1. A hydrogen atom is placed in a (time-dependent) electric field E =
E(t)k. Calculate all four matrix elements H′ij of the perturbation H' =
eEz between the ground state (n = 1) and the (quadruply degenerate)
first excited states (n = 2). Also show that H'ij = 0 for all five
states. Note: There is only one integral to be done here, if you exploit
oddness with respect to z; only one of the n = 2 states is "accessible"
from the ground state by a perturbation of this form, and therefore the
system functions as a two-state configuration—assuming transitions to
higher excited states can be ignored. Get solution
2. Solve Equation 1 for the case of a time-independent perturbation, assuming that ca (0) = 1 and cb (0) = 0. Check that |ca(t)|2 + |cb(t)|2 = 1. Comment: Ostensibly, this system oscillates between "pure ψa" and "some ψb" Doesn't this contradict my general assertion that no transitions occur for time- independent perturbations? No, but the reason is rather subtle: In this case and ψb are not, and never were, eigenstates of the Hamiltonian—a measurement of the energy never yields Ea or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, and are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time t—this doesn't affect the calculations, but it allows for a more sensible interpretation of the result.Equation 1... Get solution
3. Suppose the perturbation takes the form of a delta function (in time):...assume that Uaa = Ubb = 0, and let Uab = U*ba = α. If ca(–∞) = iandcb(–∞) = 0, find ca(t) and cb(t), and check that |ca(t)|2 + |cb(t)|2 = 1. What is the net probability (Pa→b for t→∞) that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.Answer-Pa→b = sin2(|α|/h). Get solution
4. Suppose you don't assume H'aa = H'bb = 0.(a) Find ca (t) and cb(t) in first-order perturbation theory, for the case ca(0) = 1, cb(0) = 0. Show that |ca(1)(t)|2 + |cb(1)(t)|2 = 1, to first order in H'.(b) There is a nicer way to handle this problem. Let...Show that...where...So the equations for da and db are identical in structure to Equation 1 (with an extra factor ei➢ tacked onto H').(c) Use the method in part (b) to obtain ca(t) and cb(t) in first-order perturbation theory, and compare your answer to (a). Comment on any discrepancies.Equation 1... Get solution
5. Solve Equation 1 to second order in perturbation theory, for the general case ca(0) = a, cb(0) = b.Equation 1... Get solution
6. Calculate ca(t) and cb(t), to second order, for a time-independent perturbation (Problem 1). Compare your answer with the exact result.Problem 1Solve Equation 1 for the case of a time-independent perturbation, assuming that ca (0) = 1 and cb (0) = 0. Check that |ca(t)|2 + |cb(t)|2 = 1. Comment: Ostensibly, this system oscillates between "pure ψa" and "some ψb" Doesn't this contradict my general assertion that no transitions occur for time- independent perturbations? No, but the reason is rather subtle: In this case and ψb are not, and never were, eigenstates of the Hamiltonian—a measurement of the energy never yields Ea or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, and are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time t—this doesn't affect the calculations, but it allows for a more sensible interpretation of the result.Equation 1... Get solution
7. The first term in Equation 9.25 comes from the ..., and the second from .... Thus dropping the first term is formally equivalent to writing ..., which is to say, ... (The latter is required to make the Hamiltonian matrix Hermitian-or, if you prefer, to pick out the dominant term in the formula analogous to Equation 9.25 for ...) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 9.13 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the field. ... ... Get solution
8. As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody is the source). Show that at room temperature (/ = 300 K) thermal Simulation dominates, for frequencies well below 5 × 1012 Hz, whereas spontaneous emission dominates for frequencies well above 5 × 1012 Hz. Which mechanism dominates for visible light? Get solution
9. You could derive the spontaneous emission rate (Equation 9.56) with-out the detour through Einstein's A and B coefficients if you-knew the ground-stale energy density of the electromagnetic field, ..., for then it would simply be a as f stimulated emission (Equation 9.47). To do this honestly would require quantum electrodynamics, but if you are prepared to believe that the ground state consists of one photon in each mode, then the derivation is very simple: (a) Replace Equation 5.111 by (b) Use your result, together with Equation 9.47, to obtain the spontaneous emission rate. Compare Equation 9.56. ... Get solution
10. The half-life (t1/2) of an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation between t1/2 and τ (the "lifetime" of the state). Get solution
11. Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You'll need to evaluate matrix elements of the form (ψ100|x|ψ200), (ψ100|y|ψ211), and so on. Remember that x = r sin θ cos ➢, y = r sin θ sin ➢, and z = r cos θ. Most of these integrals are zero, so scan them before you start calculating. Answer: 1.60 × 10–9 seconds for all except ψ200, which is infinite. Get solution
12. Prove the communication relation in Equation 9.74. Hint: First show that ... ... Get solution
13. Close the "loophole" in Equation 1 by showing that if l' = l = 0 then (n'l'm'|r|nlm) =0.Equation 1... Get solution
14. An electron in the n — 3, l = 0, m = 0 state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state.(a) What decay routes are open to it? Specify them in the following way:...(b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route?(c) What is the lifetime of this state? Hint: Once it's made the first transition, it's no longer in the state |300), so only the first step in each sequence is relevant in computing the lifetime. When there is more than one decay route open, the transition rates add. Get solution
15. Develop time-dependent perturbation theory for a multilevel system, starting with the generalization of Equations 9.1 and 9.2: ... ... ... ... Get solution
16. For the examples in Problem 9.15(c) and (d), calculate ..., to first order. Check the normalization condition: ... and comment on any discrepancy., Suppose you wanted to calculate the probability_ of remaining in the original state ...; would you do better to use ... or ... ... ... Get solution
17. A particle starts out (at time t = 0) in the Nth state of the infinite square well. Now the "floor" of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent: V0(t), with V0(0) = V0(T) = 0.(a) Solve for the exact cm (t) using Equation 1, and show that the wave function changes phase, but no transitions occur. Find the phase change, ➢(T), in terms of the function V0(t).(b) Analyze the same problem in first-order perturbation theory, and compare your answers.Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in t) to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.Equation 1...Problem 1Suppose you add a constant V0 to the potential energy (by “constant” I mean independent of x as well as t). In classical mechanics this doesn’t change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: exp(-i V0t/ћ). What effect does this have on the expectation value of a dynamical variable? Get solution
18. A particle of mass m is initially in the ground state of the (one-dimensional) infinite square well. At time t = 0 a "brick" is dropped into the well, so that the potential becomes...where V0 «E1. After a time T, the brick is removed, and the energy of the particle is measured. Find the probability (in first-order perturbation theory) that the result is now E2. Get solution
19. We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption? Get solution
20. Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio γ, at rest in a static magnetic field B0k, precesses at the Larmor frequencyω0 = γ B0 Now we turn on a small transverse radiofrequency (rf) field, Brf[cos(ωt)0 î – sin (ωt) j], so that the total field is...(a) Construct the 2 × 2 Hamiltonian matrix for this system.(b) If x(t) = ... is the spin state at time t, show that...where Ω. ≡ γ Brf is related to the strength of the rf field.(c) Check that the general solution for a(t) and b(t), in terms of their initial values a0 and b0, is...Where...(d) If the particle starts out with spin up (i.e., a0 = 1, b0 = 0), find the probability of a transition to spin dawn, as a function of time. Answer: P(t) = {Ω2/[(ω – ω0)2 + Ω]} sin2(ω't/2).(e) Sketch the resonance curve,...as a function of the driving frequency ω (for fixed ω0 and Ω). Note that the maximum occurs at ω = ω0. Find the "full width at half maximum," .∆ω.(f) Since ω0 = γ B0, we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the g-factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude 0.01 gauss. What will the resonant frequency be? (See Section 6.5 for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.) Get solution
21. In Equation 9.31 I assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field...can be ignored. The true electric field would be ... If the atom is centered at the origin, then k • r ... The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k • r lead to even more "forbidden" transitions, associated with higher multipole moments).21 (a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer: ... ... ... Get solution
22. Show that the spontaneous emission rate for a transition from n l to n’, l’ in hydrogen is...where...(The atom starts out with a specific value of m, and it goes to any of the states m' consistent with the selection rules: m' = m + 1, m, or m — 1. Notice that the answer is independent of m.) Hint: First calculate all the nonzero matrix elements of x, y, and z between |nlm) and |n'l'm') for the case l' = l + 1. From these, determine the quantity...Then do the same for l' = l – 1. Get solution
2. Solve Equation 1 for the case of a time-independent perturbation, assuming that ca (0) = 1 and cb (0) = 0. Check that |ca(t)|2 + |cb(t)|2 = 1. Comment: Ostensibly, this system oscillates between "pure ψa" and "some ψb" Doesn't this contradict my general assertion that no transitions occur for time- independent perturbations? No, but the reason is rather subtle: In this case and ψb are not, and never were, eigenstates of the Hamiltonian—a measurement of the energy never yields Ea or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, and are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time t—this doesn't affect the calculations, but it allows for a more sensible interpretation of the result.Equation 1... Get solution
3. Suppose the perturbation takes the form of a delta function (in time):...assume that Uaa = Ubb = 0, and let Uab = U*ba = α. If ca(–∞) = iandcb(–∞) = 0, find ca(t) and cb(t), and check that |ca(t)|2 + |cb(t)|2 = 1. What is the net probability (Pa→b for t→∞) that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.Answer-Pa→b = sin2(|α|/h). Get solution
4. Suppose you don't assume H'aa = H'bb = 0.(a) Find ca (t) and cb(t) in first-order perturbation theory, for the case ca(0) = 1, cb(0) = 0. Show that |ca(1)(t)|2 + |cb(1)(t)|2 = 1, to first order in H'.(b) There is a nicer way to handle this problem. Let...Show that...where...So the equations for da and db are identical in structure to Equation 1 (with an extra factor ei➢ tacked onto H').(c) Use the method in part (b) to obtain ca(t) and cb(t) in first-order perturbation theory, and compare your answer to (a). Comment on any discrepancies.Equation 1... Get solution
5. Solve Equation 1 to second order in perturbation theory, for the general case ca(0) = a, cb(0) = b.Equation 1... Get solution
6. Calculate ca(t) and cb(t), to second order, for a time-independent perturbation (Problem 1). Compare your answer with the exact result.Problem 1Solve Equation 1 for the case of a time-independent perturbation, assuming that ca (0) = 1 and cb (0) = 0. Check that |ca(t)|2 + |cb(t)|2 = 1. Comment: Ostensibly, this system oscillates between "pure ψa" and "some ψb" Doesn't this contradict my general assertion that no transitions occur for time- independent perturbations? No, but the reason is rather subtle: In this case and ψb are not, and never were, eigenstates of the Hamiltonian—a measurement of the energy never yields Ea or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, and are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time t—this doesn't affect the calculations, but it allows for a more sensible interpretation of the result.Equation 1... Get solution
7. The first term in Equation 9.25 comes from the ..., and the second from .... Thus dropping the first term is formally equivalent to writing ..., which is to say, ... (The latter is required to make the Hamiltonian matrix Hermitian-or, if you prefer, to pick out the dominant term in the formula analogous to Equation 9.25 for ...) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 9.13 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the field. ... ... Get solution
8. As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody is the source). Show that at room temperature (/ = 300 K) thermal Simulation dominates, for frequencies well below 5 × 1012 Hz, whereas spontaneous emission dominates for frequencies well above 5 × 1012 Hz. Which mechanism dominates for visible light? Get solution
9. You could derive the spontaneous emission rate (Equation 9.56) with-out the detour through Einstein's A and B coefficients if you-knew the ground-stale energy density of the electromagnetic field, ..., for then it would simply be a as f stimulated emission (Equation 9.47). To do this honestly would require quantum electrodynamics, but if you are prepared to believe that the ground state consists of one photon in each mode, then the derivation is very simple: (a) Replace Equation 5.111 by (b) Use your result, together with Equation 9.47, to obtain the spontaneous emission rate. Compare Equation 9.56. ... Get solution
10. The half-life (t1/2) of an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation between t1/2 and τ (the "lifetime" of the state). Get solution
11. Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You'll need to evaluate matrix elements of the form (ψ100|x|ψ200), (ψ100|y|ψ211), and so on. Remember that x = r sin θ cos ➢, y = r sin θ sin ➢, and z = r cos θ. Most of these integrals are zero, so scan them before you start calculating. Answer: 1.60 × 10–9 seconds for all except ψ200, which is infinite. Get solution
12. Prove the communication relation in Equation 9.74. Hint: First show that ... ... Get solution
13. Close the "loophole" in Equation 1 by showing that if l' = l = 0 then (n'l'm'|r|nlm) =0.Equation 1... Get solution
14. An electron in the n — 3, l = 0, m = 0 state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state.(a) What decay routes are open to it? Specify them in the following way:...(b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route?(c) What is the lifetime of this state? Hint: Once it's made the first transition, it's no longer in the state |300), so only the first step in each sequence is relevant in computing the lifetime. When there is more than one decay route open, the transition rates add. Get solution
15. Develop time-dependent perturbation theory for a multilevel system, starting with the generalization of Equations 9.1 and 9.2: ... ... ... ... Get solution
16. For the examples in Problem 9.15(c) and (d), calculate ..., to first order. Check the normalization condition: ... and comment on any discrepancy., Suppose you wanted to calculate the probability_ of remaining in the original state ...; would you do better to use ... or ... ... ... Get solution
17. A particle starts out (at time t = 0) in the Nth state of the infinite square well. Now the "floor" of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent: V0(t), with V0(0) = V0(T) = 0.(a) Solve for the exact cm (t) using Equation 1, and show that the wave function changes phase, but no transitions occur. Find the phase change, ➢(T), in terms of the function V0(t).(b) Analyze the same problem in first-order perturbation theory, and compare your answers.Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in t) to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.Equation 1...Problem 1Suppose you add a constant V0 to the potential energy (by “constant” I mean independent of x as well as t). In classical mechanics this doesn’t change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: exp(-i V0t/ћ). What effect does this have on the expectation value of a dynamical variable? Get solution
18. A particle of mass m is initially in the ground state of the (one-dimensional) infinite square well. At time t = 0 a "brick" is dropped into the well, so that the potential becomes...where V0 «E1. After a time T, the brick is removed, and the energy of the particle is measured. Find the probability (in first-order perturbation theory) that the result is now E2. Get solution
19. We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption? Get solution
20. Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio γ, at rest in a static magnetic field B0k, precesses at the Larmor frequencyω0 = γ B0 Now we turn on a small transverse radiofrequency (rf) field, Brf[cos(ωt)0 î – sin (ωt) j], so that the total field is...(a) Construct the 2 × 2 Hamiltonian matrix for this system.(b) If x(t) = ... is the spin state at time t, show that...where Ω. ≡ γ Brf is related to the strength of the rf field.(c) Check that the general solution for a(t) and b(t), in terms of their initial values a0 and b0, is...Where...(d) If the particle starts out with spin up (i.e., a0 = 1, b0 = 0), find the probability of a transition to spin dawn, as a function of time. Answer: P(t) = {Ω2/[(ω – ω0)2 + Ω]} sin2(ω't/2).(e) Sketch the resonance curve,...as a function of the driving frequency ω (for fixed ω0 and Ω). Note that the maximum occurs at ω = ω0. Find the "full width at half maximum," .∆ω.(f) Since ω0 = γ B0, we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the g-factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude 0.01 gauss. What will the resonant frequency be? (See Section 6.5 for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.) Get solution
21. In Equation 9.31 I assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field...can be ignored. The true electric field would be ... If the atom is centered at the origin, then k • r ... The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k • r lead to even more "forbidden" transitions, associated with higher multipole moments).21 (a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer: ... ... ... Get solution
22. Show that the spontaneous emission rate for a transition from n l to n’, l’ in hydrogen is...where...(The atom starts out with a specific value of m, and it goes to any of the states m' consistent with the selection rules: m' = m + 1, m, or m — 1. Notice that the answer is independent of m.) Hint: First calculate all the nonzero matrix elements of x, y, and z between |nlm) and |n'l'm') for the case l' = l + 1. From these, determine the quantity...Then do the same for l' = l – 1. Get solution
