1. Prove the following three theorems:(a) For normalizable
solutions, the separation constant E must be real. Hint: Write E (in
Equation 2.7) as E0 + iГ (with E0 andГ real), and show that if Equation
1.20 is to hold for all t, Г must be zero.(b) The time-independent wave
function ψ(x) can always be taken to be real (unlike ψ(x, t), which is
necessarily complex). This doesn't mean that every solution to the
time-independent Schrodinger equation is real; what it says is that if
you’ve got one that is not, it can always be expressed as a linear
combination of solutions (with the same energy) that are. So you might
as well stick to ψ’s that are real. Hint: If ψ(x) satisfies Equation
2.5, for a given E, so too does its complex conjugate, and hence also
the real linear combinations (ψ + ψ*) and i(ψ – ψ*)(c) If V(x) is an
even function (that is, V(–x) = v(–x)) then ψ(x) can always be taken to
be either even or odd. Hint: If ψ(x) satisfies Equation 2.5, for a given
E, so too does ψ(–x), and hence also the even and odd linear
combinations ψ(x) ± ψ (–x) Get solution
2. Show that E must exceed the minimum value of V(x), for every normalizable solution to the time-independent Schrodinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form...if EVmin, then ψ and its second derivative always have the same sign—argue that such a function cannot be normalized. Get solution
3. Show that there is no acceptable solution to the (time-independent) Schrodinger equation for the infinite square well with E = 0 or E Get solution
4. Calculate ‹x›, ‹x2› ‹p›, ‹p2›, σx, and σp, for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit? Get solution
5. A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states:...(a) Normalize ψ(x,0). (That is, find A. This is very easy, if you exploit the orthonormality of ψ1 and ψ2 Recall that, having normalized ψ at t = 0, you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).)(b) Find ψ (x, t)and |ψ(x, t)|2, Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let ω ≡ π2ħ/2ma2.(c) Compute ‹x›. Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than a/2, go directly to jail.)(d) Compute ‹p›. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”) Get solution
6. Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of ... in Problem 2.5: ... where ... is some constant. Find ... and compare your results with what you got before. Study the special cases ... (For a graphical exploration of this problem see the apples in footnote 7.) ... ... ... Get solution
7. A particle in the infinite square well has the initial wave function15...(a) Sketch ψ(x, 0), and determine the constant A.(b) Find ψ(x, t)(c) What is the probability that a measurement of energy would yield the value E1?(d) find the expectation value of the energy. Get solution
8. A particle of mass m in the infinite square well (of width a) starts out in the left half of the well, and is (at t = 0) equally likely to be found at any point in that region.(a) What is its initial wave function, ψ(x, 0)? (Assume it is real. Don't forget to normalize it.)(b) What is the probability that a measurement of the energy would yield the value π2 ħ2/2ma2? Get solution
9. For the wave function in Example 2.2, find the expectation value of H, at time t = 0, the “old fashioned” way:...Compare the result obtained in Example 2.3. Note: Because ‹H› is independent of time, there is no loss of generality in using t = 0. Get solution
10. (a) Construct ψ2 (x).(b) Sketch ψ0, ψ1, and ψ2.(c) Check the orthogonality of ψ0, ψ1, and ψ2, by explicit integration. Hint: If you exploit the even-ness odd and old-ness of the function, there is really only one integral left to do. Get solution
11. (a) Compute ‹x›, ‹p› ‹x2› and ‹p2›, for the states ψ0 (Equation 2.59) and ψ1 (Equation 2.62), by explicit integration. Comment: In this and other problems involving the harmonic oscillator it simplifies matters if you introduce the variable ... and the constant.(b) Check the uncertainty principle for these states.(c) Compute ‹T› (the average kinetic energy) and ‹V› (the average potential energy) for these states. (No new integration allowed!) Is their sum what you would expect? Get solution
12. Find ‹x›, ‹p›, ‹x2›, ‹p2›, and ‹T›, for the nth stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied. Get solution
13. A particle in the harmonic oscillator potential starts out in the state ... (a) Find A. (b) Construct ... (c) Find ... Don’t get too excited if they oscillate at the classical frequency; what would it have been had I specified ... instead of ...check that Ehrenfest’s theorem (Equation 1.38) holds for this wave function. (d) If you measured the energy of this particle, what values might you get, and with what probabilities? ... Get solution
14. A particle is in the ground state of the harmonic oscillator with classical frequency ω, when suddenly the spring constant quadruples, so ω′ without initially changing the wave function (of course, ψ will now evolve differently, because the Hamiltonian has changed). What is the probability that a measurement of the energy would still return the value hω/2? What is the probability of getting hω? [Answer: 0.943.] Get solution
15. In the ground state of the harmonic oscillator, what is the probability (correct to three significant digits) of finding the particle outside the classically allowed region? Hint: Classically, the energy of an oscillator is E = (1/2)ka2 = (1/2)mω2a2, where a is the amplitude. So the “classically allowed region” for an oscillator of energy E extends from .... Look in a math table under “Normal Distribution” or “Error Function" for the numerical value of the integral. Get solution
16. Use the recursion formula (Equation 2.84) to work out ... and ... Invoke the convention that the coefficient of the highest power of ... to fix the overall constant. ... Get solution
17. In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that ... Use it to derive H3 and H4. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: ... Use it, together with your answer in (a), to obtain H5 and H6. ... ... (c) If you differentiate an nth-order polynomial, you get a polynomial of order (n - 1). For the Hermite polynomials, in fact, ... Check this, by differentiating H5 and H6. ... Get solution
18. Show that [Aeik + Be–ikx] and [C cos kx + D sin kx] are equivalent ways of writing the same function of x, and determine the constants C and D in terms of A and B, and vice versa. Comment: In quantum mechanics, when V = 0, the exponentials represent traveling waves, and are most convenient in discussing the free particle, where as sins and cosines correspond to standing waves, which arise naturally in the case of the infinite square well. Get solution
19. Find the probability current, J (Problem 1.14) for the free particle wave function equation 2.94. which direction does the probability current flow? ... Get solution
20. This problem is designed to guide you through a "proof" of Plancherel's theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity. (a) Dirichlet's theorem says that "any" function f(x) on the interval [-a, +a] can be expanded as a Fourier series: ... (d) Take the limit a → ∞ to obtain Plancherel's theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas—one for F(k) in terms of f(x), the other for f(x) in terms of F(k)—have such a similar structure in the limit a → ∞. Get solution
21. A free particle has the initial wave function...Where A and a are positive real constants.(a) Normalize ψ(x, 0)(b) Find φ (k).(c) Construct φ(x, t), in the form of an integral.(d) Discuss the limiting cases (a very large, and a very small). Get solution
22. The Gaussian wave packet. A free particle has the initial wave function ... Where A and a are constants (a is real and positive). (a) Normalize ... (b) Find ... Hint: Integrals of the form ... Can be handled by “ completing the square”: Let ... and note that ... .Answer: ... (c) Find ... .Express your answer in terms of the quantity ... Sketch ... (as a function of x) at t = 0 and again for some very large t. Qualitatively, what happens to ... as time goes on? (d) Find ... Partial answer: ... but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time t does the system come closest to the uncertainty limit? Get solution
23. Evaluate the following integrals:... Get solution
24. Delta functions live under integral signs, and two expressions (D1(x) and D2(x)) involving delta functions are said to be equal if...for every (ordinary) function f(x)(a) Show that...where c is a real constant. (Be sure to check the case where c is negative.)(b) Let θ(x) be the step function:...(In the rare case where it actually matters, we define θ(0) to be 1/2.) Show that dθ/dx = δ(x). Get solution
25. Check the uncertainty principle for the wave function in Equation 2.129. Hint: Calculating ‹p2› is tricky, because the derivative of ψ has a step discontinuity at x = 0. Use the result in Problem 2.42(b). Partial answer: ... ... Get solution
26. What is the Fourier transform of ... Using Plancherel's theorem, show that ... Comment: This formula gives any respectable mathematician apoplexy. Although the integral is clearly infinite when x = 0, it doesn't converge (to zero or any-thing else) when ... since the integrand oscillates forever. There are ways to patch it up (for instance, you can integrate from -L to +L, and interpret Equation 2.144 to mean the average value of the finite integral, as ... The source of the problem is that the delta function doesn't meet the requirement (square-integrability) for Plancherel's theorem (see footnote 33). In spite of this, Equation 2.144 can be extremely useful, if handled with care. Get solution
27. Consider the double detla-function potential...Where α and a are positive constants.(a) Sketch this potential(b) How many bound states does it possess? Find the allowed energies, for α = ħ/ma and for a = ħ2/4ma, and sketch the wave functions. Get solution
28. Find the transmission coefficient for the potential in Problem 1Problem 1Consider the double detla-function potential...Where α and a are positive constants.(a) Sketch this potential(b) How many bound states does it possess? Find the allowed energies, for α = ħ/ma and for a = ħ2/4ma, and sketch the wave functions. Get solution
29. Analyze the odd bound state wave functions for the finite square well. Derive the transcendental equation for the allowed energies, and solve it graphically. Examine the two limiting cases. Is there always an odd bound state? Get solution
30. Normalize ... in Equation 2.151, to determine the constants D and F. ... Get solution
31. The Dirac delta function can be thought of as the limiting case of a rectangle of area 1, as the height goes to infinity and the width goes to zero. Show that the delta-function well (Equation 2.114) is a "weak" potential (even though it is infinitely deep), in the sense that z0 → 0. Determine the bound state energy for the delta-function potential, by treating it as the limit of a finite square well. Check that your answer is consistent with Equation 2.129. Also show that Equation 2.169 reduces to Equation 2.141 in the appropriate limit. ... [2.169] ... ... Get solution
32. Derive Equations 2.167 and 2.168. Hint: Use Equations 2.165 and 2.166 to solve for C and D in terms of F:...Plug these back into Equations 2.16 and 2.164. Obtain the transmission coefficient, and confirm Equation 2.169. Get solution
33. Determine the transmission coefficient for a rectangular barrier (same as Equation 2.145, only with V(x) = +V0 > 0 in the region –a Vq (note that the wave function inside the barrier is different in the three cases). Partial answer: For E0,... Get solution
34. Consider the "step" potential: ... (a) Calculate the reflection coefficient, for the case E 0, and comment on the answer. (b) Calculate the reflection coefficient for the case E > V0. (c) For a potential such as this, which does not go back to zero to the right of the barrier, the transmission coefficient is not simply |F|2 / |A|2 (with A the incident amplitude and F the transmitted amplitude), because the transmitted wave travels at a different speed. Show that ... for E > V0. Hint: You can figure it out using Equation 2.98, or—more elegantly, but less informatively—from the probability current (Problem 2.19). What is T, for E 0? (d) For E > V0, calculate the transmission coefficient for the step potential, and check that T + R = 1. ... Get solution
35. A particle of mass m and kinetic energy E > 0 approaches an abrupt potential drop ... (Figure 2.20). (a) What is the probability that it will "reflect" back, if ... Hint: This is just like Problem 2.34 except that the step now goes down, instead of up. (b) I drew the figure so as to make you think of a car approaching a Off, but obviously the probability of "bouncing back" from the edge of a cliff is far smaller than what you got in (a)—unless you're Bugs Bunny. Explain why this potential does not correctly represent a cliff. Hint: In Figure 2.20 the potential energy of the car drops discontinuously to —..., as it passes x = 0; would this be true for a falling car? (c) When a free neutron enters a nucleus, it experiences a sudden drop in potential energy, from V = 0 outside to around —12 MeV (million electron volts) inside. Suppose a neutron, emitted with kinetic energy 4 MeV by a fission event, strikes such a nucleus. What is the probability it will be absorbed, thereby initiating another fission? Hint: You calculated the probability of reflection in part (a); use T = 1 — R to get the probability of transmission through the surface. (Reference: Figure 2.20) ... Get solution
36. Solve the time-independent Schrodinger equation with appropriate boundary conditions for the "centered" infinite square well: V(x) = 0 (for -a V(x) = ∞ (otherwise). Check that your allowed energies are consistent with mine (Equation 2.27), and confirm that your ψ's can be obtained from mine (Equation 2.28) by the substitution x → (x + a)/2 (and appropriate renormalization). Sketch your first three solutions, and compare Figure 2.2. Note that the width of the well is now 2a. (Reference) ... ... ... Get solution
37. A particle in the infinite square well (Equation 2.19) has the initial wave function ... Determine A, find ...as function of time. What is the expectation value of the energy? Hint: ... can be reduced, by repeated application of the trigonometric sum formulas, to linear combinations of ... and ... ... Get solution
38. A particle of mass in is in the ground state of the infinite square well (Equation 2.19). Suddenly the well expands to twice its original size—the right wall moving from a to 2a —leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured. (a) What is the most probable result? What is the probability of getting that result? (b) What is the next most probable result, and what is its probability? (c) What is the expectation value of the energy? Hint: If you find yourself confronted with an infinite series, try another method. ... Get solution
39. (a) Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time T = 4ma2/πħ. That is: ψ(x, T) = ψ(x, 0) for any state (not just a stationary state).(b) What is the classical revival time, for a particle of energy E bouncing back and forth between the walls?(C) For what energy are the two revival times equal? Get solution
40. A particle of mass m is in the potential...(a) How many bound states are there?(b) In the highest-energy bound state, what is the probability that the particle would be found outside the well (x > a)? Answer: 0.542, so even though it is bound by the well it is more likely to be found outside than inside!. Get solution
41. A particle of mass m in the harmonic oscillator potential (Equation 2.43) starts out in the state ... for some constant A. (a) What is the expectation value of the energy? (b) At some later time T the wave function is 2 ... for some constant B. What is the smallest possible value of T? ... Get solution
42. Find the allowed energies of the half harmonic oscillator...(This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual computation. Get solution
43. In Problem 2.22 you analyzed the stationary Gaussian free particle wave packet. Now solve the same problem for the traveling Gaussian wave packet, starting with the initial wave function...where l is a real constant. Get solution
44. Solve the time-independent Schrodinger equation for a centered infinite square well with a delta-function barrier in the middle:...Treat the even and odd wave functions separately. Don't bother to normalize them. Find the allowed energies (graphically, if necessary). How do they compare with the corresponding energies in the absence of the delta function? Explain why the odd solutions are not affected by the delta function. Comment on the limiting cases α → 0 and α → ∞. Get solution
45. If two (or more) distinct solutions to the (time-independent) Schrodinger equation have the same energy E, these states are said to be degenerate. For example, the free particle states are doubly degenerate—one solution representing motion to the right, and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension there are no degenerate bound states. Hint: Suppose there are two solutions, ψ1 and ψ2, with the same energy E. Multiply the Schrodinger equation for ψ1 by ψ1, and the Schrodinger equation for ψ2 by ψ1, and subtract, to show that (ψ1dψ2 – ψ1ψ2/dx) is a constant. Use the fact that for normalizable solutions i/r 0 at ± ∞ to demonstrate that this constant is in fact zero. Conclude that ψ2 is a multiple of ψ1, and hence that the two solutions are not distinct. Get solution
46. Imagine a bead of mass m that slides frictionless 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 E„—corresponding to clockwise and counter-clockwise circulation; call them if£(x) and 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
47. Attention: This is a strictly qualitative problem—no calculations allowed! Consider the "double square well" potential (Figure 2.21). Suppose the depth V0 and the width a are fixed, and large enough so that several bound states occur. ... FIGURE 2.21: The double square well (Problem 2.47). (a) Sketch the ground state wave function ψ1 and the first excited state ψ2, (i) for the case b = 0, (ii) for b ≈ a, and (iii) for b » a. (b) Qualitatively, how do the corresponding energies (E1 and E2) vary, as b goes from 0 to ∞? Sketch E1(b) and E2(b) on the same graph. (c) The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). If the nuclei are free to move, they will adopt the configuration of minimum energy. In view of your conclusions in (b), does the electron tend to draw the nuclei together, or push them apart? (Of course, there is also the internuclear repulsion to consider, but that's a separate problem.) Get solution
48. In Problem 2.7(d) you got the expectation value of the energy by summing the series in Equation 2.39, but I warned you (in footnote 15) not to try it the "old fashioned way," ... because the discontinuous first derivative of ... renders the second derivative problematic. Actually, you could have done it using integration by parts, but the Dirac delta function affords a much cleaner way to handle such anomalies. (a) Calculate the first derivative of ... (in Problem 2.7), and express the answer in terms of the step function, ... defined in Equation 2.143. (Don't worry about the end points—just the interior region 0 (b) Exploit the result of Problem 2.24(b) to write the second derivative of ... in terms of the delta function. (c) Evaluate the integral ... and check that you get the same answer as before. ... ... ... Get solution
49. (a) Show that...satisfies the time-dependent Schrodinger equation for the harmonic oscillator potential (Equation 2.43). Here a is any real constant with the dimensions of length. Get solution
50. Consider the moving delta-function well:V(x, t) = –αδ(x – vt)where v is the (constant) velocity of the well(a) show that the-dependent Schrödinger equation admits the exact solution...(b) Find the expectation value of the Hamiltonian in this state, and comment on the result. Get solution
51. consider the potential...Where a is a positive constant, and “such” stand for the hyperbolic secant.(a) Graph this potential(b) Check that this potential has the ground state Get solution
52. The scattering matrix. The theory of scattering generalizes in a pretty obvious way to localized potentials...To the right (Region II), of course, I can’t tell you what ψ is until you specify the form. Get solution
53. ... ... ... ... Get solution
54. Find the ground state energy of the harmonic oscillator, to five significant digits, by the "wag-the-dog" method. That is, solve Equation 2.72 numerically, varying K until you get a wave function that goes to zero at large ...In Mathematica, appropriate input code would be ... (Here (a, b) is the horizontal range of the graph, and (c, d) is the vertical range start with a = 0, b = 10, c = —10, d = 10.) We know that the correct solution is K = 1, so you might start with a "guess" of K = 0.9. Notice what the "tail" of the wave function does. Now try K = 1.1, and note that the tail flips over. Somewhere in between those values lies the correct solution. Zero in on it by bracketing K tighter and tighter. As you do so, you may want to adjust a, b, c, and d, to zero in on the cross-over point. ... Get solution
2. Show that E must exceed the minimum value of V(x), for every normalizable solution to the time-independent Schrodinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form...if EVmin, then ψ and its second derivative always have the same sign—argue that such a function cannot be normalized. Get solution
3. Show that there is no acceptable solution to the (time-independent) Schrodinger equation for the infinite square well with E = 0 or E Get solution
4. Calculate ‹x›, ‹x2› ‹p›, ‹p2›, σx, and σp, for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit? Get solution
5. A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states:...(a) Normalize ψ(x,0). (That is, find A. This is very easy, if you exploit the orthonormality of ψ1 and ψ2 Recall that, having normalized ψ at t = 0, you can rest assured that it stays normalized—if you doubt this, check it explicitly after doing part (b).)(b) Find ψ (x, t)and |ψ(x, t)|2, Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let ω ≡ π2ħ/2ma2.(c) Compute ‹x›. Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than a/2, go directly to jail.)(d) Compute ‹p›. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”) Get solution
6. Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of ... in Problem 2.5: ... where ... is some constant. Find ... and compare your results with what you got before. Study the special cases ... (For a graphical exploration of this problem see the apples in footnote 7.) ... ... ... Get solution
7. A particle in the infinite square well has the initial wave function15...(a) Sketch ψ(x, 0), and determine the constant A.(b) Find ψ(x, t)(c) What is the probability that a measurement of energy would yield the value E1?(d) find the expectation value of the energy. Get solution
8. A particle of mass m in the infinite square well (of width a) starts out in the left half of the well, and is (at t = 0) equally likely to be found at any point in that region.(a) What is its initial wave function, ψ(x, 0)? (Assume it is real. Don't forget to normalize it.)(b) What is the probability that a measurement of the energy would yield the value π2 ħ2/2ma2? Get solution
9. For the wave function in Example 2.2, find the expectation value of H, at time t = 0, the “old fashioned” way:...Compare the result obtained in Example 2.3. Note: Because ‹H› is independent of time, there is no loss of generality in using t = 0. Get solution
10. (a) Construct ψ2 (x).(b) Sketch ψ0, ψ1, and ψ2.(c) Check the orthogonality of ψ0, ψ1, and ψ2, by explicit integration. Hint: If you exploit the even-ness odd and old-ness of the function, there is really only one integral left to do. Get solution
11. (a) Compute ‹x›, ‹p› ‹x2› and ‹p2›, for the states ψ0 (Equation 2.59) and ψ1 (Equation 2.62), by explicit integration. Comment: In this and other problems involving the harmonic oscillator it simplifies matters if you introduce the variable ... and the constant.(b) Check the uncertainty principle for these states.(c) Compute ‹T› (the average kinetic energy) and ‹V› (the average potential energy) for these states. (No new integration allowed!) Is their sum what you would expect? Get solution
12. Find ‹x›, ‹p›, ‹x2›, ‹p2›, and ‹T›, for the nth stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied. Get solution
13. A particle in the harmonic oscillator potential starts out in the state ... (a) Find A. (b) Construct ... (c) Find ... Don’t get too excited if they oscillate at the classical frequency; what would it have been had I specified ... instead of ...check that Ehrenfest’s theorem (Equation 1.38) holds for this wave function. (d) If you measured the energy of this particle, what values might you get, and with what probabilities? ... Get solution
14. A particle is in the ground state of the harmonic oscillator with classical frequency ω, when suddenly the spring constant quadruples, so ω′ without initially changing the wave function (of course, ψ will now evolve differently, because the Hamiltonian has changed). What is the probability that a measurement of the energy would still return the value hω/2? What is the probability of getting hω? [Answer: 0.943.] Get solution
15. In the ground state of the harmonic oscillator, what is the probability (correct to three significant digits) of finding the particle outside the classically allowed region? Hint: Classically, the energy of an oscillator is E = (1/2)ka2 = (1/2)mω2a2, where a is the amplitude. So the “classically allowed region” for an oscillator of energy E extends from .... Look in a math table under “Normal Distribution” or “Error Function" for the numerical value of the integral. Get solution
16. Use the recursion formula (Equation 2.84) to work out ... and ... Invoke the convention that the coefficient of the highest power of ... to fix the overall constant. ... Get solution
17. In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that ... Use it to derive H3 and H4. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: ... Use it, together with your answer in (a), to obtain H5 and H6. ... ... (c) If you differentiate an nth-order polynomial, you get a polynomial of order (n - 1). For the Hermite polynomials, in fact, ... Check this, by differentiating H5 and H6. ... Get solution
18. Show that [Aeik + Be–ikx] and [C cos kx + D sin kx] are equivalent ways of writing the same function of x, and determine the constants C and D in terms of A and B, and vice versa. Comment: In quantum mechanics, when V = 0, the exponentials represent traveling waves, and are most convenient in discussing the free particle, where as sins and cosines correspond to standing waves, which arise naturally in the case of the infinite square well. Get solution
19. Find the probability current, J (Problem 1.14) for the free particle wave function equation 2.94. which direction does the probability current flow? ... Get solution
20. This problem is designed to guide you through a "proof" of Plancherel's theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity. (a) Dirichlet's theorem says that "any" function f(x) on the interval [-a, +a] can be expanded as a Fourier series: ... (d) Take the limit a → ∞ to obtain Plancherel's theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas—one for F(k) in terms of f(x), the other for f(x) in terms of F(k)—have such a similar structure in the limit a → ∞. Get solution
21. A free particle has the initial wave function...Where A and a are positive real constants.(a) Normalize ψ(x, 0)(b) Find φ (k).(c) Construct φ(x, t), in the form of an integral.(d) Discuss the limiting cases (a very large, and a very small). Get solution
22. The Gaussian wave packet. A free particle has the initial wave function ... Where A and a are constants (a is real and positive). (a) Normalize ... (b) Find ... Hint: Integrals of the form ... Can be handled by “ completing the square”: Let ... and note that ... .Answer: ... (c) Find ... .Express your answer in terms of the quantity ... Sketch ... (as a function of x) at t = 0 and again for some very large t. Qualitatively, what happens to ... as time goes on? (d) Find ... Partial answer: ... but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time t does the system come closest to the uncertainty limit? Get solution
23. Evaluate the following integrals:... Get solution
24. Delta functions live under integral signs, and two expressions (D1(x) and D2(x)) involving delta functions are said to be equal if...for every (ordinary) function f(x)(a) Show that...where c is a real constant. (Be sure to check the case where c is negative.)(b) Let θ(x) be the step function:...(In the rare case where it actually matters, we define θ(0) to be 1/2.) Show that dθ/dx = δ(x). Get solution
25. Check the uncertainty principle for the wave function in Equation 2.129. Hint: Calculating ‹p2› is tricky, because the derivative of ψ has a step discontinuity at x = 0. Use the result in Problem 2.42(b). Partial answer: ... ... Get solution
26. What is the Fourier transform of ... Using Plancherel's theorem, show that ... Comment: This formula gives any respectable mathematician apoplexy. Although the integral is clearly infinite when x = 0, it doesn't converge (to zero or any-thing else) when ... since the integrand oscillates forever. There are ways to patch it up (for instance, you can integrate from -L to +L, and interpret Equation 2.144 to mean the average value of the finite integral, as ... The source of the problem is that the delta function doesn't meet the requirement (square-integrability) for Plancherel's theorem (see footnote 33). In spite of this, Equation 2.144 can be extremely useful, if handled with care. Get solution
27. Consider the double detla-function potential...Where α and a are positive constants.(a) Sketch this potential(b) How many bound states does it possess? Find the allowed energies, for α = ħ/ma and for a = ħ2/4ma, and sketch the wave functions. Get solution
28. Find the transmission coefficient for the potential in Problem 1Problem 1Consider the double detla-function potential...Where α and a are positive constants.(a) Sketch this potential(b) How many bound states does it possess? Find the allowed energies, for α = ħ/ma and for a = ħ2/4ma, and sketch the wave functions. Get solution
29. Analyze the odd bound state wave functions for the finite square well. Derive the transcendental equation for the allowed energies, and solve it graphically. Examine the two limiting cases. Is there always an odd bound state? Get solution
30. Normalize ... in Equation 2.151, to determine the constants D and F. ... Get solution
31. The Dirac delta function can be thought of as the limiting case of a rectangle of area 1, as the height goes to infinity and the width goes to zero. Show that the delta-function well (Equation 2.114) is a "weak" potential (even though it is infinitely deep), in the sense that z0 → 0. Determine the bound state energy for the delta-function potential, by treating it as the limit of a finite square well. Check that your answer is consistent with Equation 2.129. Also show that Equation 2.169 reduces to Equation 2.141 in the appropriate limit. ... [2.169] ... ... Get solution
32. Derive Equations 2.167 and 2.168. Hint: Use Equations 2.165 and 2.166 to solve for C and D in terms of F:...Plug these back into Equations 2.16 and 2.164. Obtain the transmission coefficient, and confirm Equation 2.169. Get solution
33. Determine the transmission coefficient for a rectangular barrier (same as Equation 2.145, only with V(x) = +V0 > 0 in the region –a Vq (note that the wave function inside the barrier is different in the three cases). Partial answer: For E0,... Get solution
34. Consider the "step" potential: ... (a) Calculate the reflection coefficient, for the case E 0, and comment on the answer. (b) Calculate the reflection coefficient for the case E > V0. (c) For a potential such as this, which does not go back to zero to the right of the barrier, the transmission coefficient is not simply |F|2 / |A|2 (with A the incident amplitude and F the transmitted amplitude), because the transmitted wave travels at a different speed. Show that ... for E > V0. Hint: You can figure it out using Equation 2.98, or—more elegantly, but less informatively—from the probability current (Problem 2.19). What is T, for E 0? (d) For E > V0, calculate the transmission coefficient for the step potential, and check that T + R = 1. ... Get solution
35. A particle of mass m and kinetic energy E > 0 approaches an abrupt potential drop ... (Figure 2.20). (a) What is the probability that it will "reflect" back, if ... Hint: This is just like Problem 2.34 except that the step now goes down, instead of up. (b) I drew the figure so as to make you think of a car approaching a Off, but obviously the probability of "bouncing back" from the edge of a cliff is far smaller than what you got in (a)—unless you're Bugs Bunny. Explain why this potential does not correctly represent a cliff. Hint: In Figure 2.20 the potential energy of the car drops discontinuously to —..., as it passes x = 0; would this be true for a falling car? (c) When a free neutron enters a nucleus, it experiences a sudden drop in potential energy, from V = 0 outside to around —12 MeV (million electron volts) inside. Suppose a neutron, emitted with kinetic energy 4 MeV by a fission event, strikes such a nucleus. What is the probability it will be absorbed, thereby initiating another fission? Hint: You calculated the probability of reflection in part (a); use T = 1 — R to get the probability of transmission through the surface. (Reference: Figure 2.20) ... Get solution
36. Solve the time-independent Schrodinger equation with appropriate boundary conditions for the "centered" infinite square well: V(x) = 0 (for -a V(x) = ∞ (otherwise). Check that your allowed energies are consistent with mine (Equation 2.27), and confirm that your ψ's can be obtained from mine (Equation 2.28) by the substitution x → (x + a)/2 (and appropriate renormalization). Sketch your first three solutions, and compare Figure 2.2. Note that the width of the well is now 2a. (Reference) ... ... ... Get solution
37. A particle in the infinite square well (Equation 2.19) has the initial wave function ... Determine A, find ...as function of time. What is the expectation value of the energy? Hint: ... can be reduced, by repeated application of the trigonometric sum formulas, to linear combinations of ... and ... ... Get solution
38. A particle of mass in is in the ground state of the infinite square well (Equation 2.19). Suddenly the well expands to twice its original size—the right wall moving from a to 2a —leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured. (a) What is the most probable result? What is the probability of getting that result? (b) What is the next most probable result, and what is its probability? (c) What is the expectation value of the energy? Hint: If you find yourself confronted with an infinite series, try another method. ... Get solution
39. (a) Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time T = 4ma2/πħ. That is: ψ(x, T) = ψ(x, 0) for any state (not just a stationary state).(b) What is the classical revival time, for a particle of energy E bouncing back and forth between the walls?(C) For what energy are the two revival times equal? Get solution
40. A particle of mass m is in the potential...(a) How many bound states are there?(b) In the highest-energy bound state, what is the probability that the particle would be found outside the well (x > a)? Answer: 0.542, so even though it is bound by the well it is more likely to be found outside than inside!. Get solution
41. A particle of mass m in the harmonic oscillator potential (Equation 2.43) starts out in the state ... for some constant A. (a) What is the expectation value of the energy? (b) At some later time T the wave function is 2 ... for some constant B. What is the smallest possible value of T? ... Get solution
42. Find the allowed energies of the half harmonic oscillator...(This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual computation. Get solution
43. In Problem 2.22 you analyzed the stationary Gaussian free particle wave packet. Now solve the same problem for the traveling Gaussian wave packet, starting with the initial wave function...where l is a real constant. Get solution
44. Solve the time-independent Schrodinger equation for a centered infinite square well with a delta-function barrier in the middle:...Treat the even and odd wave functions separately. Don't bother to normalize them. Find the allowed energies (graphically, if necessary). How do they compare with the corresponding energies in the absence of the delta function? Explain why the odd solutions are not affected by the delta function. Comment on the limiting cases α → 0 and α → ∞. Get solution
45. If two (or more) distinct solutions to the (time-independent) Schrodinger equation have the same energy E, these states are said to be degenerate. For example, the free particle states are doubly degenerate—one solution representing motion to the right, and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension there are no degenerate bound states. Hint: Suppose there are two solutions, ψ1 and ψ2, with the same energy E. Multiply the Schrodinger equation for ψ1 by ψ1, and the Schrodinger equation for ψ2 by ψ1, and subtract, to show that (ψ1dψ2 – ψ1ψ2/dx) is a constant. Use the fact that for normalizable solutions i/r 0 at ± ∞ to demonstrate that this constant is in fact zero. Conclude that ψ2 is a multiple of ψ1, and hence that the two solutions are not distinct. Get solution
46. Imagine a bead of mass m that slides frictionless 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 E„—corresponding to clockwise and counter-clockwise circulation; call them if£(x) and 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
47. Attention: This is a strictly qualitative problem—no calculations allowed! Consider the "double square well" potential (Figure 2.21). Suppose the depth V0 and the width a are fixed, and large enough so that several bound states occur. ... FIGURE 2.21: The double square well (Problem 2.47). (a) Sketch the ground state wave function ψ1 and the first excited state ψ2, (i) for the case b = 0, (ii) for b ≈ a, and (iii) for b » a. (b) Qualitatively, how do the corresponding energies (E1 and E2) vary, as b goes from 0 to ∞? Sketch E1(b) and E2(b) on the same graph. (c) The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). If the nuclei are free to move, they will adopt the configuration of minimum energy. In view of your conclusions in (b), does the electron tend to draw the nuclei together, or push them apart? (Of course, there is also the internuclear repulsion to consider, but that's a separate problem.) Get solution
48. In Problem 2.7(d) you got the expectation value of the energy by summing the series in Equation 2.39, but I warned you (in footnote 15) not to try it the "old fashioned way," ... because the discontinuous first derivative of ... renders the second derivative problematic. Actually, you could have done it using integration by parts, but the Dirac delta function affords a much cleaner way to handle such anomalies. (a) Calculate the first derivative of ... (in Problem 2.7), and express the answer in terms of the step function, ... defined in Equation 2.143. (Don't worry about the end points—just the interior region 0 (b) Exploit the result of Problem 2.24(b) to write the second derivative of ... in terms of the delta function. (c) Evaluate the integral ... and check that you get the same answer as before. ... ... ... Get solution
49. (a) Show that...satisfies the time-dependent Schrodinger equation for the harmonic oscillator potential (Equation 2.43). Here a is any real constant with the dimensions of length. Get solution
50. Consider the moving delta-function well:V(x, t) = –αδ(x – vt)where v is the (constant) velocity of the well(a) show that the-dependent Schrödinger equation admits the exact solution...(b) Find the expectation value of the Hamiltonian in this state, and comment on the result. Get solution
51. consider the potential...Where a is a positive constant, and “such” stand for the hyperbolic secant.(a) Graph this potential(b) Check that this potential has the ground state Get solution
52. The scattering matrix. The theory of scattering generalizes in a pretty obvious way to localized potentials...To the right (Region II), of course, I can’t tell you what ψ is until you specify the form. Get solution
53. ... ... ... ... Get solution
54. Find the ground state energy of the harmonic oscillator, to five significant digits, by the "wag-the-dog" method. That is, solve Equation 2.72 numerically, varying K until you get a wave function that goes to zero at large ...In Mathematica, appropriate input code would be ... (Here (a, b) is the horizontal range of the graph, and (c, d) is the vertical range start with a = 0, b = 10, c = —10, d = 10.) We know that the correct solution is K = 1, so you might start with a "guess" of K = 0.9. Notice what the "tail" of the wave function does. Now try K = 1.1, and note that the tail flips over. Somewhere in between those values lies the correct solution. Zero in on it by bracketing K tighter and tighter. As you do so, you may want to adjust a, b, c, and d, to zero in on the cross-over point. ... Get solution
