IPhO 2009 · Theory problem 2

Doppler Laser Cooling and Optical Molasses

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The purpose of this problem is to develop a simple theory to understand the so-called “laser cooling” and “optical molasses” phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997.

Figure 1

The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called “optical molasses” because the dissipative optical force resembles the viscous drag on a body moving through molasses.

In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension.

PART I: BASICS OF LASER COOLING

Consider an atom of mass \(m\) moving in the \(+x\) direction with velocity \(v\). For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the \(y\) and \(z\) directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be \(\hbar\omega_0\), where \(\hbar=h/2\pi\). The atom is initially in the lowest state. A laser beam with frequency \(\omega_L\) in the laboratory is directed in the \(-x\) direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy \(\hbar\omega_L\) and momentum \(-\hbar q\). A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the \(+x\) and \(-x\) directions. Since the atom moves at non-relativistic speeds, \(v/c\ll1\) (with \(c\) the speed of light) keep terms up to first order in this quantity only. Consider also \(\hbar q/mv\ll1\), namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities.

Figure 2

Fig.1 Sketch of an atom of mass \(m\) with velocity \(v\) in the \(+x\) direction, colliding with a photon with energy \(\hbar\omega_L\) and momentum \(-\hbar q\). The atom has two internal states with energy difference \(\hbar\omega_0\).

Assume that the laser frequency \(\omega_L\) is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions:

1. Absorption.

1a Write down the resonance condition for the absorption of the photon. 0.2
1b Write down the momentum \(p_{at}\) of the atom after absorption, as seen in the laboratory. 0.2
1c Write down the total energy \(\varepsilon_{at}\) of the atom after absorption, as seen in the laboratory. 0.2

2. Spontaneous emission of a photon in the \(-x\) direction.

At some time after the absorption of the incident photon, the atom may emit a photon in the \(-x\) direction.

2a Write down the energy of the emitted photon, \(\varepsilon_{ph}\), after the emission process in the \(-x\) direction, as seen in the laboratory. 0.2
2b Write down the momentum of the emitted photon \(p_{ph}\), after the emission process in the \(-x\) direction, as seen in the laboratory. 0.2
2c Write down the momentum of the atom \(p_{at}\), after the emission process in the \(-x\) direction, as seen in the laboratory. 0.2
2d Write down the total energy of the atom \(\varepsilon_{at}\), after the emission process in the \(-x\) direction, as seen in the laboratory. 0.2

3. Spontaneous emission of a photon in the \(+x\) direction.

At some time after the absorption of the incident photon, the atom may instead emit a photon in the \(+x\) direction.

3a Write down the energy of the emitted photon, \(\varepsilon_{ph}\), after the emission process in the \(+x\) direction, as seen in the laboratory. 0.2
3b Write down the momentum of the emitted photon \(p_{ph}\), after the emission process in the \(+x\) direction, as seen in the laboratory. 0.2
3c Write down the momentum of the atom \(p_{at}\), after the emission process in the \(+x\) direction, as seen in the laboratory. 0.2
3d Write down the total energy of the atom \(\varepsilon_{at}\), after the emission process in the \(+x\) direction, as seen in the laboratory. 0.2

4. Average emission after the absorption.

The spontaneous emission of a photon in the \(-x\) or in the \(+x\) directions occurs with the same probability. Taking this into account, answer the following questions.

4a Write down the average energy of an emitted photon, \(\varepsilon_{ph}\), after the emission process. 0.2
4b Write down the average momentum of an emitted photon \(p_{ph}\), after the emission process. 0.2
4c Write down the average total energy of the atom \(\varepsilon_{at}\), after the emission process. 0.2
4d Write down the average momentum of the atom \(p_{at}\), after the emission process. 0.2

5. Energy and momentum transfer.

Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom.

5a Write down the average energy change \(\Delta\varepsilon\) of the atom after a complete one-photon absorption-emission process. 0.2
5b Write down the average momentum change \(\Delta p\) of the atom after a complete one-photon absorption-emission process. 0.2

6. Energy and momentum transfer by a laser beam along the \(+x\) direction.

Consider now that a laser beam of frequency \(\omega'_L\) is incident on the atom along the \(+x\) direction, while the atom moves also in the \(+x\) direction with velocity \(v\). Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions:

6a Write down the average energy change \(\Delta\varepsilon\) of the atom after a complete one-photon absorption-emission process. 0.3
6b Write down the average momentum change \(\Delta p\) of the atom after a complete one-photon absorption-emission process. 0.3

PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES

Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams \(\omega_L\) and \(\omega'_L\) may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by \(\Gamma^{-1}\).

Consider a collection of \(N\) atoms at rest in the laboratory frame of reference, and a laser beam of frequency \(\omega_L\) incident on them. The atoms absorb and emit continuously such that there is, on average, \(N_{exc}\) atoms in the excited state (and therefore, \(N-N_{exc}\) atoms in the ground state). A quantum mechanical calculation yields the following result:

\[ N_{exc} = N\frac{\Omega_R^2} {(\omega_0-\omega_L)^2+\dfrac{\Gamma^2}{4}+2\Omega_R^2} \]

where \(\omega_0\) is the resonance frequency of the atomic transition and \(\Omega_R\) is the so-called Rabi frequency; \(\Omega_R^2\) is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency \(\omega_0\) is different from the frequency of the laser beam \(\omega_L\). An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is \(N_{exc}\Gamma\).

Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency \(\omega_L\) are incident on a gas of \(N\) atoms that move in the \(+x\) direction with velocity \(v\).

Figure 3

Figure 2. Two counter propagating laser beams with the same but arbitrary frequency \(\omega_L\) are incident on a gas of \(N\) atoms that move in the \(+x\) direction with velocity \(v\).

7. Force on the atomic beam by the lasers.

7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that \(mv\gg\hbar q\). 1.5

8. Low velocity limit.

Assume now that the velocity of the atoms is small enough such that you can expand the force up to first order in \(v\).

8a Find an expression for the force found in Question (7a), in this limit. 1.5

Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation.

8b Write down the condition to obtain a positive force (speeding up the atoms). 0.25
8c Write down the condition to obtain a zero force. 0.25
8d Write down the condition to obtain a negative force (slowing down the atoms). 0.25
8e Consider now that the atoms are moving with a velocity \(-v\) (in the \(-x\) direction). Write down the condition to obtain a slowing down force on the atoms. 0.25

9. Optical molasses.

In the case of a negative force, one obtains a frictional dissipative force. Assume that initially, at \(t=0\), the gas of atoms has velocity \(v_0\).

9a In the limit of low velocities, find the velocity of the atoms after the laser beams have been on for a time \(\tau\). 1.5
9b Assume now that the gas of atoms is in thermal equilibrium at a temperature \(T_0\). Find the temperature \(T\) after the laser beams have been on for a time \(\tau\). 0.5

This model does not allow you to go to arbitrarily low temperatures.

Official English text of the problem.

Answers

Define \[ \beta=\frac vc,\qquad q=\frac{\omega_L}{c},\qquad q'=\frac{\omega'_L}{c},\qquad K=\frac12mv^2. \] For Part II, define \[ \Delta=\omega_0-\omega_L,\qquad D=\Delta^2+\frac{\Gamma^2}{4}+2\Omega_R^2,\qquad \alpha=\frac{4\hbar q^2\Gamma\Omega_R^2\Delta}{D^2}. \] The two beams are taken to have the same \(\Omega_R\).

Photon quantities below retain their first Doppler correction. Atomic quantities retain the leading recoil correction, omitting products of \(v/c\) and \(\hbar q/(mv)\). The small momentum corrections before this truncation are displayed in the solution. Atomic energies omit the constant rest energy.

  • 1a: \(\displaystyle \omega_L(1+\beta)=\omega_0\), equivalently \(\omega_L+qv=\omega_0\).
  • 1b: \(\displaystyle p_{at}=mv-\hbar q.\)
  • 1c: \(\displaystyle \varepsilon_{at}=K+\hbar\omega_L=K+\hbar\omega_0-\hbar qv.\)
  • 2a: \(\displaystyle \varepsilon_{ph}=\hbar\omega_0(1-\beta)\simeq\hbar\omega_L.\)
  • 2b: \(\displaystyle p_{ph}=-\frac{\hbar\omega_0}{c}(1-\beta)\simeq-\hbar q.\)
  • 2c: \(\displaystyle p_{at}=mv.\)
  • 2d: \(\displaystyle \varepsilon_{at}=K.\)
  • 3a: \(\displaystyle \varepsilon_{ph}=\hbar\omega_0(1+\beta)\simeq\hbar\omega_L(1+2\beta).\)
  • 3b: \(\displaystyle p_{ph}=\frac{\hbar\omega_0}{c}(1+\beta)\simeq\hbar q(1+2\beta).\)
  • 3c: \(\displaystyle p_{at}\simeq mv-2\hbar q.\)
  • 3d: \(\displaystyle \varepsilon_{at}\simeq K-2\hbar qv.\)
  • 4a: \(\displaystyle \langle\varepsilon_{ph}\rangle=\hbar\omega_0.\)
  • 4b: \(\displaystyle \langle p_{ph}\rangle=\frac{\hbar\omega_0v}{c^2}\simeq\hbar q\beta.\)
  • 4c: \(\displaystyle \langle\varepsilon_{at}\rangle\simeq K-\hbar qv.\)
  • 4d: \(\displaystyle \langle p_{at}\rangle\simeq mv-\hbar q.\)
  • 5a: \(\displaystyle \Delta\varepsilon=-\frac{\hbar\omega_0v}{c}\simeq-\hbar qv.\)
  • 5b: \(\displaystyle \Delta p\simeq-\hbar q.\)
  • 6a: \(\displaystyle \Delta\varepsilon=+\frac{\hbar\omega_0v}{c}\simeq+\hbar q'v.\)
  • 6b: \(\displaystyle \Delta p\simeq+\hbar q'.\)
  • 7a: \[ \boxed{ F(v)=N\hbar q\Gamma\Omega_R^2 \left[ \frac{1}{(\Delta+qv)^2+\Gamma^2/4+2\Omega_R^2} - \frac{1}{(\Delta-qv)^2+\Gamma^2/4+2\Omega_R^2} \right]. } \]
  • 8a: \[ \boxed{F(v)\simeq-N\alpha v =-\frac{4N\hbar q^2\Gamma\Omega_R^2(\omega_0-\omega_L)} {\left[(\omega_0-\omega_L)^2+\Gamma^2/4+2\Omega_R^2\right]^2}\,v.} \]
  • 8b: \(\displaystyle \omega_L>\omega_0.\)
  • 8c: \(\displaystyle \omega_L=\omega_0.\)
  • 8d: \(\displaystyle \omega_L<\omega_0.\)
  • 8e: \(\displaystyle \omega_L<\omega_0\); the force then points in \(+x\).
  • 9a: \(\displaystyle \boxed{v(\tau)=v_0e^{-\alpha\tau/m}}\), with \(\alpha>0\).
  • 9b: \(\displaystyle \boxed{T(\tau)=T_0e^{-2\alpha\tau/m}}\).

Solution

The photon energy–momentum relation gives \[ \hbar\omega_L=c\hbar q,\qquad q=\frac{\omega_L}{c}. \] Write the two small parameters as \[ \beta=\frac vc\ll1,\qquad \eta=\frac{\hbar q}{mv}\ll1. \]

The recoil energy \((\hbar q)^2/(2m)\) is quadratic in \(\eta\) relative to the initial atomic kinetic energy, so it is omitted. The change in atomic velocity following absorption is \(-\hbar q/m\); its contribution to a Doppler factor is \[ \frac{\hbar q}{mc}=\eta\beta, \] which is also omitted.

Thus, when calculating emitted photon frequencies, the emitting atom can be assigned velocity \(v\). Momentum corrections of order \(\hbar q\beta=mv\,\eta\beta\) are shown where useful, but omitted from the leading atomic recoil results.

1a. Resonance condition

For a photon propagating in \(-x\), the frequency measured by an atom moving in \(+x\) is, to first order, \[ \omega_{\mathrm{atom}} =\omega_L-\mathbf{k}\cdot\mathbf{v} =\omega_L+qv. \] Resonance requires this frequency to equal \(\omega_0\): \[ \boxed{\omega_L+qv=\omega_0}, \qquad \boxed{\omega_L\left(1+\frac vc\right)=\omega_0}. \]

This also follows from conservation of energy: \[ \frac12mv^2+\hbar\omega_L = \frac{(mv-\hbar q)^2}{2m}+\hbar\omega_0. \] Discarding the recoil energy gives the same condition.

1b. Atomic momentum after absorption

Conservation of momentum gives \[ \boxed{p_{at}=mv-\hbar q}. \]

1c. Atomic energy after absorption

The atom acquires the incident photon’s energy: \[ \boxed{\varepsilon_{at}=\frac12mv^2+\hbar\omega_L}. \] Equivalently, adding its kinetic and internal energies, \[ \varepsilon_{at} = \frac{(mv-\hbar q)^2}{2m}+\hbar\omega_0 \simeq \boxed{\frac12mv^2-\hbar qv+\hbar\omega_0}. \] The resonance condition makes these expressions equal to the retained order.

For the emission questions, let \(s=+1\) denote a photon emitted in \(+x\), and \(s=-1\) a photon emitted in \(-x\). Neglecting recoil in the transition frequency, its energy in the emitting atom’s rest frame is \(\hbar\omega_0\). The first-order Doppler transformation gives \[ \varepsilon_{ph}^{(s)} =\hbar\omega_0(1+s\beta), \qquad p_{ph}^{(s)} =s\frac{\hbar\omega_0}{c}(1+s\beta). \]

2a. Photon energy for emission in \(-x\)

Taking \(s=-1\), \[ \boxed{\varepsilon_{ph}^{(-)} =\hbar\omega_0(1-\beta)}. \] Since \[ \omega_L=\frac{\omega_0}{1+\beta} \simeq\omega_0(1-\beta), \] this is \[ \boxed{\varepsilon_{ph}^{(-)}\simeq\hbar\omega_L}. \]

2b. Photon momentum for emission in \(-x\)

Using \(p_{ph}=-\varepsilon_{ph}/c\), \[ \boxed{p_{ph}^{(-)} =-\frac{\hbar\omega_0}{c}(1-\beta) \simeq-\hbar q}. \]

2c. Atomic momentum after emission in \(-x\)

Conservation of momentum over the complete process gives \[ p_{at}^{(-)} =mv-\hbar q-p_{ph}^{(-)} \simeq mv-\hbar q+\hbar q. \] Therefore \[ \boxed{p_{at}^{(-)}=mv}. \]

2d. Atomic energy after emission in \(-x\)

The emitted photon carries away the energy supplied by the absorbed photon: \[ \varepsilon_{at}^{(-)} =K+\hbar\omega_L-\varepsilon_{ph}^{(-)} \simeq K. \] Thus \[ \boxed{\varepsilon_{at}^{(-)}=\frac12mv^2}. \]

3a. Photon energy for emission in \(+x\)

For \(s=+1\), \[ \boxed{\varepsilon_{ph}^{(+)} =\hbar\omega_0(1+\beta)}. \] Using \(\omega_0=\omega_L(1+\beta)\), \[ \boxed{\varepsilon_{ph}^{(+)} \simeq\hbar\omega_L(1+2\beta)}. \]

3b. Photon momentum for emission in \(+x\)

The momentum is positive: \[ \boxed{p_{ph}^{(+)} =\frac{\hbar\omega_0}{c}(1+\beta) \simeq\hbar q(1+2\beta)}. \]

3c. Atomic momentum after emission in \(+x\)

Momentum conservation, retaining the photon’s Doppler correction temporarily, gives \[ p_{at}^{(+)} =mv-\hbar q-\hbar q(1+2\beta) =mv-2\hbar q-2\hbar q\beta. \] The last term is of relative order \(\eta\beta\) in the atomic momentum. With that mixed correction omitted, \[ \boxed{p_{at}^{(+)}\simeq mv-2\hbar q}. \]

3d. Atomic energy after emission in \(+x\)

The atom returns to its ground state. Conservation of energy gives \[ \begin{aligned} \varepsilon_{at}^{(+)} &=K+\hbar\omega_L-\hbar\omega_0(1+\beta)\\ &\simeq K+\hbar\omega_0(1-\beta) -\hbar\omega_0(1+\beta)\\ &=K-2\hbar\omega_0\beta. \end{aligned} \] At the retained atomic accuracy, \[ \boxed{\varepsilon_{at}^{(+)} \simeq\frac12mv^2-2\hbar qv}. \]

This also follows by expanding \[ \frac{(mv-2\hbar q)^2}{2m} = K-2\hbar qv+\frac{2\hbar^2q^2}{m} \] and discarding the quadratic recoil term.

4a. Average emitted photon energy

The two directions have equal probability: \[ \begin{aligned} \langle\varepsilon_{ph}\rangle &=\frac12\hbar\omega_0(1-\beta) +\frac12\hbar\omega_0(1+\beta)\\ &=\boxed{\hbar\omega_0}. \end{aligned} \]

4b. Average emitted photon momentum

Averaging the signed momenta, \[ \begin{aligned} \langle p_{ph}\rangle &=\frac12\left[ -\frac{\hbar\omega_0}{c}(1-\beta) +\frac{\hbar\omega_0}{c}(1+\beta) \right]\\ &=\boxed{\frac{\hbar\omega_0v}{c^2}} \simeq\boxed{\hbar q\beta}. \end{aligned} \]

Although the directions are equally probable, the two photon energies differ in the laboratory. Their mean laboratory momentum is consequently small and positive.

Relative to the atomic momentum, this mean is of order \[ \frac{\langle p_{ph}\rangle}{mv}\simeq\eta\beta. \] It is therefore omitted in the leading atomic momentum changes below.

4c. Average atomic energy

Using the results of 2d and 3d, \[ \langle\varepsilon_{at}\rangle =\frac12K+\frac12(K-2\hbar qv). \] Hence \[ \boxed{\langle\varepsilon_{at}\rangle \simeq K-\hbar qv}. \]

Equivalently, \[ \langle\varepsilon_{at}\rangle =K+\hbar\omega_L-\hbar\omega_0. \]

4d. Average atomic momentum

Before discarding the mixed correction, momentum conservation gives \[ \langle p_{at}\rangle =mv-\hbar q-\langle p_{ph}\rangle =mv-\hbar q-\frac{\hbar\omega_0v}{c^2}. \] Thus, to the requested leading atomic recoil order, \[ \boxed{\langle p_{at}\rangle\simeq mv-\hbar q}. \]

5a. Average atomic energy change

Subtracting the initial energy, \[ \Delta\varepsilon =\langle\varepsilon_{at}\rangle-K =\hbar(\omega_L-\omega_0). \] The resonance condition yields \[ \boxed{\Delta\varepsilon =-\frac{\hbar\omega_0v}{c} \simeq-\hbar qv}. \]

The emitted photon has, on average, more energy than the absorbed photon. That difference comes from the atom’s kinetic energy.

5b. Average atomic momentum change

Subtracting the initial momentum from 4d gives \[ \boxed{\Delta p\simeq-\hbar q}. \] The smaller correction before truncation is \[ -\langle p_{ph}\rangle =-\frac{\hbar\omega_0v}{c^2}. \] At leading recoil order, the energy and momentum changes obey \[ \Delta\varepsilon\simeq v\,\Delta p. \]

6a. Energy change for a laser propagating in \(+x\)

The atom now moves away from the incident light. The resonance condition is \[ \omega'_L\left(1-\frac vc\right)=\omega_0, \] so \[ \omega'_L\simeq\omega_0(1+\beta). \]

The mean emitted photon energy remains \(\hbar\omega_0\). Therefore \[ \Delta\varepsilon =\hbar(\omega'_L-\omega_0) =\boxed{\frac{\hbar\omega_0v}{c}} \simeq\boxed{\hbar q'v}, \qquad q'=\frac{\omega'_L}{c}. \] The atom gains kinetic energy.

6b. Momentum change for a laser propagating in \(+x\)

The absorbed photon supplies momentum \(+\hbar q'\). Consequently, \[ \Delta p =\hbar q'-\frac{\hbar\omega_0v}{c^2} \simeq\boxed{+\hbar q'}. \] The omitted second term is again a mixed Doppler–recoil correction in the atomic momentum.

7a. Force from the two laser beams

Use the supplied scattering law separately for the two beams, taking their intensities, and hence their Rabi frequencies, to be equal.

Let \(+\) label the beam propagating in \(+x\), and \(-\) the beam propagating in \(-x\). The frequencies seen by the moving atoms are \[ \omega_{\mathrm{atom},+}=\omega_L-qv, \qquad \omega_{\mathrm{atom},-}=\omega_L+qv. \] Thus their detunings from the atomic transition are \[ \omega_0-\omega_{\mathrm{atom},+}=\Delta+qv, \qquad \omega_0-\omega_{\mathrm{atom},-}=\Delta-qv. \]

The total numbers of scattering events per unit time are \[ R_+(v) = \frac{N\Gamma\Omega_R^2} {(\Delta+qv)^2+\Gamma^2/4+2\Omega_R^2}, \] \[ R_-(v) = \frac{N\Gamma\Omega_R^2} {(\Delta-qv)^2+\Gamma^2/4+2\Omega_R^2}. \]

At leading recoil order, each event transfers \(+\hbar q\) or \(-\hbar q\), respectively. Therefore \[ F(v)=\hbar qR_+(v)-\hbar qR_-(v), \] giving \[ \boxed{ F(v)=N\hbar q\Gamma\Omega_R^2 \left[ \frac{1}{(\Delta+qv)^2+\Gamma^2/4+2\Omega_R^2} - \frac{1}{(\Delta-qv)^2+\Gamma^2/4+2\Omega_R^2} \right]. } \]

This is the force on all \(N\) atoms. The force on one atom is \(F/N\).

The Doppler shift \(qv\) is retained in the scattering probabilities: it can be significant compared with the transition linewidth even when \(v/c\) is very small.

8a. Low-velocity expansion

Put \[ D=\Delta^2+\frac{\Gamma^2}{4}+2\Omega_R^2. \] For sufficiently small \(qv\), \[ \frac{1}{(\Delta\pm qv)^2+\Gamma^2/4+2\Omega_R^2} \simeq \frac1D\mp\frac{2\Delta qv}{D^2}. \] For example, \(|qv|\ll\sqrt D\) is a sufficient small-velocity condition.

Subtracting the two expansions, \[ F(v) \simeq N\hbar q\Gamma\Omega_R^2 \left(-\frac{4\Delta qv}{D^2}\right). \] Hence \[ \boxed{F(v)\simeq-N\alpha v}, \qquad \boxed{\alpha=\frac{4\hbar q^2\Gamma\Omega_R^2\Delta}{D^2}}. \]

The coefficient \(\alpha\) has units of mass per time, so \(\alpha v\) has units of force.

8b. Positive force

For \(v>0\), all factors in \(\alpha\) except \(\Delta\) are positive. A positive force requires \[ \Delta<0. \] Therefore \[ \boxed{\omega_L>\omega_0}. \] This is blue detuning.

8c. Zero force

For nonzero intensity and a moving atom, the detuning condition is \[ \Delta=0, \] or \[ \boxed{\omega_L=\omega_0}. \] The two scattering rates are then equal. Independently, symmetry also gives zero force at \(v=0\).

8d. Negative force

For \(v>0\), a negative force requires \(\alpha>0\), hence \[ \boxed{\omega_L<\omega_0}. \] This is red detuning. The counterpropagating beam is Doppler shifted closer to resonance and is scattered more strongly.

8e. Slowing atoms moving in \(-x\)

For velocity \(-v\), with \(v>0\), \[ F(-v)=+N\alpha v. \] Slowing requires a force in \(+x\), so again \(\alpha>0\). Thus \[ \boxed{\omega_L<\omega_0}. \]

Red detuning produces a force opposing motion in either direction.

9a. Velocity after time \(\tau\)

For red detuning, \(\alpha>0\). The gas has total mass \(Nm\), so Newton’s second law gives \[ Nm\frac{dv}{dt}=-N\alpha v. \] The factor \(N\) cancels: \[ \frac{dv}{dt}=-\frac{\alpha}{m}v. \] Integrating with \(v(0)=v_0\), \[ \boxed{v(\tau)=v_0e^{-\alpha\tau/m}}. \]

Explicitly, \[ \boxed{ v(\tau)=v_0 \exp\!\left[ -\frac{4\hbar q^2\Gamma\Omega_R^2(\omega_0-\omega_L)\tau} {m\left[(\omega_0-\omega_L)^2+\Gamma^2/4+2\Omega_R^2\right]^2} \right]. } \]

The velocity decay time is \(m/\alpha\).

9b. Temperature after time \(\tau\)

In this deterministic friction model, each atomic velocity is multiplied by \[ e^{-\alpha\tau/m}. \] Therefore the variance of the velocities is multiplied by \[ e^{-2\alpha\tau/m}. \]

For one-dimensional thermal motion, \[ m\left\langle (v-\langle v\rangle)^2\right\rangle=k_BT. \] Consequently, \[ \boxed{T(\tau)=T_0e^{-2\alpha\tau/m}}. \]

Explicitly, \[ \boxed{ T(\tau)=T_0 \exp\!\left[ -\frac{8\hbar q^2\Gamma\Omega_R^2(\omega_0-\omega_L)\tau} {m\left[(\omega_0-\omega_L)^2+\Gamma^2/4+2\Omega_R^2\right]^2} \right]. } \]

This exponential cannot be extrapolated to arbitrarily low temperatures. The calculation omitted quadratic recoil contributions and the fluctuations caused by random scattering events. In particular, the assumption \(mv\gg\hbar q\), applied to a typical thermal speed, requires \[ k_BT\gg\frac{(\hbar q)^2}{m}. \] As the atoms cool, the terms omitted from this model become relevant.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.