### Chapter #11 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

1. Rutherford scattering. An incident particle of charge q1 and kinetic energy E scatters off a heavy stationary particle of charge q2.(a) Derive the formula relating the impact parameter to the scattering angle. Answer: b = (q1q2/8π∑0E) cot(θ/2).(a) Determine the differential scattering cross-section. Answer:...(c) Show that the total cross-section for Rutherford scattering is infinite. We say that the 1/r potential has "infinite range"; you can't escape from a Coulomb force. Get solution

2. Construct the analogs to Equation 11.12 for one-dimensional and two-dimensional scattering. Get solution

3. Prove Equation 11.33, starting with Equation 11.32. Hint: Exploit the orthogonality of the Legendre polynomials to show that the coefficients with different values of l must separately vanish. Get solution

4. Consider the case of low-energyscattering from a spherical delta function shell:V(r) = αδ(r-a)Where α and a are constants. Calculate the scattering amplitude, f(θ), the differenential cross-section, d(θ), and the total cross-section, σ. Assume ka « 1, so that only the l = 0 term contributes significantly. (To simplify matters, throw out all I ≠ 0 terms right from the start.) The main problem, of course, is to determine ao. Express your answer in terms of the dimensionless quantity ß ≡ 2maa/h2. Answer: σ = 4πa2ß2/(l + ß). Get solution

5. A particle of mass m and energy E is incident from the left on the potential...(a) If the incoming wave is Ae'kx (where ...), find the reflected wave.Answer..., where ....(b) Confirm that the reflected wave has the same amplitude as the incident wave.(b) Find the phase shift δ (Equation 1) for a very deep well (E « v0)Answer: δ = —ka.Equation 1Ψo(x) = A(eikx – e-ikx) (V(x) = 0). Get solution

6. What are the partial wave phase shifts ... for hard-sphere scattering (Example 11.3)? ... ... That's the exact answer, but it's not terribly illuminating, so let's consider the limiting case of low-energy scattering: ..., this amounts to saying that the wavelength is much greater than the radius of the sphere.) Referring to Table 4.4, we note that ..., for small z, so ... Get solution

7. Find the S-wave (l = 0) partial wave phase shift δ0(k) for scattering from a delta-function shell (Problem 1). Assume that the ratial wave function u(r) goes to o as r →∞ Answer:...Problem 1Consider the case of low-energyscattering from a spherical delta function shell:V(r) = αδ(r-a)Where α and a are constants. Calculate the scattering amplitude, f(θ), the differenential cross-section, d(θ), and the total cross-section, σ. Assume ka « 1, so that only the l = 0 term contributes significantly. (To simplify matters, throw out all I ≠ 0 terms right from the start.) The main problem, of course, is to determine ao. Express your answer in terms of the dimensionless quantity ß ≡ 2maa/h2. Answer: σ = 4πa2ß2/(l + ß). Get solution

8. Check that Equation 11.65 satisfies Equation 11.52, by direct substitution. Hint: ... ... Get solution

9. Show that the ground state of hydrogen (Equation 4.80) satisfies the integral form of the Schrodinger equation, for the appropriate V and E (note that E is negative, so .... ... Get solution

10. Find the scattering amplitude, in the Born approximation, for soft-sphere scattering at arbitrary energy. Show that your formula reduces to Equation 11.82 in the low-energy limit. ... Get solution

11. Evaluate the integral in Equation 11.91, to confirm the expression on the right. ... Get solution

12. Calculate the total cross-section for scattering from a Yukawa potential, in the Born approximation. Express your answer as a function of E Get solution

13. For the potential in Problem 1,(a) calculate f(θ), D(θ), and σ, in the low-energy Born approximation;(b) calculate f (θ) for arbitrary energies, in the Born approximation;(c) show that your results are consistent with the answer to Problem 11.4, in the appropriate regime.Problem 1Calculate θ (as a function of the impact parameter) for Rutherford scattering, in the impulse approximation. Show that your result is consistent with the exact expression (Problem 1), in the appropriate limit.Problem 1Rutherford scattering. An incident particle of charge q1 and kinetic energy E scatters off a heavy stationary particle of charge q2.(a) Derive the formula relating the impact parameter to the scattering angle. Answer: b = (q1q2/8π∑0E) cot(θ/2). Get solution

14. Calculate θ (as a function of the impact parameter) for Rutherford scattering, in the impulse approximation. Show that your result is consistent with the exact expression (Problem 1), in the appropriate limit.Problem 1Rutherford scattering. An incident particle of charge q1 and kinetic energy E scatters off a heavy stationary particle of charge q2.(a) Derive the formula relating the impact parameter to the scattering angle. Answer: b = (q1q2/8π∑0E) cot(θ/2). Get solution

15. Find the scattering amplitude for low-energy soft-sphere scattering in the second Born approximation. Answer: —(2m V­­oa3/3h2)[1 - (4mVoa2/5h2)]. Get solution

16. Find the Green's function for the one-dimensional Schrodinger equation, and use it to construct the integral form (analogous to Equation 1). Answer:...Equation 1... Get solution

17. Use your result in problem 1 to develop the Born approximation for one-dimensional scattering (on the interval -∞ ...problem 1Find the Green's function for the one-dimensional Schrodinger equation, and use it to construct the integral form (analogous to Equation 1). Answer:...Equation 1... Get solution

18. Use the one-dimensional Born approximation (Problem 11.17) to compute the transmission coefficient (T = 1 — R) for scattering from a delta function (Equation 2.114) and from a finite square well (Equation 2.145). Compare your results with the exact answers (Equations 2.141 and 2.169). V (x) = -αδ(x) Get solution

19. Prove the optical theorem, which relates the total cross-section to the imaginary part of the forward scattering amplitude....Hint : Use Equations 1 and 2.Equation 1...Equation 2... Get solution

20. Use the Bom approximation to determine the total cross-section for scattering from a gaussian potential...Express your answer in terms of the constants A, μ, and m (the mass of the incident particle), and k ≡ ..., where E is the incident energy. Get solution

### Chapter #10 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

1. The case of an infinite square well whose right wall expands at a constant velocity (v) can be solved exactly2. A complete set of solutions isEquation 1...where w(t) ≡ a + vt is the (instantaneous) width of the well and E'n ≡ n2π2h2/2ma2 is the nth allowed energy of the original well (width a). The general solution is a linear combination of the Φ's:...the coefficients cn are independent of t.(a) Check that Equation 1 satisfies the time-dependent Schròdinger equation, with the appropriate boundary conditions.(b) Suppose a particle starts out (t = 0) in the ground state of the initial well:...Show that the expansion coefficients can be written in the form...where α = mva/2 π 2h is a dimensionless measure of the speed with which the well expands. (Unfortunately, this integral cannot be evaluated in terms of elementary functions.)(c) Suppose we allow the well to expand to twice its original width, so the "external" time is given by w(Te) = 2a. The "internal" time is the period of the time-dependent exponential factor in the (initial) ground state. Determine Te and Ti, and show that the adiabatic regime corresponds to a « 1, so that exp(—iαz2) ≡ 1 over the domain of integration. Use this to determine the expansion coefficients, cn. Construct Ψ(x, t), and confirm that it is consistent with the adiabatic theorem.(d) Show that the phase factor in Ψ (x, t) can be written in the form...where En(t) = n2π2h2/2mw2 is the instantaneous eigenvalue, at time t. Comment on this result. Get solution

2. Check that Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33 and show that the sum-of the squares of the coefficients is 1, as required for normalization. ... Get solution

3. (a) Use Equation 10.42 to calculate the geometric phase change when the infinite square well expands adiabatically from width W1 to width W2. Comment on this result.(b) If the expansion occurs at a constant rate (dw/dt = v), what is the dynamic phase change for this process?(c) If the well now contracts back to its original size, what is Berry's phase for the cycle? Get solution

4. The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change when α gradually increases from ..., what is the dynamic phase change for this process? V (x) = -αδ(x) Get solution

5. Show that if ... is real, the geometric phase vanishes. (Problems 10.3and 10.4 are examples of this.) You might try to beat the rap by tracking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: ... where ... is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions. V(x) = -αδ(x) ... Get solution

6. Work out the analog to Equation 1 for a particle of spin 1.Equation 1... Get solution

7. (a) Derive Equation 10.67 from Equation 10.65. (b) Derive Equation 10.79, starting with Equation 10.78. ... ... Get solution

8. A particle starts out in the ground state of the infinite square well (on the interval 0 ≤ x ≤ a). Now a wall is slowly erected, slightly off-center:...Where f(t) rises gradually from 0 to ∞. Accoroding to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.(a) Find (and sketch) the ground state at t → ∞. Hint: This should be the ground state of the infinite square well with an impenetrable barrier at a/2 + ∑. Note that the particle is confined to the (slightly) larger left “half” of the well.(b) find the (transcendental) equation for the ground state of the Hamiltonian at time t. Answer:...Where z ≡ ka, T ≡ maf(t)/h2, δ ≡ 2∑/a, and k ≡ ....(c) Setting δ = 0, solve grahically for z, and show that the smallest z goes from π to 2 π as T goes from 0 to ∞. Explain this result.(d) Now set δ = 0.01, solve numerically foe z, using T = 0,1,5,20,100. And 1000.(e) Find the prbalility pr that the particle is in the right “half” of the well, as a function of z and δ. Answer : Pr = 1/[1+(I+/I-)], where I± ≡ [1±δ-(1/z) sin (z(1±δ))] sin2[z(1 ... δ)/2]. Evalute this expression numerically for the T’s in part (d). Comment on your results.(f) Plot the ground state wave function for those same values of T and δ. Note how it gets squeezed into the left half of the well, as the barrier grows. Get solution

9. Suppose the one-dimensional harmonic oscillator (mass m, frequency ω) is subjected to a driving force of the form ..., where f(t) is some specified function (I have factored out ...for notational convenience; f (t) has the dimensions of length). The Hamiltonian is ... Assume that the force was first turned on at time .... This system can be solved exactly both in classical mechanics and in quantum mechanics.2t (a) Determine the classical position of the oscillator, assuming it started from rest at the origin ... Answer: ... (b) Show that the solution to the (time-dependent) Schrodinger equation for this oscillator, assuming it started out in the nth state of the undriven oscillator ... is given by Equation 2.61), can be written as ... ... Get solution

10. The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficients cm(t) in Equation 10.12. Suppose the system starts out in the nth state; in the adiabatic approximation, it remains in the nth state, picking up only a time-dependent geometric phase factor (Equation 10.21):...(a) Substitute this into the right side of Equation 10.16 to obtain the "first correction" to adiabaticity:...  (1)This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equation 1 on the right side of Equation 10.16, and so on.(b) As an example, apply Equation 1 to the driven oscillator (Problem 10.9). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which...(The transition probabilities are the absolute squares of these, of course.) Get solution

### Chapter #9 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

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

### Chapter #8 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

1. Use the WKB approximation to find the allowed energies (En) of an infinite square well with a “shelf,” of height V0 extending half-way across (Figure 6.3):...Express your answer in terms of V0 and ... (the nthallowedenergy for the infinite square well with no shelf). Assume that ..., but do not assume that En >> V0. Compare your result with what we got in Example 6.1 using first-order perturbation theory. Note that they are in agreement if either V0 is very small (the perturbation theory regime) or n is very large (the WKB—semi-classical—regime). Get solution

2. An illuminating alternative derivation of the WKB formula (Equation 1) is based on an expansion in powers of ћ. Motivated by the free-particle wave function, ψ= A exp(± ipx/ћ), we write...where f(x) is some complex function. (Note that there is no loss of generality here—any nonzero function can be written in this way.)(a) Put this into Schrödinger's equation (in the form of Equation 8.1), and shov that...(b) Write f(x) as a power series in ћ:...and, collecting like powers of ћ, show that... etc.(c) Solve for f0(x) and f1(x), and show that—to first order in ћ—you recover Equation 1.Note: The logarithm of a negative number is defined by ln(‒z) = ln(z) + inπ, where n is an odd integer. If this formula is new to you, try exponentiating both sides, and you'll see where it comes from.Equation 1... Get solution

3. . Potential well with one vertical wall. Imagine a potential well that has one vertical side (at x = 0) and one sloping side (Figure 8.10). In this case ..., so Equation 8.46 says ... ... Get solution

4. Calculate the lifetimes of U238 and Po212, using Equations 1 and 8.25. Hint: The density of nuclear matter is relatively constant (i.e., the same for all nuclei), so (r1)3 is proportional to A (the number of neutrons plus protons). Empirically,...[8.29]...FIGURE 8.6: Graph of the logarithm of the lifetime versus 1/√E (where E is the energy of the emitted alpha particle), for uranium and thorium.The energy of the emitted alpha particle can be deduced by using Einstein's formula (E = mc2):... [8.30]where mp is the mass of the parent nucleus, md is the mass of the daughter nucleus, and mα is the mass of the alpha particle (which is to say, the He4 nucleus). To figure out what the daughter nucleus is, note that the alpha particle carries off two protons and two neutrons, so Z decreases by 2 and A by 4. Look up the relevant nuclear masses. To estimate v, use E = (1/2)mav2; this ignores the (negative) potential energy inside the nucleus, and surely underestimates v, but it's about the best we can do at this stage. Incidentally, the experimental lifetimes are 6 x 109 yrs and 0.5 μs, respectively.Equations 1... Get solution

5. Consider the quantum mechanical analog to the classical problem of a ball (mass m) bouncing elastically on the floor.13(a) What is the potential energy, as a function of height x above the floor? (For negative x, the potential is infinite—the ball can't get there at all.)(b) Solve the Schrödinger equation for this potential, expressing your answer in terms of the appropriate Airy function (note that Bi(z) blows up for large z, and must therefore be rejected). Don’t bother to normalize ψ(x).(c) Using g = 9.80 m/s2 and m = 0.100 kg, find the first four allowed energies, in joules, correct to three significant digits. Hint: See Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions, Dover, New York (1970), page 478; the notation is defined on page 450.(d) What is the ground state energy, in eV, of an electron in this gravitational field? How high off the ground is this electron, on the average? Hint: Use the virial theorem to determine ...x.... Get solution

6. Analyze the bouncing ball (Problem 1) using the WKB approximation.(a) Find the allowed energies, En, in terms of m, g, and h.(b) Now put in the particular values given in Problem 8.5(c), and compare the WKB approximation to the first four energies with the “exact” results.(c) About how large would the quantum number n have to be to give the ball an average height of, say, 1 meter above the ground?Problem 1Consider the quantum mechanical analog to the classical problem of a ball (mass m) bouncing elastically on the floor.13(a) What is the potential energy, as a function of height x above the floor? (For negative x, the potential is infinite—the ball can't get there at all.)(b) Solve the Schrödinger equation for this potential, expressing your answer in terms of the appropriate Airy function (note that Bi(z) blows up for large z, and must therefore be rejected). Don’t bother to normalize ψ(x).(c) Using g = 9.80 m/s2 and m = 0.100 kg, find the first four allowed energies, in joules, correct to three significant digits. Hint: See Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions, Dover, New York (1970), page 478; the notation is defined on page 450.(d) What is the ground state energy, in eV, of an electron in this gravitational field? How high off the ground is this electron, on the average? Hint: Use the virial theorem to determine ...x.... Get solution

7. Use the WKB approximation to find the allowed energies of the harmonic oscillator. Get solution

8. Consider a particle of mass m in the nth stationary state of the harmonic oscillator (angular frequency-ω) (a) Find the turning point, x2. (b) How far (d) could you go above the turning point before the error in the linearized potential (Equation 8.32, but with the turning point at x2) reaches 1%? That is, if ... (c) The. asymptotic form of Ai (z) is accurate to 1% as long as z > 5. For the d in part (b), determine the smallest it such that ad > 5. (For any it larger than this there exists an overlap region in which the linearized potential is good to 1% and the large-z form of the Airy function is good to 1%.) ... Get solution

9. Derive the connection formulas at a downward-sloping turning point, and confirm Equation 1.Equation 1... Get solution

10. Use appropriate connection formulas to analyze the problem of scattering from a barrier with sloping walls (Figure 8.12). Hint: Begin by writing the WKB wave function in the form ... Do not assume C = 0. Calculate the tunneling probability, ..., and show that your result reduces to Equation 8.22 in the case of a broad, high bather. ... ... Get solution

11. Use the WKB approximation to find the allowed energies of the general power-law potential:...where v is a positive number. Check your result for the case v = 2. Answer:14... [8.53] Get solution

12. Use the WKB approximation to find the bound state energy for the potential in Problem 2.51. Compare the exact answer. Answer: ‒[(9/8) ‒ (1/√2)] ћ2a2/m. Get solution

13. For spherically symmetrical potentials we can apply the WKB approximation to the radial part (Equation 1). In the case l = 0 it is reasonable15 to use Equation 8.47 in the form... [8.54]where r0 is the turning point (in effect, we treat r = 0 as an infinite wall) Exploit this formula to estimate the allowed energies of a particle in the logarithmic potential...(for constants V0 and a). Treat only the case l = 0. Show that the spacing between the levels is independent of mass. Partial answer:...Equation 1... Get solution

14. Use the WKB approximation in the form... [8.55]to estimate the bound state energies for hydrogen. Don’t forget the centrifugal term in the effective potential (Equation 1). The following integral may help:... [8.56]Note that you recover the Bohr levels when n >> l and n >> 1/2. Answer:... [8.57]Equation 1... Get solution

15. Consider the case of a symmetrical double well, such as the one pictured in Figure 8.13. We are interested in bound states with E (a) Write down the WKB wave functions in regions ..., and .... Impose the appropriate connection formulas at xi and x2 (this has already been done, in Equation 8.46, for x2; you will have to work out x1 for yourself), to show that ... ... Get solution

16. Tunneling in the Stark effect. When you turn on an external electric field, the electron in an atom can, in principle, tunnel out, ionizing the atom. Question: Is this likely to happen in a typical Stark effect experiment? We can estimate the probability using a crude one-dimensional model, as follows. Imagine a particle in a very deep finite square well (Section 2.6). (a) What is the energy of the ground state, measured up from the bottom of the well? Assume .... Hint: This is just the ground state energy of the infinite square well (of width 2a). (b) Now introduce a perturbation H' = -aα (for an electron in an electric field ...we would have .... Assume it is relatively weak .... Sketch the total potential, and note that the particle can now tunnel out, in the direction of poTitiecr (c) Calculate the tunnelisfactor y (Equation 8.22), and estimate the time it would take itirthe particle to escape (Equation 8.28). Answer: γ = .... ... Get solution

17. About how long would it take for a can of beer at room temperature to topple over spontaneously, as a result of quantum tunneling? Hint: Treat it as a uniform cylinder of mass in, radius R, and length h. As the can tips, let x be the height of the center above its equilibrium position (h/2). The potential enemy is mgx, and it topples when x reaches the critical value ... Calculate the tunneling probability (Equation 8.22), for E = 0. Use Equation 8.28, with the thermal energy ... to estimate the velocity. Put in reasonable numbers, and give your final answer in years.17 ... Get solution

### Chapter #7 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

1. Use a gaussian trial function (Equation 7.2) to obtain the lowest upper bound you can on the ground state energy of (a) the linear potential: V(x) ... (b) the quartic potential: .... ... Get solution

2. Find the best bound on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form...where A is determined by normalization and b is an adjustable parameter. Get solution

3. Find the best bound on ...for the delta-function potential V(x) = ..., using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter. ... Get solution

4. (a) Prove the following corollary to the variational principle: If 〈ψ | ψgs〉 = 0, then 〈H〉 ≥ Efe, where Efe is the energy of the first excited state.Thus, if we can find a trial function that is orthogonal to the exact ground state, we can get an upper bound on the first excited state. In general, it's difficult to be sure that is orthogonal to ψgs, since (presumably) we don't know the latter. However, if the potential V(x) is an even function of x, then the ground state is likewise even, and hence any odd trial function will automatically meet the condition for the corollary.(b) Find the best bound on the first excited state of the one-dimensional harmonic oscillator using the trial functionΨ (x)= Axe–bx. Get solution

5. (a) Use the variational principle to prove that first-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy. (b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Confirm that this is indeed the case, by examining Equation 6.15. ... Get solution

6. Using Egs = – 79.0 eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies. Get solution

7. Apply the techniques of this Section to the H− and Li+ ions (each has two electrons, like helium, but nuclear charges Z = 1 and Z = 3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on Egs, for each case. Comment: In the case of H− you should find that 〈H〉>–13.6 eV, which would appear to indicate that there is no bound state at all, since it would be energetically favorable for one electron to fly off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 7.18) it can be shown that Egs–13.6 eV, and hence that a bound state does exist. It's only barely bound, however, and there are no excited bound states,11 so H− has no discrete spectrum (all transitions are to and from the continuum). As a result, it is difficult to study in the laboratory, although it exists in great abundance on the surface of the sun12 Get solution

8. Evaluate D and X (Equations 7.45 and 7.46). Check your answers against Equations 7.47 and 7.48. Get solution

9. Suppose we used a minus sign in our trial wave function (Equation 7.37) ... Without doing any new integrals, find F(x) (the analog to Equation 7.51) for this case, and construct the graph. Show that there is no evidence of bonding.13 (Since the variational principle only gives an upper bound, this doesn't prove that bonding cannot occur for such a state, but it certainly doesn't look promising). Comment: Actually, any function of the form ... has the desired property that the electron is equally likely to be associated with either proton. However, since the Hamiltonian (Equation 7.35) is invariant under the interchange ... its eigenfunctions can be chosen to be 'simultaneously eigenfunctions of P. The plus sign (Equation 7.37) goes with the eigenvalue +1, and the minus sign (Equation 7.52) with the eigenvalue -1; nothing is to be gained by considering the ostensibly more general case (Equation 7.53), though you're welcome to try it, if you're interested. ... ... ... Get solution

10. The second derivative of F(x), at the equilibrium point, can be used to estimate the natural frequency of vibration (ω) of the two protons in the hydrogen molecule ion (see Section 2.3). If the ground state energy (hω/2) of this oscillator exceeds the binding energy of the system, it will fly apart. Show that in fact the oscillator energy is small enough that this will not happen, and estimate how many bound vibrational levels there are. Note: You're not going to be able to obtain the position of the minimum—still less the second derivative at that point—analytically. Do it numerically, on a computer. Get solution

11. (a) Use a trial wave function of the form ... ... Example 7.3 Find an upper bound on the ground state energy of the one-dimensional infinite square well (Equation 2.19), using the "triangular" trial wave function (Figure 7.1).3 ... Get solution

12. (a) Generalize Problem 7.2, using the trial wave function14...for arbitrary n. Partial answer: The best value of b is given by...(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form...Partial answer: The best value of b is given by...(c) Notice that the bounds approach the exact energies as n → ∞ Why is that? Hint: Plot the trial wave functions for n = 2,n = 3, and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity... Get solution

13. Find the lowest bound on the ground state of hydrogen you can get using a gaussian trial wave function...where A is determined by normalization and b is an adjustable parameter. Answer:–11.5 eV. Get solution

14. If the photon had a nonzero mass (mγ ≠ 0), the Coulomb potential would be replaced by the Yukawa potential,...'Where µ = mγc/h. With a trial wave function of your own devising, estimate the binding energy of a "hydrogen" atom with this potential. Assume µa ≪ 1, and give your answer correct to order (µa)2. Get solution

15. Suppose you’re given a quantum system whose Hamiltonian H0 admits just two eigenstates ψa (with energy Ea), and ψb (with energy Eb). They are orthogonal, normalized, and nondegenerate (assume Ea is the smaller of the two energies). Now we turn on a perturbation H', with the following matrix elements:... [7.55]where h is some specified constant.(a) Find the exact eigenvalues of the perturbed Hamiltonian.(b) Estimate the energies of the perturbed system using second-order perturbation theory.(c) Estimate the ground state energy of the perturbed system using the variational principle, with a trial function of the form [7.56]...where ϕ is an adjustable parameter. Note: Writing the linear combination in this way is just a neat way to guarantee that Ψ is normalized.(d) Compare your answers to (a), (b), and (c). Why is the variational principle so accurate, in this case? Get solution

16. (As an explicit example of the method developed in Problem 7.15, consider an electron at rest in a uniform magnetic field ..., for which the Hamiltonian is (Equation 4.158): ... The eigenspinors, ..., and the corresponding energies..., are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction: ... (a) Find the matrix elements of H', and confirm that they have the structure of Equation 735. What is h? (b) Using your result in Problem 7.15(b), find the new ground state energy, in second-order perturbation theory. (c) Using your result in Problem 7.15(c), find the variational principle bound on the ground state energy. ... ... ... Get solution

17. Although the Schrodinger equation for helium itself cannot be solved exactly, there exist "helium-like" systems that do admit exact solutions. ... Get solution

18. In Problem 7.7 we found that the trial wave function with shielding (Equation 7.27), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion. Chandraselchar16 used a trial wave function of the form ... Where ... In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetrized with respect to interchange. The spin state—which is irrelevant to the calculation—is evidently ... ... Get solution

19. The fundamental problem in harnessing nuclear fusion is getting the two particles (say, two deuterons) close enough together for the attractive (but short-range) nuclear force to overcome the Coulomb repulsion. The "bulldozer" method is to heat the particles up to fantastic temperatures, and allow the random collisions to bring them together. A more exotic proposal is muon catalysis, in which we construct a "hydrogen molecule ion," only with deuterons in place of protons, and a muon in place of the electron. Predict the equilibrium separation distance between the deuterons in such a structure, and explain why muons are superior to electrons for this purpose. Get solution

20. Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region shown in Figure 7.8. The "arms" of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state.(a) Show that the lowest energy that can propagate off to infinity is...any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say, x ≫ a), and solve the Schrödinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the form exp(ikxx) with kx > 0....FIGURE 7.8: The cross-shaped region for Problem 7.20.(b) Now use the variational principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Krishna Rajagopal):...Normalize it to determine A, and calculate the expectation value of H. Answer:...Now minimize with respect to α, and show that the result is less than Ethreshold. Hint: Take full advantage of the symmetry of the problem—yon only need to integrate over 1/8 of the open region, since the other 7 integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not—there are "roof-lines" at x = 0, y = 0, x = ± a, and y = ± a, where you will need to exploit the technique of Example 7.3. Get solution

### Chapter #6 Solutions - Introduction to Quantum Mechanics - David J. Griffiths - 2nd Edition

1. Suppose we put a delta-function bump in the center of the infinite square well: ... where α is a constant. (a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even n. (b) Find the first three nonzero terms in the expansion (Equation 6.13) of the correction to the ground state, .... ... Get solution

2. For the harmonic oscillator ... the allowed energies are ... where ... in is the classical frequency. Now suppose the spring constant increases slightly: ... (Perhaps we cool the spring, so it becomes less flexible.) (a) Find the exact new energies (trivial, in this case). Expand your formula as a power series in ?, up to second order. (b) Now calculate the first-order perturbation– in the energy, using Equation 6.9. What is H' here? Compare your result with part (a). Hint: It is not necessary—in fact, it is not permitted —to calculate a single integral in doing this problem. (Reference: Equation) ... Get solution

3. Two identical bosons are placed in an infinite square well (Equation 2.19). They interact weakly with one another, via the potential ... (where V0 is a constant with the dimensions of energy, and a is the width of the well). (a) First, ignoring the interaction between the particles, find the ground state and the first excited state—both the wave functions and the associated energies. (b) Use first-order perturbation theory to estimate the effect of the particle-particle interaction on the energies of the ground state and the first excited state. ... Get solution

4. (a) Find the second-order correction to the energies ...for odd n. (b) Calculate the second-order correction to the ground state energy ... for the potential in Problem 6.2. Check that your result is consistent with the exact solution. (Reference: Problems 6.1) Suppose we put a delta-function bump in the center of the infinite square well: ... where α is a constant. (a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even n. (b) Find the first three nonzero terms in the expansion (Equation 6.13) of the correction to the ground state, ... (Reference: Equation) ... (Reference: Problems 6.2) For the harmonic oscillator ... the allowed energies are ... where ... in is the classical frequency. Now suppose the spring constant increases slightly: ... (Perhaps we cool the spring, so it becomes less flexible.) (a) Find the exact new energies (trivial, in this case). Expand your formula as a power series in ?, up to second order. (b) Now calculate the first-order perturbation– in the energy, using Equation 6.9. What is H' here? Compare your result with part (a). Hint: It is not necessary—in fact, it is not permitted —to calculate a single integral in doing this problem. (Reference: Equation) ... Get solution

5. Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric field (E), so that the potential energy is shifted by an amount H' = —qEx.(a) Show that there is no first-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.33.(b) The Schrodinger equation can be solved directly in this case, by a change of variables: x′ = x — (qE/mω2). Find the exact energies, and show that they are consistent with the perturbation theory approximation. Get solution

6. Let the two “good” unperturbed states be ... ... ... Get solution

7. Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as in Problem 2.46). (a) Show that the stationary states can be written in the form ... ... Notice that – with the exception of ground state (n =0) – these are all doubly degenerate. (b) Now suppose we introduce the perturbation ... (Reference: Problem 2.46 & Equations 6.9, 6.27) Problem 2.46 Imagine a bead of mass m that slides frictionlessly around a circular wire ring of circumference L. (This is just like a free particle, except that ψ (x + L) = ψ (x).) Find the stationary states (with appropriate normalization) and the corresponding allowed energies. Note that there are two independent solutions for each energy ... —corresponding to clockwise and counter-clockwise circulation; call them ... How do you account for this degeneracy, in view of the theorem in Problem 2.45 (why does the theorem fail, in this case)? ... ... Get solution

8. Problem 6.8 Suppose we perturb the infinite cubical well (Equation 6.30) by putting a delta function "bump" at the point ...: ... Find the first-order corrections to the energy of the ground state and the (triply degenerate) first excited states. ... Get solution

9. Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian, in matrix form, is...where V0 is a constant, and e is some small number (e 1).(a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian= 0).(b) Solve for the exact eigenvalues of H. Expand each of them as a power series in e, up to second order.(c) Use first- and second-order nondegenerate perturbation theory to find the approximate eigenvalue for the state that grows out of the nondegenerate eigenvector of H0. Compare the exact result, from (a). Get solution

10. In the text I asserted that the first-order corrections to an ≪-fold degenerate energy are the eigenvalues of the W matrix, and I justified this claim as the "natural" generalization of the case n = 2. Prove it, by reproducing the steps in Section 6.2.1, starting with...(generalizing Equation 6.17), and ending by showing that the analog to Equation 6.22 can be interpreted as the eigenvalue equation for the matrix W. Get solution

11. (a)Express the Bohr energies in terms of the fine structure constant and the rest energy (mc2) of the electron.(b) Calculate the fine structure constant from first principles (i.e., without recourse to the empirical values of ... Comment: The fine structure con­stant is undoubtedly the most fundamental pure (dimensionless) number in all of physics. It relates the basic constants of electromagnetism (the charge of the electron), relativity (the speed of light), and quantum mechanics (Planck's constant). If you can solve part (b), you have the most certain Nobel Prize in history waiting for you. But I wouldn't recommend spending a lot of time on it right now; many smart people have tried, and all (so far) have failed. Get solution

12. Use the virial theorem (Problem 4.40) to prove Equation 6.55. ... Get solution

13. In Problem 4.43 you calculated the expectation value of ...in the state .... Check your answer for the special cases s = 0 (trivial), s =-1 (Equation 6.55), s = -2 (Equation 6.56), and s = -3 (Equation 6.64). Coalmen% on the case s = -7. ... Get solution

14. Find the (lowest-order) relativistic correction to the energy levels of the one-dimensional harmonic oscillator. Hint: Use the technique in Example 2. Get solution

15. Show that ...is not, for hydrogen states with l = 0. Hint: For such states ... so ... Using integration by parts, show that ... Check that the boundary term vanishes for ..., which goes like ... ... Get solution

16. Evaluate the following commutators: (a) [L • S, L], (b) [L • S S] (c) [L • S, J], (d) [L • S, L2], (e) [L • S, S2], (f) [L • S, J2]. Hint: L and S satisfy the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134), but they commute with each other. Get solution

17. Derive the fine structure formula (Equation 6.66) from the relativis- tic correction (Equation 6.57) and the spin-orbit coupling (Equation 6.65). Hint. Note that j = I ± 1/2; treat the plus sign and the minus sign separately, and you'll find that you get the same final answer either way. Get solution

18. The most prominent feature of the hydrogen spectrum in the visible region is the red Balmer line, coming from the transition n — 3 to n = 2. First of all, determine the wavelength and frequency of this line according to the Bohr theory. Fine structure splits this line into several closely spaced lines; the question is: How many, and what is their spacing ? Hint: First determine how many sublevels the n = 2 level splits into, and find E1fs for each of these, in eV. Then do the same for n = 3. Draw an energy level diagram showing all possible transitions from n = 3 to n = 2. The energy released (in the form of a photon) is (E3 — E2) + A E, the first part being common to all of them, and the ∆ E (due to fine structure) varying from one transition to the next. Find ∆ E (in eV) for each transition. Finally, convert to photon frequency, and determine the spacing between adjacent spectral lines (in Hz)—not the frequency interval between each line and the unperturbed line (which is, of course, unobservable), but the frequency interval between each line and the next one. Your final answer should take the form: "The red Balmer line splits into (???) lines. In order of increasing frequency, they come from the transitions (1) j = (???) to j = (???), (2) j = (???) to j = (???), .... The frequency spacing between line (1) and line (2) is (???) Hz, the spacing between line (2) and line (3) is (???) Hz…….. Get solution

19. The exact fine-structure formula for hydrogen (obtained from the Dirac equation without recourse to perturbation theory) is 16 ... ... ... Get solution

20. Use Equation 6.59 to estimate the internal field in hydrogen, and characterize quantitatively a "strong" and "weak" Zeeman field. ... Get solution

21. Consider the (eight) n = 2 states, |2ljmj>. Find the energy, of each state, under weak-field Zeeman splitting, and construct a diagram like Figure 6.11 to show how the energies evolve as Bext increases. Label each line clearly, and indicate its slope. Get solution

22. Starting with Equation 6.80, and using Equations 6.57, 6.61, 6.64 and 6.81, derive Equation 6.82. ... Get solution

23. Consider the (eight) n — 2 states, |2l ml ms〉. Find the energy of each state, under strong-field Zeeman splitting. Express each answer as the sum of three terms: the Bohr energy, the fine-structure (proportional to α2), and the Zeeman contribution (proportional toμBbext).If you ignore fine structure altogether, how many distinct levels are there, and what are their degeneracies? Get solution

24. If l = 0, then j = s, ... for weak and strong fields. Determine ... (from Equation 6.72) and the fine structure energies (Equation 6.67), and write down the general result for the I = 0 Zeeman effect—regardless of the strength of the field. Show that the strong-field formula (Equation 6.82) reproduces this result, provided that we interpret the indeterminate term in square brackets as 1. ... Get solution

25. Work out the matrix elements of H'z and H'is, and construct the W-matr given in the text, for n = 2. Get solution

26. Analyze the Zeeman effect for the n = 3 states of hydrogen, in the weak, strong, and intermediate field regimes. Construct a table of energies (analo­gous to Table 6.2), plot them as functions of the external field (as in Figure 6.12), and check that the intermediate-field results reduce properly in the two limiting cases. Get solution

27. Let a and b be two constant vectors. Show that...(the integration is over the usual range: O...for states with l = 0. Hint: r = sin θ cosϕ i + sin θ sin ϕ j + cos θk. Get solution

28. By appropriate modification of the hydrogen formula, determine the hyperfine splitting in the ground state of (a) muonic hydrogen (in which a muon—same charge and g-factor as the electron, but 207 times the mass—substi­tutes for the electron), (b) positronium (in which a positron—same mass and g-factor as the electron, but opposite charge—substitutes for the proton), and (c) muonium (in which an anti-muon—same mass and g-factor as a muon, but opposite charge—substitutes for the proton). Hint: Don't forget to use the reduced mass (Problem 5.1) in calculating the "Bohr radius" of these exotic "atoms." Inci­dentally, the answer you get for positronium (4.82 x 10-4 eV) is quite far from the experimental value (8.41 x 10-4 eV); the large discrepancy is due to pair anni­hilation (e+ + e- → y + y), which contributes an extra (3/4)AE, and does not occur (of course) in ordinary hydrogen, muonic hydrogen, or muonium. Get solution

29. Estimate the collection to the ground state energy of hydrogen due to the finite size of the nucleus. Treat the proton as a uniformly charged spherical shell of radius b, so the potential energy of an electron inside the shell is constant: —e2/(4neob)-, this isn't very realistic, but it is the simplest model, and it will give us the right order of magnitude. Expand your result in powers of the small parameter (b/a), where a is the Bohr radius, and keep only the leading term, so your final answer takes the form...Your business is to determine the constant A and the power n. Finally, put in b ~ 10"15 m (roughly the radius of the proton) and work out the actual number. How does it compare with fine structure and hyperfine structure? Get solution

30. Consider the isotropic three-dimensional harmonic oscillator (Problem 4.38). Discuss the effect (in first order) of the perturbation...(for some constant X)(a) on the ground state;(b) (b) the (triply degenerate) first excited state. Hint: Use the answers to Problems 2.12 and 3.33 Get solution

31. Van der Waals interaction. Consider two atoms a distance R apart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact .a weak attraction. To model this system, picture each atom as an electron (mass in, charge -e) attached by a spring (spring constant k) to the nucleus (charge +e), as in Figure 6.14. We'll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is ... ... Get solution

32. Suppose the Hamiltonian If, for a particular quantum system, is a function of some parameter ... be the eigenvalues and ... ... ... Get solution

33. The Feynman-Hellmann theorem (Problem 6.32) can be used to determine the expectation values of ... for hydrogen.23 The effective Hamiltonian for the radial wave functions is (Equation 4.53) ... ... ... ... Get solution

34. Prove Kramers' relation... Get solution

35. (a) Plug s = 0, s = 1, s = 2, and s = 3 into Kramers' relation (Equation 6.104) to obtain formulas for .... Note that you could continue indefinitely, to find any positive power. (b) In the other direction, however, you hit a snag. Put in s = -1, and show that all you get is a relation between .... (c) But if you can get ..., and check your answer against Equation 6.64. ... Get solution

36. When an atom is placed in a uniform external electric field ..., the energy levels are shifted—a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyze the Stark effect for the n = 1 and n = 2 states of hydrogen. Let (the field point in the z direction, so the potential energy of the electron is ... Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.) (a) Show that the ground state energy is not affected by this perturbation, in first order. (b) The first excited state is 4-fold degenerate: .... Using degenerate perturbation theory, determine the first-order corrections to the energy. Into-how many levels does E2 split? (c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment ... Get solution

37. Consider the Stark effect (Problem 6.36) for the n = 3 states of hydrogen. There are initially nine degenerate states, ψ31m(neglecting spin, as before), and we turn on an electric field in the z direction.(a) Construct the 9 × 9 matrix representing the perturbing Hamiltonian. Partial answer: ⟨300|z|310⟩ = −3..., ⟨310|z|320⟩ = −3..., ⟨31±l|z|32±l⟩ = −(9/2)a.(b) Find the eigenvalues, and their degeneracies. Get solution

38. Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the ground state (n = 1) of deuterium. Deuterium is "heavy" hydrogen, with an extra neutron in the nucleus; the proton and neutron bind together to form a deuteron, with spin 1 and magnetic moment...the deuteron g-factor is 1.71. Get solution

39. In a crystal, the electric field of neighboring ions perturbs the energy levels of an atom. As a crude model, imagine that a hydrogen atom is surrounded by three pairs of point charges, as shown in Figure 6.15. (Spin is irrelevant to this problem, so ignore it.)(a) Assuming that r ≪ d1, r ≪ d2, and r ≪ d3, show thatH′ = V0 + 3(β1x2 + β2y2 + β3z2) – (β1 + β2 + β3)r2,where...(b) Find the lowest-order correction to the ground state energy.(c) Calculate the first-order corrections to the energy of the first excited states (n = 2). Into how many levels does this four-fold degenerate system split, (i) in the case of cubic symmetry, β1 = β2 = β3; (iii) in the case of tetragonal symmetry, β1 = β2 ≠ β3, (iii) in the general case of orthorhombic symmetry (all three different)?... Get solution

40. Sometimes it is possible to solve Equation 6.10 directly, without having to expand ...in terms of the unperturbed wave functions (Equation 6.11). Here are two particularly nice examples. (a) Stark effect in the ground state of hydrogen. (i) Find the first-order correction to the ground state of hydrogen in the presence of a uniform external electric field ... (the Stark effect—see Problem 6.36)) Hint: Try a solution of the form ... your problem is to find the constants A, B, and C that solve Equation 6.10. (ii) Use Equation 6.14 to determine the second-order correction to the ground state energy the first-order correction is zero, as you found in Problem 6.36(a))) Answer: ... (b) If the proton had an electric dipole moment p, the potential energy of the electron in hydrogen would be perturbed in the amount H, ep cos 0 4re eor2 (i) Solve Equation 6.10 for the first-order correction to the ground state wave function. ... ... Get solution