OII na Teorica

Trapping Ions and Cooling Atoms In recent decades, trapping and cooling atoms and ions has been a fascinating topic for physicists, with several Nobel prizes awarded for work in this area. In the rst part of this question, we will explore a technique for trapping ions, known as the “Paul trap”. Wolfgang Paul and Hans Dehmelt received one half of the 1989 Nobel Prize in Physics for this work. Next, we investigate the Doppler cooling technique, one of the works cited in the press release for the 1997 Nobel Prize in Physics awarded to Steven Chu, Claude Cohen-Tannoudji, and William Daniel Phillips “for developments of methods to cool and trap atoms with laser light”. A. The Paul Trap It is known that with electrostatic elds, it is not possible to create a stable equilibrium for a charged particle. Therefore, creating a stable equilibrium point for ions requires more sophisticated techniques. The Paul trap is one of these techniques. Consider a ring of charge with a radius and a uniform positive linear charge density . A positive point charge with mass is placed at the center of the ring. Figure 1 - A positively charged ring with a uniform linear charge density and radius : the origin of the coordinate system is at the center of the ring. R $\lambda$ Q m $\lambda$ R Points: 30 Time: 5.0 Hours IPhO 2024 Theory, English (Official) Page 6 of 16

A-1 a) In cartesian coordinates , obtain the electric eld due to the charged ring in the vicinity of the ring’s center to the rst order in , , and . b) Find the angular frequency of small oscillations of the charged particle around the center of the ring in the directions for which a stable equilibrium exists. 1.5 pt In order to trap the charge fully, we would like to apply alternating elds to produce a dynamic equilibrium. Assume that the charge density is in which , , and are adjustable. We shall ignore radiative e ects. Then the equation of motion for small displacements from the center of the ring, along the direction perpendicular to the plane of the ring will turn out to be: (1) A-2 Write and in terms of the known parameters. 0.4 pt We would like to obtain an approximate solution to Equation (1) by making the following simplifying assumptions: , , and . With these assumptions, it can be shown that the solution of this equation can be split into two parts: , where is a slowly varying component and is a small-amplitude rapidly-varying component with a mean value of zero. In other words, may be assumed constant over a few oscillations of (see Figure 2). (x, y, z) x/R y/R z/R Q $\lambda$ = $\lambda 0$ + u cos $\text{Omegat}$ $\lambda 0$ u $\Omega$ ̈z = (+k2 + $a\Omega 2$ cos $\text{Omegat})z$ a k a $\ll 1\Omega \text{ggk}$ $a\Omega 2\text{ggk}2$ z(t) = p(t) + q(t) p(t) q(t) p(t) q(t) Points: 30 Time: 5.0 Hours IPhO 2024 Theory, English (Official) Page 7 of 16

Figure 2 - A typical solution for the equation of motion of the charged particle: gives the overall motion, and represents small oscillations around this trajectory. The ellipse on the right is a magni cation of a part of this trajectory. A-3 a) Using the approximations stated above, nd the equation of motion for in terms of , , and . b) Find the solution of this equation by considering appropriate initial conditions corresponding to the required properties of this function. 1.8 pt A-4 a) Using the mean e ect of the rapidly varying component and obtain an e ective equation of motion for . b) Investigate the stability of the equilibrium point and nd the condition for a stable equilibrium. 1.5 pt Assume that and . We would like to use this device to trap a singly ionized atom times heavier than a hydrogen atom. A-5 Calculate . Assume and estimate the smallest frequency required to stabilize the motion of this ion. Use the data given at the end of the question. 0.4 pt p(t) q(t) q(t) a $\Omega$ p p(t) $\lambda 0$ = 8 $\times 10-9$ C/m R = 10 cm 100 k a = 0. 04 Points: 30 Time: 5.0 Hours IPhO 2024 Theory, English (Official) Page 8 of 16

B. Doppler Cooling It may be necessary to cool a trapped atom or ion. Assume that a trapped atom of mass , has two energy levels with an energy di erence of . Electrons in the lower level may absorb a photon and jump to the higher level, but after a period they will return to the lower level and emit a photon with a frequency predominantly within . B-1 Use the Heisenberg’s uncertainty principle to nd . 0.5 pt With a similar reasoning, when we shine a laser light on the trapped atom, if the angular frequency of the laser, , falls in the interval , the atom may absorb the photon. Assume that the frequency of the laser light is slightly lower than . For a particular device, the rate of photon absorption by an atom in the reference frame of the atom is given in Figure 3. The absorbed photon is then re-emitted in a random direction. To make things simple, we consider the problem in one dimension, i.e. we assume that the atoms can only move in the

direction and the laser light shines on them both from the left and from the right. In the atom’s reference frame, the light has a higher or lower frequency due to the motion of the atoms. Since the velocity of the atoms is very small, we only include terms of order and ignore all the higher-order terms. Moreover, we have so that the velocity of the atom nearly does not change after absorbing the photon. Also, the change in frequency due to the Doppler e ect is so small compared to , that the function for in the diagram of Figure 3 may be approximated by the following linear function: where is the number of absorbed photons per unit of time, is the value of for , and is the slope of the line tangent to the curve at . The frequency of the re-emitted photon is almost equal to the frequency of the incident photon, but it is emitted with equal probability in the positive or negative -direction. In fact, up to the order considered here the two frequencies are identical. Note that we are considering the whole process in the atom’s reference frame. m E0 = $\hslash \text{omegaA}$ $\tau$ $[\text{omegaA}-\Gamma$, $\text{omegaA}$ + $\Gamma ]$ $\Gamma$ $\text{omegaL}$ $[\text{omegaA}-\Gamma$, $\text{omegaA}$ + $\Gamma ]$ $\text{omegaL}$ $\text{omegaA}$ x v v/c m $\gg ħ\text{omegaA}/c2$ $\text{omegaA}-\text{omegaL}$ s $s(\omega$) = sL + $\alpha (\omega -\text{omegaL}$) s sL s $\omega$ = $\text{omegaL}$ $\alpha$ $\text{omegaL}$ x Points: 30 Time: 5.0 Hours IPhO 2024 Theory, English (Official) Page 9 of 16

Figure 3 - The rate of photon absorption as a function of the frequency for a particular trap: the frequency corresponding to the energy di erence between the two atomic levels is indicated by and the slighlty smaller frequency of the laser is indicated by . B-2 a) Assume that the trapped atom is moving with a velocity, in the lab frame. In the frame of reference of the atom, calculate the collision rate of the photons, incident from each of the two directions, with the atoms (denoted by and ) and the rate of absorption of momentum in each direction (denoted by and ). b) Determine the e ective force on the atom as a function of , , , and , in the reference frame of the laboratory. Assume , 1.7 pt We would like to nd the lowest temperature that can be achieved using this technique. Assume that the velocity of a particular atom has been reduced to zero exactly, and at this very moment it absorbs a photon (incident from any of the two directions), and re-emits the photon randomly in any of the two directions, with almost the same frequency. Assume that this process happens once every units of time. B-3 Considering the momentum of the atom after such a process for the two possible outcomes, calculate the average power absorbed by the atom. 1.0 pt $\text{omegaA}$ $\text{omegaL}$ v = vx s+ $s-$ $\pi +$ $\pi -$ v kL = $\text{omegaL}/c$ ħ $\alpha$ sL $\ll \alpha \text{omegaL}$ $\tau$ Points: 30 Time: 5.0 Hours IPhO 2024 Theory, English (Official) Page 10 of 16

B-4 Consider the force calculated in Task B-2 and calculate the output power. Then, calculate the average value of at equilibrium. Using your knowledge of the kinetic theory of gases estimate the temperature of the atoms. 0.8 pt B-5 Estimate this temperature, for an atom 100 times heavier than a hydrogen atom. Assume that , , and . 0.4 pt mass of hydrogen atom: charge of an electron:

Topic: Electrostatics, Oscillations & Waves, Modern-Quantum Physics Metodi: Coulomb’s Law, Differential Equations, Simple Harmonic Motion Analysis, Approximation & Series Expansion, Photon Energy Relation, Kinetic Theory of Gases Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Objects: Point Charge, Atom, Photon Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)

Trapping ions and cooling atoms In recent decades, trapping and cooling atoms and ions has been a fascinating topic for physicists, with Several Nobel Prizes awarded for work in this area. In the first part of this question, we will explore a technique for trapping ions, known as the “Paul trap”. Wolfgang Paul and Hans Dehmelt received one half of the 1989 Nobel Prize in Physics for this work. Next, we investigate the Doppler cooling technique, one of the works cited in the press release for the 1997 Nobel Prize in Physics awarded to Steven Chu, Claude Cohen-Tannoudji, and William Daniel Phillips “for developments of methods to cool and trap atoms with laser light”. A. The Paul Trap It is known that with electrostatic The Commission has already decided to take a decision on the charged particles. Therefore, creating a stable equilibrium point for ions requires more sophisticated techniques. The Paul trap is one of those techniques. Consider a ring of charge with a radius and a uniform positive linear charge density . A positive point charge with mass is placed at the center of the ring. Figure 1 - A positively charged ring with a uniform linear charge density and radius : the origin of the coordinate system is at the center of the ring. R $\lambda$ Q m $\lambda$ R The score: 30 Time: 5.0 hours The following information shall be provided: Theory, English (Official) Page 6 of 16

A-1 (a) In cartesian coordinates , obtain the electricald due to the charged ring in the vicinity of the rings center to the rst order in , , and . (b) Find the angular frequency of small oscillations of the charged particle around the center of the ring in the directions for which a stable The balance exists. 1.5 pt In order to trap the charge fully, we would like to apply alternating elds to produce a dynamic The balance. Assume that the charge density is in which , , and are adjustable. We shall ignore radiatives and ects. Then the equation of motion for small displacements From the center of the ring, along the direction perpendicular to the plane of the ring will turn out to be: (1) A-2 Write and in terms of the known parameters. 0.4 pt We would like to obtain an approximate solution to Equation (1) by making the following The following information shall be provided: , , and . With these assumptions, it can be shown that the solution of this equation can be divided into two parts: , where is a slowly varying component and is a small-amplitude rapidly-varying component with a mean The value of zero. In other words, may be assumed constant over a few oscillations of (see The following table shows the results of the study: (x, y, z) x/R y/R z/R Q $\lambda$ = $\lambda 0$ + u cos $\text{Omegat}$ $\lambda 0$ u $\Omega$ ̈z = (+k2 + $a\Omega 2$ cos $\text{Omegat})z$ a k a $\ll 1\Omega \text{ggk}$ $a\Omega 2\text{ggk}2$ z(t) = p(t) + q(t) p(t) q(t) p(t) q(t) The score: 30 Time: 5.0 hours The following information shall be provided: Theory, English (Official) Page 7 of 16

Figure 2 - A typical solution for the equation of motion of the charged particle: gives the overall motion, and represents small oscillations around this trajectory. The ellipse on the right is a magnification of a part of this trajectory. A-3 (a) Using the approximations stated above, nd the equation of motion for in terms of , And , and . (b) Find the solution of this equation by considering appropriate initial conditions corresponding to the required properties of this function. 1.8 pt A-4 (a) Using the mean and ect of the rapidly varying component and obtain an and ective equation of motion for . (b) Investigate the stability of the equilibrium point and the condition The Commission will take the necessary measures to ensure that the measures taken are in a stable equilibrium. 1.5 pt Assumes that and . We would like to use this device to trap a single ionized atom times heavier than a hydrogen atom. A-5 Calculate . He assumes and estimate the smallest frequency required to stabilize the motion of this ion. Use the date given at the end of the question. 0.4 pt p(t) q(t) q(t) a $\Omega$ p p(t) $\lambda 0$ = 8 $\times 10-9$ C/m R = 10 cm 100 k a = 0. 04 The score: 30 Time: 5.0 hours The following information shall be provided: Theory, English (Official) Page 8 of 16

B. Doppler cooling It may be necessary to cool a trapped atom or ion. Assumes that a trapped atom of mass , has two energy levels with an energy of erence of . Electrons in the lower level may absorb a photon and jump to the higher level, but after a period they will return to the lower level and emit a photon with a frequency predominantly within . B-1 Use the Heisenberg uncertainty principle to . 0.5 pt With a similar reasoning, when we shine a laser light on the trapped atom, if the angular frequency of the laser, , falls in the interval , the atom may absorb the The photon. Assumes that the frequency of the laser light is slightly lower than . For a particular device, the rate of photon absorption by an atom in the reference frame of the atom is given in Figure 3 is shown below. The absorbed photon is then re-emitted in a random direction. To make things simple, we Consider the problem in one dimension, i.e. We assume that the atoms can only move in the

direction and the laser light shines on them both from the left and from the right. In the atoms reference frame, the light has a higher or lower frequency due to the motion of the atoms. Since the speed of the atoms is very small, we only include terms of order and ignore all the higher order terms. Moreover, we have So that the velocity of the atom almost does It doesn’t change after absorbing the photon. Also, the change in frequency due to the Doppler ect is So small compared to , that the function for in the diagram of Figure 3 may be approximated by the following linear function: where is the number of absorbed photons per unit time, is the value of for , and is the slope of the line tangent to the curve at . The frequency of the re-emitted photon is almost equal to the frequency of the incident photon, but it is emitted with equal probability in the Positive or negative direction. In fact, up to the order considered here the two frequencies are It’s identical. Note that we are considering the whole process in the atoms reference frame. m E0 = $\hslash \text{omegaA}$ $\tau$ $[\text{omegaA}-\Gamma$, $\text{omegaA}$ + $\Gamma ]$ $\Gamma$ $\text{omegaL}$ $[\text{omegaA}-\Gamma$, $\text{omegaA}$ + $\Gamma ]$ $\text{omegaL}$ $\text{omegaA}$ x v v/c m $\gg ħ\text{omegaA}/c2$ $\text{omegaA}-\text{omegaL}$ s $s(\omega$) = sL + $\alpha (\omega -\text{omegaL}$) s sL s $\omega$ = $\text{omegaL}$ $\alpha$ $\text{omegaL}$ x The score: 30 Time: 5.0 hours The following information shall be provided: Theory, English (Official) Page 9 of 16

Figure 3 - The rate of photon absorption as a function of the frequency for a particular trap: the frequency corresponding to the energy of erence between the two atomic levels is indicated by and the slighlty smaller frequency of the laser is indicated by . B-2 (a) Assumes that the trapped atom is moving with a velocity, in the lab frame. In the frame of reference of the atom, calculate the collision rate of the photons, incident from each of the two directions, with the atoms (denoted by and ) and the rate of absorption of momentum in each direction (denoted by and ). (b) Determine the ective force on the atom as a function of , , , and , in the reference frame of the laboratory. He assumes , 1.7 pt We would like to have the lowest temperature that can be achieved using this technique. He assumes That the velocity of a particular atom has been reduced to zero exactly, and at this very moment it absorbs a photon (incident from any of the two directions), and re-emits the photon randomly in Any of the two directions, with almost the same frequency. Assume that this process happens once every units of time. B-3 Considering the momentum of the atom after such a process for the two The average power absorbed by the atom. 1.0 pt $\text{omegaA}$ $\text{omegaL}$ v = vx s+ $s-$ $\pi +$ $\pi -$ v kL = $\text{omegaL}/c$ ħ $\alpha$ sL $\ll \alpha \text{omegaL}$ $\tau$ The score: 30 Time: 5.0 hours The following information shall be provided: Theory, English (Official) Page 10 of 16

B-4 Consider the force calculated in Task B-2 and calculate the output power. Then, calculate the average value of The Commission is not in a position to take any further action. Using your knowledge of the kinetic theory of gases estimate the temperature of the The atomic number is not a single atom. 0.8 pt B-5 Estimate this temperature, for an atom 100 times heavier than a hydrogen. The atom. Assumes that , , and . 0.4 pt mass of hydrogen atom: charge of an electron:

Topic: Electrostatics, Oscillations & Waves, Modern-Quantum Physics Metodi: Coulomb’s Law, Differential Equations, Simple Harmonic Motion Analysis, Approximation & Series Expansion, Photon Energy Relation, Kinetic Theory of Gases Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Objects: Point Charge, Atom, Photon Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)