IPhO 1991 · Theory problem 3

Cooling Atoms by laser

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To study the properties of isolated atoms with a high degree of precision they must be kept almost at rest for a length of time. A method has recently been developed to do this. It is called “laser cooling” and is illustrated by the problem below.

In a vacuum chamber a well collimated beam of \(\mathrm{Na}^{23}\) atoms (coming from the evaporation of a sample at \(10^3\ \mathrm{K}\)) is illuminated head-on with a high intensity laser beam (fig. 3.1). The frequency of laser is chosen so there will be resonant absorption of a photon by those atoms whose velocity is \(v_0\). When the light is absorbed, these atoms are exited to the first energy level, which has a mean value \(E\) above the ground state and uncertainty of \(\Gamma\) (fig. 3.2).

Figure 1

Figure 3.1

Figure 2

Figure 3.2

For this process, the atom’s decrease in velocity \(\Delta v_1\) is given by \(\Delta v_1=v_1-v_0\). Light is then emitted by the atom as it returns to its ground state. The atom’s velocity changes by \(\Delta v'=v'_1-v_1\) and its direction of motion changes by an angle \(\varphi\) (fig. 3.3).

Figure 3

Figure 3.3

This sequence of absorption and emission takes place many times until the velocity of the atoms has decreased by a given amount \(\Delta v\) such that resonant absorption of light at frequency \(\nu\) no longer occurs. It is then necessary to change the frequency of laser so as to maintain resonant absorption. The atoms moving at the new velocity are further slowed down until some of them have a velocity close to zero.

As first approximation we may ignore any atomic interaction processes apart from the spontaneous absorption and emission light described above.
Furthermore, we may assume the laser to be so intense that the atoms spend practically no time in the ground state.

Questions

a) Find the laser frequency needed ensure the resonant absorption of the light by those atoms whose kinetic energy of the atoms inside the region behind the collimator. Also find the reduction in the velocity of these atoms, \(\Delta v_1\), after the absorption process.

b) Light of the frequency calculated in question a) is absorbed by atoms which velocities lie within a range \(\Delta v_0\). Calculate this velocity range.

c) When an atom emits light, its direction of motion changes by \(\varphi\) from initial direction. Calculate \(\varphi\).

d) Find the maximum possible velocity decrease \(\Delta v\) for a given frequency.

e) What is the approximate number \(N\) of absorption-emission events necessary to reduce the velocity of an atom from is initial value \(v_0\)
-found in question a) above- almost to zero? Assume the atom travels in a straight line.

f) Find the time \(t\) that the process in question e takes. Calculate the distance \(\Delta S\) an atom travels in this time.

Data

\[ \begin{aligned} E&=3{,}36\cdot10^{-19}\ \mathrm{J}\\ \Gamma&=7{,}0\cdot10^{-27}\ \mathrm{J}\\ c&=3\cdot10^8\ \mathrm{ms^{-1}}\\ m_p&=1{,}67\cdot10^{-27}\ \mathrm{kg}\\ h&=6{,}62\cdot10^{-34}\ \mathrm{Js}\\ k&=1{,}38\cdot10^{-23}\ \mathrm{JK^{-1}} \end{aligned} \]

where \(c\) is speed of light, \(h\) is Planck’s constant, \(k\) is the Boltzmann constant, and \(m_p\) is the mass of proton.

Official English text of the problem.

Answers

The numerical results use the assumed intended condition in (a), \(\frac12mv_0^2=\frac32kT\), giving \(v_0=1.04\times10^3\ \mathrm{m\,s^{-1}}\).

  • a)
    \[ \boxed{\nu\simeq\frac{E}{h(1+v_0/c)} \simeq5.08\times10^{14}\ \mathrm{Hz}} \] This is approximately \(1.76\times10^9\ \mathrm{Hz}\) below the stationary-atom resonance. \[ \boxed{\Delta v_1\simeq-2.92\times10^{-2}\ \mathrm{m\,s^{-1}}} \] The positive magnitude of the speed reduction is \(2.92\times10^{-2}\ \mathrm{m\,s^{-1}}\).

  • b) The full resonant velocity range is \[ \boxed{\Delta v_0=\frac{c\Gamma}{h\nu}\simeq6.3\ \mathrm{m\,s^{-1}}}, \] extending approximately from \(v_0-3.1\ \mathrm{m\,s^{-1}}\) to \(v_0+3.1\ \mathrm{m\,s^{-1}}\).

  • c) For the emission angle \(\theta\) shown in figure 3.3, let \(v_r=E/(mc)\). Then \[ \boxed{\varphi\simeq\frac{v_r}{v_1}\sin\theta \simeq2.8\times10^{-5}\sin\theta\ \mathrm{rad}}. \] The maximum initial deflection is approximately \(2.8\times10^{-5}\ \mathrm{rad}\).

  • d) Starting at the central resonant velocity \(v_0\), \[ \boxed{\Delta v\simeq\frac{c\Gamma}{2h\nu}\simeq3.1\ \mathrm{m\,s^{-1}}}. \] An atom starting at the upper edge of the resonant range can traverse the full range, approximately \(6.3\ \mathrm{m\,s^{-1}}\).

  • e) With the laser retuned during slowing, \[ \boxed{N\simeq\frac{mcv_0}{E}\simeq3.6\times10^4}. \]

  • f) Taking \(\Gamma\) as the full natural energy linewidth, \[ \boxed{t\simeq5.4\times10^{-4}\ \mathrm{s}=0.54\ \mathrm{ms}}, \qquad \boxed{\Delta S\simeq0.28\ \mathrm{m}}. \]

Solution

a) Resonant frequency and absorption recoil

The sentence specifying the initial kinetic energy is incomplete in the supplied text. We use the likely intended condition that the selected atoms have the mean thermal kinetic energy: \[ \frac12mv_0^2=\frac32kT. \] Approximating the sodium atom’s mass by \(23m_p\), \[ m=23(1.67\times10^{-27}) =3.841\times10^{-26}\ \mathrm{kg}, \] and therefore \[ v_0=\sqrt{\frac{3kT}{m}} =\sqrt{\frac{3(1.38\times10^{-23})(10^3)} {3.841\times10^{-26}}} =1.038\times10^3\ \mathrm{m\,s^{-1}}. \] Thus \(v_0/c\simeq3.46\times10^{-6}\), so nonrelativistic mechanics and the first-order Doppler formula suffice.

Choose the positive direction along the incident atomic beam. The counterpropagating photon has momentum \(-h\nu/c\). Conservation of momentum during absorption gives \[ mv_1=mv_0-\frac{h\nu}{c}, \] hence \[ \Delta v_1=v_1-v_0=-\frac{h\nu}{mc}. \]

Energy conservation gives \[ \frac12mv_0^2+h\nu=\frac12mv_1^2+E. \] Substituting the momentum relation, \[ h\nu\left(1+\frac{v_0}{c}\right) =E+\frac{(h\nu)^2}{2mc^2}. \] The final term is the recoil energy. Its value is approximately \[ E_r\simeq\frac{E^2}{2mc^2} =1.6\times10^{-29}\ \mathrm{J}, \] which is small even compared with \(\Gamma=7.0\times10^{-27}\ \mathrm{J}\). Neglecting it, \[ \boxed{\nu\simeq\frac{E}{h(1+v_0/c)}}. \] This is also the Doppler resonance condition: the approaching atom sees a frequency \(\nu(1+v_0/c)\), which must equal \(E/h\).

For clarity, retaining extra digits temporarily, \[ \nu_{\mathrm{rest}}=\frac Eh=5.0755287\times10^{14}\ \mathrm{Hz}, \] \[ \nu\simeq5.0755111\times10^{14}\ \mathrm{Hz}. \] The required laser frequency is therefore below the stationary-atom resonance by \[ \nu_{\mathrm{rest}}-\nu \simeq\nu_{\mathrm{rest}}\frac{v_0}{c} =1.76\times10^9\ \mathrm{Hz}. \]

The absorption recoil is \[ \boxed{\Delta v_1=-\frac{h\nu}{mc} \simeq-2.92\times10^{-2}\ \mathrm{m\,s^{-1}}}. \] Its sign is negative under the definition printed in the question.

b) Resonant velocity range

Following the energy interval shown in figure 3.2, we treat resonance as occurring when the Doppler-shifted photon energy lies between \(E-\Gamma/2\) and \(E+\Gamma/2\): \[ E-\frac{\Gamma}{2} \leq h\nu\left(1+\frac vc\right) \leq E+\frac{\Gamma}{2}. \]

The limiting velocities are consequently \[ v_\pm=c\left(\frac{E\pm\Gamma/2}{h\nu}-1\right). \] Since the laser is tuned to the centre of the line for velocity \(v_0\), \[ v_\pm=v_0\pm\frac{c\Gamma}{2h\nu}. \] Thus the full velocity range is \[ \boxed{\Delta v_0=v_+-v_-=\frac{c\Gamma}{h\nu}}. \] Using \(h\nu\simeq E\), \[ \Delta v_0\simeq \frac{(3.0\times10^8)(7.0\times10^{-27})} {3.36\times10^{-19}} =6.25\ \mathrm{m\,s^{-1}} \simeq6.3\ \mathrm{m\,s^{-1}}. \]

The half-width is approximately \(3.1\ \mathrm{m\,s^{-1}}\).

c) Deflection caused by spontaneous emission

The emitted photon has momentum approximately \(E/c\). Define the corresponding recoil speed \[ v_r=\frac{E}{mc}\simeq2.92\times10^{-2}\ \mathrm{m\,s^{-1}}. \] The small Doppler and recoil corrections to the emitted photon’s energy are negligible here.

For the geometry in figure 3.3, the photon travels at angle \(\theta\) below the initial direction. Conservation of momentum parallel and perpendicular to that direction gives \[ mv_1=mv'_1\cos\varphi+\frac Ec\cos\theta, \] \[ mv'_1\sin\varphi=\frac Ec\sin\theta. \] Dividing, \[ \boxed{\tan\varphi =\frac{v_r\sin\theta}{v_1-v_r\cos\theta}}. \]

Initially \(v_r\ll v_1\), so \[ \varphi\simeq\frac{v_r}{v_1}\sin\theta. \] With \(v_1\simeq1038.16\ \mathrm{m\,s^{-1}}\), \[ \boxed{\varphi\simeq2.8\times10^{-5}\sin\theta\ \mathrm{rad}}. \]

The largest deflection occurs for emission approximately perpendicular to the atomic velocity. More precisely, for fixed photon momentum, \[ \varphi_{\max}=\arcsin\left(\frac{v_r}{v_1}\right) \simeq2.8\times10^{-5}\ \mathrm{rad}. \] Thus one emission changes the initial direction only very slightly. A unique value of \(\varphi\) requires specifying the random emission angle \(\theta\).

d) Maximum slowing at fixed laser frequency

Consider first an atom starting at the central resonant velocity \(v_0\), as in part (a). As it slows, the photon energy in its rest frame decreases. The lower boundary of the resonance interval is reached when \[ h\nu\left(1+\frac{v_0-\Delta v}{c}\right) =E-\frac{\Gamma}{2}. \] Subtracting this from \[ h\nu\left(1+\frac{v_0}{c}\right)=E \] gives \[ \frac{h\nu}{c}\Delta v=\frac{\Gamma}{2}. \] Therefore, to the linewidth accuracy used in the problem, \[ \boxed{\Delta v=\frac{c\Gamma}{2h\nu} \simeq3.1\ \mathrm{m\,s^{-1}}}. \]

If “maximum possible” includes all initial velocities that can absorb this fixed laser frequency, an atom may begin at \(v_+\) and slow to \(v_-\). The maximum span is then \[ \boxed{v_+-v_-=\frac{c\Gamma}{h\nu} \simeq6.3\ \mathrm{m\,s^{-1}}}. \]

Both values are much smaller than \(v_0\). Continued slowing therefore requires increasing the laser frequency toward \(E/h\) as the atoms slow.

e) Number of absorption–emission events

Every absorption gives the atom a directed momentum change approximately \[ \Delta p_{x,\mathrm{abs}}=-\frac Ec. \] Spontaneous emission has no preferred forward or backward direction in the atom’s rest frame. Consequently, to the present accuracy, \[ \langle\cos\theta\rangle=0, \qquad \langle\Delta p_{x,\mathrm{em}}\rangle=0. \] The mean longitudinal speed change per complete cycle is therefore \[ \langle\Delta v_x\rangle=-v_r=-\frac{E}{mc}. \]

Using the straight-line approximation requested in the question and retuning the laser to maintain resonance, \[ Nv_r\simeq v_0. \] Hence \[ \boxed{N\simeq\frac{v_0}{v_r} =\frac{mcv_0}{E}}. \] Numerically, \[ N\simeq \frac{(3.841\times10^{-26})(3.0\times10^8)(1038.19)} {3.36\times10^{-19}} =3.56\times10^4. \] Thus approximately \[ \boxed{N\simeq3.6\times10^4} \] cycles are required. This estimates the reduction of the mean directed velocity; individual atoms undergo random emission recoils.

f) Cooling time and distance

For an excited state with mean lifetime \(\tau\), the full natural energy linewidth satisfies \[ \Gamma=\frac{\hbar}{\tau}. \] Here \(\Gamma\) is taken as the full width, consistent with the boundaries \(E\pm\Gamma/2\). Thus \[ \tau=\frac{\hbar}{\Gamma} =\frac{h}{2\pi\Gamma}. \] Numerically, \[ \tau= \frac{6.62\times10^{-34}} {2\pi(7.0\times10^{-27})} =1.51\times10^{-8}\ \mathrm{s}. \]

By the explicit assumption that the atoms spend practically no time in the ground state, the mean duration of one cycle is \(\tau\). Therefore \[ t=N\tau =(3.56\times10^4)(1.51\times10^{-8}) \simeq5.36\times10^{-4}\ \mathrm{s}, \] or \[ \boxed{t\simeq5.4\times10^{-4}\ \mathrm{s}}. \]

The mean recoil per cycle and mean cycle duration are approximately constant. Thus the mean deceleration has constant magnitude \[ a=\frac{v_r}{\tau}\simeq1.94\times10^6\ \mathrm{m\,s^{-2}}. \] For slowing from \(v_0\) almost to zero, the mean speed over the process is \(v_0/2\), giving \[ \Delta S\simeq\frac{v_0t}{2} =\frac{(1038.19)(5.36\times10^{-4})}{2} \simeq0.278\ \mathrm{m}. \] Therefore \[ \boxed{\Delta S\simeq0.28\ \mathrm{m}}. \]

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.