CHAPTER 36
Applications of the Schrödinger Equation
1* ∙ True or false: Boundary conditions on the wave function lead to energy quantization. True
2 ∙ Sketch (a) the wave function and (b) the probability distribution for the n = 4 state for the finite square- well potential. (a) The wave function is shown below (b) The probability density is shown below
3 ∙ Sketch (a) the wave function and (b) the probability distribution for the n = 5 state for the finite square- well potential. (a) The wave function is shown below (b) The probability density is shown below Chapter 36 Applications of the Schrödinger Equation
4 ∙∙ Show that the expectation value
∞ 2 The integral ∫ xψ dx = 0 because the integrand is an odd function of x for the ground state as well as any −∞ excited state of the harmonic oscillator.
5* ∙∙ Use the procedure of Example 36-1 to verify that the energy of the first excited state of the harmonic _ ω ω 3 _ 0. (Note: Rather than solve for a again, use the result a = m 0/2h obtained in Example oscillator is E1 = 2 h 36-1.)
2 dψ 2 2 ψ −ax −ax − 2 −ax The wave function is = A1 x e (see Equ. 36-25). Then = A1 e 2 ax A1 e and dx 2ψ d − −ax2 − −ax2 2 3 −ax2 2 3 − −ax2 = 2 axA1 e 4 axA1 e + 4 a x A1 e = (4 a x 6ax) A1 e . We now substitute this into the dx2 _ 2 2 3 2 3 Schrödinger equation. The exponentials and the constant A1 cancel, so (h /2m)(4a x – 6ax) + 1/2mω0 x = E1x. _ 3 _ 2 _ With a = 1/2mω0/h, the terms in x cancel, and solving for the energy E1 we find E1 = 6h a/2m = 3hω0/2 = 3E0.
1/4 6 ∙∙∙ Show that the normalization constant A0 of Equation 36-23 is A0 = (2mω0/h) . ∞ ∞ ∞ 1/4 _ 1/4 ψ 2 _ 1/2 _ 2 π −1/ 2 −s2 = Let A0 = (2mω0/h) = (mω0/πh) . Then ∫ 0 (x) dx = (mω0/πh) ∫ exp[(–mω0/h)x ] dx = ∫ e ds 1 . −∞ −∞ −∞
7 ∙∙∙ Find the normalization constant A1 for the wave function of the first excited state of the harmonic oscillator, Equation 36-25. ∞ 2 2 −2ax2 = _ 2 2 We require that A1 ∫ x e dx 1 where a = mω0/2h. Let y = 2ax ; then the integral becomes −∞ ∞ 1/4 3 3 3 2 2 π −3/ 2 2 − y 32 a 32 m ω 0π (2a) y e dy = . Consequently, 2 = and = . ∫ 3 / 2 A1 π A1 3 −∞ 2(2a) h
8 ∙∙∙ Find the expectation value
2 _ 2 2 _ A0 given in Problem 36-6, 〈x 〉 = h/2mω0. The average potential energy of the oscillator is 1/2mω0 〈x 〉 = hω0/4
= E0/2.
−ax2 9* ∙∙∙ Verify that Ψ1(x) = A1 xe is the wave function corresponding to the first excited state of a harmonic oscillator by substituting it into the time-independent Schrödinger equation and solving for a and E.
From Problem 5 we know that the Schrödinger equation for Ψ1 gives –
_ 2 2 3 2 3 (h /2m)(4a x – 6ax) + 1/2mω0 x = E1x.
3 _ If we now set the coefficients of x = 0 and solve for a we find that a = 1/2mω0/h, and using this expression and _ 2 _ solving for the energy E1 we find E1 = 6h a/2m = 3hω0/2 = 3E0.
10 ∙∙∙ Find the expectation value
11 ∙∙∙ Classically, the average kinetic energy of the harmonic oscillator equals the average potential energy. We may assume that this is also true for the quantum mechanical harmonic oscillator. Use this condition to determine the expectation value of p2 for the ground state of the harmonic oscillator. 2 2 2 2 2 2 According to the problem statement, 〈p 〉/2m = k〈x 〉/2, from which 〈p 〉 = m ω0 〈x 〉. From Problem 36-7, 2 _ 2 _ 〈x 〉 = h/2mω0. Thus, 〈p 〉 = hω0m/2.
12 ∙∙∙ We know that for the classical harmonic oscillator, pav = 0. It can be shown that for the quantum mechanical harmonic oscillator,
= 0. Use the results of Problems 4, 6, and 11 to determine the uncertainty product ∆x ∆p for the ground state of the harmonic oscillator. We associate (∆p)2 and (∆x)2 with the standard deviations as in the statement of Problem 36-41. Then, 2 2 2 2 2 2 2 2 (∆p) = [(p –2ppav – pav )]av and likewise for (∆x) . Since pav = xav = 0, (∆p) = 〈p 〉 and (∆x) = 〈x 〉. Now _ _ from Problems 36-8 and 36-11, (∆p)2(∆x)2 = h2/4 in the ground state of the harmonic oscillator, so ∆p∆x = h/2.
13* ∙∙ A free particle of mass m with wave number k1 is traveling to the right. At x = 0, the potential jumps from zero _ 2 2 to U0 and remains at this value for positive x. (a) If the total energy is E = h k1 /2m = 2U0, what is the wave
number k2 in the region x > 0? Express your answer in terms of k1 and in terms of U0. (b) Calculate the reflection coefficient R at the potential step. (c) What is the transmission coefficient T? (d) If one million particles with wave Chapter 36 Applications of the Schrödinger Equation
number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? _ 2 2 _ 2 2 _ (a) We are given that E = h k1 /2m = 2U0. For x > 0, h k2 /2m + U0 = 2U0. So k2 = 2mU 0 /h, whereas _ k1 = 4mU 0 /h. k2 = k1/ 2 . − 2 ( k1 k 2 ) (b) The reflection coefficient is given by Equ. 36-27. So R = 2 = 0.0294. ( k1 + k 2 ) (c) T = 1 – R = 0.971. 5 6 (d) The number of particles that continue beyond the step is N0T = 9.71×10 ; classically, 1×10 would continue to move past the step.
14 ∙∙ Suppose that the potential jumps from zero to –U0 at x = 0 so that the free particle speeds up instead of
slowing down. The wave number for the incident particle is again k1, and the total energy is 2U0. (a) What is the wave number for the particle in the region of positive x? (b) Calculate the reflection coefficient R at x = 0.
(c) What is the transmission coefficient T? (d) If one million particles with wave number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? _2 2 _2 2 (a) Proceed as in the preceding problem. Now E = h k2 /2m – U0 = 2U0 = h k1 /2m. Consequently,
_2 1/2 k2 = (6mU0/h ) or k 2 = k1 3/2 . (b) Use Equ. 36-27 R = 0.0102 (c) Use Equ. 36-28 T = 0.99 (d) Number of particles transmitted = NT NT = 9.9×105; classically 106 particles are transmitted
15 ∙∙ Work Problem 13 for the case in which the energy of the incident particle is 1.01U0 instead of 2U0.
(a) See Problem 36-13; write k1 and k2 _2 1/2 _2 1/2 k1 = (2.02mU0/h ) ; k2 = (0.02mU0/h ) ; k2 = 0.0995k1 2 ()2.02 − 0.02 (b) Use Equ. 36-27 R = 2 = 0.6710 ()2.02 + 0.02 (c) T = 1 – R T = 0.329 5 6 (d) Number of particles transmitted = NT NT = 3.29×10 ; classically 10 particles are transmitted
16 ∙∙ A particle of energy E approaches a step barrier of height U. What should be the ratio E/U so that the 1 reflection coefficient is 2 ? − k 2 2 1 1. Find r = k2/k1 using Equ. 36-27 r = = = 0.1716 k1 2 + 1 2 2 –1 2. Use the result of Problem 36-38 U/E = 1 – r ; E/U = (1 – r ) = 1.03
17* ∙∙ Use Equation 36-29 to calculate the order of magnitude of the probability that a proton will tunnel out of a nucleus in one collision with the nuclear barrier if it has energy 6 MeV below the top of the potential barrier and the barrier thickness is 10-15 m. Chapter 36 Applications of the Schrödinger Equation
2 − _ _ –13 1. Rewrite α of Equ. 36-29 in units of MeV α = 2/mc (U 0 E) hc; hc = 1.974×10 MeV.m –2αa 2 –15 1/2 –13 2. T = e ; mpc = 938 MeV T = exp[–2×10 (2×938×6) /1.974×10 ] = 0.341
18 ∙∙ A 10-eV electron is incident on a potential barrier of height 25 eV and width of 1 nm. (a) Use Equation 36- 29 to calculate the order of magnitude of the probability that the electron will tunnel through the barrier. (b) Repeat your calculation for a width of 0.1 nm. (a) 1. Evaluate α α = 2×1010 m–1 2. Estimate T using Equ. 36-29 T ≈ 5×10–18 (b) Repeat with a = 10–10 m T ≈ 2×10–2
19 ∙ A particle is confined to a three-dimensional box that has sides L1, L2 = 2L1, and L3 = 3L1. Give the
quantum numbers n1, n2, n3 that correspond to the lowest ten quantum states of this box. 2 2 2 2 2 2 2 2 2 2 Use Equ. 36-34 to write E = (h /8mL1 )(n1 + n2 /4 + n3 /9); E = (h /288mL1 )(36n1 + 9n2 + n3 ). The energies 2 2 in units of h /288mL1 are listed in the following table.
n1 n2 n3 E 111 49 112 61 121 76 113 81 122 88 123 108 114 109 131 121 132 133 124 136
20 ∙ Give the wave functions for the lowest ten quantum states of the particle in Problem 19.
The wave functions are of the form ψ= A sin(n1πx/L1) sin (n2πy/2L1) sin(n3πz/3L1).
21* ∙ (a) Repeat Problem 19 for the case L2 = 2L1 and L3 = 4L1. (b) What quantum numbers correspond to degenerate energy levels? 2 2 2 2 2 2 2 2 2 2 (a) From Equ. 36-32, E = (h /8mL1 )(n1 + n2 /4 + n3 /16) = (h /128mL1 )(16n1 + 4n2 + n3 ). The table below 2 2 lists the ten lowest energy levels in units of h /128mL1 .
n1 n2 n3 E 11121 11224 11329 12133 11436 Chapter 36 Applications of the Schrödinger Equation
12236 12341 11545 12448 13153 11656 13256
(b) There are two degenerate levels, namely, (1,1,4) and (1,2,2) and (1,1,6) and (1,3,2).
22 ∙ Give the wave functions for the lowest ten quantum states of the particle in Problem 21.
The wave functions are of the form ψ = A sin(n1πx/L1) sin (n2πy/2L1) sin(n3πz/4L1).
23 ∙ A particle moves in a potential well given by U(x, y, z) = 0 for –L/2 < x < L/2, 0 < y < L, and 0 < z < L, and U = ∞ outside these ranges. (a) Write an expression for the ground-state wave function for this particle. (b) How do the allowed energies compare with those for a box having U = 0 for 0 < x < L, rather than for –L/2 < x < L/2? (a) The boundary conditions in the y and z directions are as in Figure 36-1. In the x direction, we require that ψ = 0 at –L/2 and at L/2. Using the solution given by Equ. 36-9, we see that the boundary conditions are satisfied
for A ≠ 0, B = 0 if n1 is an even integer, and are satisfied for A = 0, B ≠ 0 if n1 is an odd integer. The normalization constants A and B are both equal to 8/ L3 . Thus,
ψ (x, y, z) = B cos(n1πx/L) sin(n2πy/L) sin(n3πz/L), n1 = 2n +1;
ψ (x, y, z) = A sin(n1πx/L) sin(n2πy/L) sin(n3πz/L), n1 = 2n
The ground-state wave function is ψ111 = A cos(πx/L) sin(πy/L) sin(πz/L). (b) The allowed energies are the same as those for the box with U = 0 for 0 < x < L.
24 ∙∙ A particle moves freely in the two-dimensional region defined by 0 ≤ x ≤ L and 0 ≤ y ≤ L. (a) Find the wave function satisfying Schrödinger’s equation. (b) Find the corresponding energies. (c) Find the lowest two states that are degenerate. Give the quantum numbers for this case. (d) Find the lowest three states that have the same energy. Give the quantum numbers for the three states having the same energy.
(a) In Equ. 36-31, set k3 = 0. Thus, ψ(x, y) = A sin(nπx/L) sin(mπy/L), where n and m are integers. 2 2 2 2 (b) En, m = (h /8mL )(n + m )
(c) E1, 2 = E2, 1. These are the two lowest degenerate states. (d) The lowest triply degenerate states are for n = 1, m = 7; n = 7, m = 1; n = m = 5. The energy of that state is E = 25h2/4mL2.
25* ∙∙ What is the next energy level above those found in Problem 24c for a particle in a two-dimensional square box for which the degeneracy is greater than 2? We need to find the least integral values for n and m such that n2 + m2 are the same for more than two choices of n and m. For any pair of values, e.g., n = 1, m = 2, and n = 2, m = 1 we have double degeneracy. Therefore, we must find two different sets for which the sum of the squares are the same. For n = 1, m = 7 and for n = 5, m = 5, the sum of the squares equals 50. Consequently, the states n = 1, m = 7; n = 7, m = 1; and n = 5, m = 5 are Chapter 36 Applications of the Schrödinger Equation
degenerate (triple degeneracy); the energy of this triply degenerate state is 50h2/8mL2. The next higher degeneracy is for n = 4, m = 7; n = 7, m = 4; n = 1, m = 8; and n = 8, m = 1. These states are four-fold degenerate.
26 ∙ Show that Equation 36-37 satisfies Equation 36-35 with U = 0, and find the energy of this state. _2 2 2 2 2 Substitute Equ. 36-37 into Equ. 36-35. One obtains (h /2m)[(π /L ) + (4π /L )]ψ 1, 2 = Eψ1, 2. That equation is _ satisfied if E = 5h2π2/2mL2 = 5h2/8mL2.
27 ∙ What is the ground-state energy of ten noninteracting bosons in a one-dimensional box of length L? The ten bosons can occupy the same ground state. Consequently, E = 10(h2/8mL2) = 5h2/4mL2.
28 ∙ What is the ground-state energy of ten noninteracting fermions, such as neutrons, in a one-dimensional box of length L? (Because the quantum number associated with spin can have two values, each spatial state can hold two neutrons.) For fermions, such as neutrons for which the spin quantum number is 1/2, two particles can occupy the same
spatial state. Consequently, the lowest total energy for the 10 fermions is E = 2E1(1 + 4 + 9 + 16 + 25) = 55h2/4mL2.
The integral of two functions over some space interval is somewhat analogous to the dot product of two vectors. If this integral is zero, the functions are said to be orthogonal, which is analogous to two vectors being perpendicular. The following problems illustrate the general principle that any two wave functions corresponding to different energy levels in the same potential are orthogonal.
29* ∙∙ Show that the ground-state wave function and that of the first excited state of the harmonic oscillator are
orthogonal; i.e., show that ∫ ψ0(x)ψ 1(x) dx = 0.
∞ ψ ψ = We need to show that ∫ 0 (x) 1 (x)dx 0 , where ψ0(x) and ψ1(x) are given by Equs. 36-23 and 36-25, −∞
respectively. Note that ψ0(x) is an even function of x and ψ1(x) is an odd function of x. It follows that the integral from –∞ to ∞ must vanish.
ψ 2 1 −ax2 30 ∙∙ The wave function for the state n = 2 of the harmonic oscillator is 2(x) = A2(2ax – 2 ) e , where A2 is the normalization constant for this wave function. Show that the wave functions for the states n = 1 and n = 2 of the harmonic oscillator are orthogonal.
We note that ψ2(x) is an even function of x, whereas ψ(x), given by Equ. 36-25, is an odd function of x. Therefore,
∞ ψ ψ = ∫ 1 (x) 2 (x)dx 0 . Q.E.D. −∞
ψ π 31 ∙∙∙ For the wave functions n (x) = 2/L sin(n x/L) corresponding to a particle in an infinite square well Chapter 36 Applications of the Schrödinger Equation
∫ ψ ψ ψ ψ potential from 0 to L, show that n(x) m(x) dx = 0 , that is, n and m are orthogonal. L nπx mπx We need to show that ∫sin sin dx = 0. The product of the two sine functions can be rewritten as L L 0 the sum of two cosines, i.e., sin(nπx/L) sin(mπx/L) = 1/2{cos[(n – m)π x/L] – cos[(n + m)π x/L]}. The integral of L sin[(n − m)π x/L] the first term is ; a similar expression holds for the second term with (n – m) replaced by π (n − m) (n + m). Since n and m are integers and n ≠ m, the sine functions vanish at the two limits x = 0 and x = L.
Therefore, ∫ψn(x)ψ m(x)dx = 0 for n ≠ m.
32 ∙∙ Consider a particle in a one-dimensional box of length L that is centered at the origin. (a) What are the 2 values of ψ1(0) and ψ2(0)? (b) What are the values of
33* ∙∙ Eight identical noninteracting fermions (such as neutrons) are confined to a two-dimensional square box of side length L. Determine the energies of the three lowest states. (See Problem 26.) Each n, m state can accommodate only 2 particles. Therefore, in the ground state of the system of 8 fermions, the four lowest quantum states are occupied. These are (1,1), (1,2), (2,1) and (2,2). [Note that the states (1,2) and (2,1) are distinctly different states since the x and y directions are distinguishable.] The energy of the 2 2 2 2 ground state is E0 = 2(h /8mL )(2 + 5 + 5 + 8) = 5h /mL . The next higher state is achieved by taking one fermion from the (2,2) state and raising it into the next higherunoccupied state. That state is the (1,3) state. The energy difference between the ground state and this state is (h2/8mL2)(10 – 8) = h2/4mL2. The (3,1) is another excited state that is accessible, and it is degenerate with the (1,3) state. The three lowest energy levels are 2 2 2 2 therefore E0 = 5h /mL , and two states of energy E1 = E2 = 21h /4mL .
34 ∙∙ A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x,y) = 0 for –L/2 ≤ x ≤ L/2 and –3L/2 ≤ y ≤ 3L/2; and U(x,y) = ∞ elsewhere. (a) Determine the energies of the lowest three bound states. Are any of these states degenerate? (b) Identify the lowest doubly degenerate bound state by appropriate quantum numbers and determine its energy. (a) The energy levels are the same as for a two-dimensional box of widths L and 3L, i.e., 2 2 2 2 2 2 2 2 2 2 En, m = (h /8mL )(n + m /9) = (h /72mL )(9n + m ). The three lowest states are E1, 1 = 5h /36mL , Chapter 36 Applications of the Schrödinger Equation
2 2 2 2 E1, 2 = 13h /72mL , and E1, 3 = h /4mL . None of these states are degenerate. 2 2 2 2 (b) For two states to be degenerate we must have 9(n1 –n2 ) = (m2 – m1 ). That condition is first satisfied for 2 2 n1 = 2, m1 = 3 and n2 = 1, m2 = 6. The energy of that doubly degenerate state is E2, 3 = E1, 6 = 5h /8mL .
35 ∙∙ A particle moves in a potential given by U(x) = A|x|. Without attempting to solve the Schrödinger equation, sketch the wave function for (a) the ground-state energy of a particle inside this potential and (b) the first excited state for this potential. The wave functions for the ground state and first excited state are sketched below.
36 ∙∙∙ The classical probability distribution function for a particle in a one-dimensional box of length L is P = 1/L. (See Example 17-5) . (a) Show that the classical expectation value of x2 for a particle in a one-dimensional box of length L centered at the origin (Problem 32) is L2/12. (b) Find the quantum expectation value of x2 for the nth state of a particle in the one-dimensional box of Problem 32 and show that it approaches the classical limit L2/12 for n >> 1. 1 L / 2 L2 (a) The classical expectation value is given by ∫ x 2 dx = . L 12 −L / 2 (b) As shown in Problem 36-32, for any value of the quantum number n, 〈x2〉 = (L2/12)[1 + (6/n2π2)]. In the limit n >> 1, 〈x2〉 = L2/12.
37* ∙∙∙ Show that Equations 36-27 and 36-28 imply that the transmission coefficient for particles of energy E
incident on a step barrier U0 < E is given by
4 k1 k 2 4r T = 2 = 2 ( k1 + k 2 ) (1 + r )
where r = k2/k1.
− 2 2 − − 2 ( k1 k 2 ) ( k1 + k 2 ) ( k1 k 2 ) 4 k1 k 2 4r R = 2 and T = 1 – R = 2 = 2 = 2 where r = k2/k1. ( k1 + k 2 ) ( k1 + k 2 ) ( k1 + k 2 ) (1 + r )
38 ∙∙∙ (a) Show that for the case of a particle of energy E incident on a step barrier U0 < E, the wave numbers k1
and k2 are related by Chapter 36 Applications of the Schrödinger Equation
k 2 U 0 = r = 1 − k1 E Use this and the results of Problem 37 to calculate the transmission coefficient T and the reflection coefficient
R for the case (b) E = 1.2U0, (c) E = 2.0U0, and (d) E = 10.0U0. _2 2 _2 2 (a) In the region U = 0, E = h k1 /2m. In the region U = U0, h k2 /2m = (E – U0). It follows that
− E U 0 r = k 2 = = 1 − U 0 k1 E E
(b), (c), (d) Find r and use the result of Problem (b) T = 0.823, R = 0.177; (c) T = 0.9706, R = 0.0294; 36-37 to determine T and R (d) T = 0.9993, R = 0.000693
39 ∙∙∙ Determine the normalization constant A2 in Problem 30. ∞ ∞ 2 2 − 2 −2ax2 = 2 2 − 2 −2ax2 = We require that A2 ∫∫(2ax 1/ 2) e dx 2A2 (2ax 1/ 2) e dx 1 . Expand the integrand to the −∞ 0 − 2 − 2 − 2 three terms 4 a2 x4 e 2ax , − 2 ax2 e 2ax , and e 2ax / 4, and integrate term by term using the definite integrals ∞ ∞ 2 2 π 2 ⋅ ⋅ ⋅⋅⋅ − π −b x = 2n −bx = 1 3 5 (2n 1) ≥ ∫ e dx and ∫ x e dx + , n 1. Evaluating the three integrals one obtains 2b n 1 n b 0 0 2 b 1/4 1 π 2a 8mω 2 0 A2 = 1 and A2 = 2 = . 2 2a π h
40 ∙∙∙ Consider the time-independent one-dimensional Schrödinger equation when the potential function is symmetric about the origin, i.e., when U(x) = U(–x). (a) Show that if ψ(x) is a solution of the Schrödinger equation with energy E, then ψ(–x) is also a solution with the same energy E, and that, therefore, ψ(x) and ψ(– x) can differ by only a multiplicative constant. (b) Write ψ(x) = Cψ (–x), and show that C = ±1. Note that C = +1 means that ψ(x) is an even function of x, and C = –1 means that ψ(x) is an odd function of x. (a) Do a spacial inversion, i.e., let x = –x. The second derivative is an even operator, that is to say, d2ψ (–x)/d(–x)2 = d2ψ (x)/dx2. Therefore, if U(–x) = U(x), the Schrödinger equation for ψ(–x) is exactly the same as for ψ(x) and must give the same values for the energy E. Then if ψ(–x) does differ from ψ(x), the ratio ψ(–x)/ψ (x) cannot be a function of x but must be a constant. So ψ(x) = Cψ (–x). (b) The previous result means that replacing the argument of the wave function by its negative is equivalent to multiplication by C. Thus, if Cψ (–x) is a good wave function and we now replace its argument by its negative, that is, by x, we must multiply by C again. Thus ψ(x) = C 2ψ (x), C 2 = 1, and C = ±1.
41*∙∙∙ In this problem you will derive the ground-state energy of the harmonic oscillator using the precise form of _ the uncertainty principle, ∆x ∆p ≥ h/2, where ∆x and ∆p are defined to be the standard deviations (∆x)2 = [(x – 2 2 2 xav) ]av and (∆p) = [(p – pav) ]av (see Equation 18-31). Proceed as follows: 1 ω2 2 2 1. Write the total classical energy in terms of the position x and momentum p using U(x) = 2 m x and K = p /2m. ∆ 2 − 2 2 − 2 2. Use the result of Equation 18–35 to write ( x ) = [(x xav ) ]av = (x )av xav and ∆ 2 − 2 2 − 2 ( p ) = [( p pav ) ]av = (p )av pav . Chapter 36 Applications of the Schrödinger Equation
2 3. Use the symmetry of the potential energy function to argue that xav and pav must be zero, so that (∆x) = (x)av 2 2 and (∆p) = (p )av. _ ∆ ∆ 2 2 1 ω2 2 4. Assume that p = h/2 x to eliminate (p )av from the average energy Eav = (p )av/2m + 2 m (x )av and write _ 2 1 ω2 2 Eav as Eav = h /8mZ + 2 m Z, where Z = (x )av. 5. Set dE/dZ = 0 to find the value of Z for which E is a minimum. _ 1 ω 6. Show that the minimum energy is given by (Eav)min = + 2 h . 2 2 2 1. Eav = Uav + Kav = 1/2mω (x )av + (1/2m)(p )av . 2 2 2 2 2 2 2 2, 3. (∆p) = [(p – pav) ]av = [p – 2ppav + pav ]av = (p )av since pav = 0; likewise, (∆x) = (x )av . 2 2 _ 2 2 2 _ 2 4. Eav = 1/2mω (x )av + (h /8m)/(x )av = 1/2mω Z + h /8mZ. 2 _ 2 2 _ 5. dEav /dZ = 1/2mω – h /8mZ = 0; Z = h/2mω. 2_ _ 2 6. (Eav)min = 1/2mω h/2mω + 2mωh /8mω = 1/2hω.
42 ∙∙∙ A particle of mass m near the earth’s surface at z = 0 can be described by the potential energy U = mgz, z > 0 U = ∞, z < 0 For some positive value of total energy E, indicate the classically allowed region on a sketch of U(z) versus z. Sketch also the classical kinetic energy versus z. The Schrödinger equation for this problem is quite difficult to solve. Using arguments similar to those in Section 36-2 about the curvature of the wave function as given by the Schrödinger equation, sketch your "guesses" for the shape of the wave function for the ground state and the first two excited states. The classically allowed region is for E ≥U(z). In the figure below, this region extends from z = 0 to
z = zmax. The kinetic energy is E – U(z). In this case, K(z) is a straight line extending from E at z = 0 to 0 at z = zmax. A sketch of the wave functions for the lowest three energy states is shown in the third figure.