d) Determine numerically the square value of the change of the momentum of the atom, (\Delta p)^2 , as the result of one absorption or emission event.

This follows directly from the photon momentum ( \Delta p)^2= \left( \frac{\hbar \omega_0 }{c} \right)^2 = 1.26 \times 10^{-54} J m

e) Because of the recoil effect, average temperature of the gas after long time doesn’t become an absolute zero, but reaches some finite value. The evolution of the momentum of the atom can be represented as a random walk in the momentum space with an average step < ( \Delta p )^2 > , and a cooling due to the dissipative force. The steady-state temperature is determined by the combined effect of these two different processes. Show that the steady state temperature T_d is of the form: T_d=\frac{\hbar \gamma}{4k_B} \left(x + \frac{1}{x} \right) . Determine x . Assume that T_d is much larger than \frac{< ( \Delta p) ^2 >}{2k_B m} .

Lets assume that we are already in the steady-state temperature with an average momentum of P_d . After a certain time t the atom will go through N process of absoption-emission that may excite the movement of the atoms. However, the atoms will feel a damping force that relaxes them back a steady value. The temperature that we are looking for will be characterized by the equilibrium state of this two process.

For each absorption-emission process, we will have a change of the square mometum by the same quantity as we calculated before. Also notice that, since there are 6 lasers interacting with the atoms N=6\gamma_p t . Since each of those nudges are uncorrelated to the others we can consider its contributions to the mean square of the momentun as independent, yielding an increase of 2N < ( \Delta p )^2 > = 12 \gamma_p t  \left( \frac{\hbar \omega_0 }{c} \right)^2 . The extra factor two comes from the fact that \gamma_p <<\gamma , which says that the atom does not stays excited for a long time before reemitting the absorbed photon, receiving second kick.

To study the effect of the damping force, consider a particle moving with an average velocity of \frac{ P_d }{ m } during a short time t . During this time it will have a displacement of  \frac{ P_d t }{ m } feeling an opposing force of \alpha\frac{P_d}{m} as we showed in question b). The work done on the particle yields a change in the kinetic energy of around -\alpha\frac{P_d^2}{m^2}t .

Putting together both contribution for the square mean momentum, we can express the following steady-state condition:

\displaystyle 12 \gamma_p  \left( \frac{\hbar \omega_0 }{c} \right)^2 = 2\alpha\frac{P_d^2}{m}

To simplify the analysis we can consider that \gamma_p = \frac{s_0 \gamma}{2}\frac{1}{1+\left( \frac{2\delta}{\gamma} \right)^2 } . This leads to:

\displaystyle P_d^2= \frac{6 \gamma_p m}{\alpha} \left( \frac{\hbar \omega_0 }{c} \right)^2 = \frac{6 \gamma_p m c^2\gamma}{8\hbar \omega_0^2 \delta s_0} \left( 1+\frac{4\delta^2}{\gamma^2} \right)^2 \left( \frac{\hbar \omega_0 }{c} \right)^2 = \frac{3 \gamma_p \hbar m \gamma}{4 \delta s_0} \left( 1+\frac{4\delta^2}{\gamma^2} \right)^2

\displaystyle P_d^2= \frac{3 \hbar m \gamma}{4 \delta s_0} \frac{s_0 \gamma}{2}\frac{1}{1+\left( \frac{2\delta}{\gamma} \right)^2 } \left( 1+\frac{4\delta^2}{\gamma^2} \right)^2 = \frac{3 \hbar m \gamma^2}{8 \delta } \left( 1+\frac{4\delta^2}{\gamma^2} \right)= \frac{3 \hbar m \gamma}{4 } \left( \frac{\gamma}{2\delta}+\frac{2\delta}{\gamma} \right)

With the same argument as before we see that this yields a temperature of:

\displaystyle T_d= \frac{ \hbar \gamma}{4 \delta s_0} \frac{s_0 \gamma}{2}\frac{1}{1+\left( \frac{2\delta}{\gamma} \right)^2 } \left( 1+\frac{4\delta^2}{\gamma^2} \right)^2 = \frac{3 \hbar m \gamma^2}{8 \delta } \left( 1+\frac{4\delta^2}{\gamma^2} \right)= \frac{3 \hbar m \gamma}{4 k_B} \left( \frac{\gamma}{2|\delta|}+\frac{2|\delta|}{\gamma} \right)

where we find x=\frac{\gamma}{2|\delta|} .

f) Find numerically the minimal possible value of the temperature due to
recoil effect. For what ratio \frac{\delta}{\gamma} is it achieved?

From the last expression it is easy to see that the minimum is reached when x=1 , implying \delta=-\gamma/2 , with a value of T_d =  \frac{3 \hbar m \gamma}{2 k_B}=2.35 \times 10^{-4} K . We have actually freeze those atoms down.

A little epilogue.

After trying to solve this problem I felt in love. There were many things I didn’t understood at that time, for example, the “natural width” of the excited state or where did they got the expression for the probability of interaction. Some years latter this has come a little bit clearer, but it still amazes me how a such a simple idea leads to so counterintuitive results. It was, for the first time a little mix of cute math with some misterious physics giving rise to unexpected outcomes, that they I thought that, maybe, there was something for me there. That night I dreamed a little dream of small spheres slowing down like inmersed in some kind of shiny honey, later I knew that physics even has a name for this picture, they call them optical molasses.
Some months later I discover that a professor in my future university had done the first laser cooling trap in Mexico some year ago , just after the Nobel prizes I mentioned above were awarded and for many years he was one of the few that was interested on this in my country.  Rodolfo Rodriguez  y Masegosa later became my optics professor and kept inspiring me to find those beautiful curiosities hidden out there that make you dream of unconscionable realities.
trap

Rubidium atoms trap designed and constructed by Rodolfo Rodriguez y Masegosa

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