CHAPTER 11 The Need for Quantum Mechanics SECTION 11.1 11.1 Equation (11.3) states λ = (91.176 nm) n m n m, n > m. The energy of a photon of this wavelength is E = hν = hc/λ. Substituting the expression for λ into this equation gives us our desired expression: E = hc λ = hc 91.176 nm n m n m = (6.661 10 34 J Hz 1 ) (.9979 10 8 m s 1 ) (91.176 nm) (10 9 m nm 1 ) =.1787 10 18 1 J m 1 n. n m n m 11. The wavelength expression for He + is λ = (.794 nm) n m, m = 1,, 3,, n > m, n m and for n = 4 and m =, it predicts λ = 11.568 nm. If we substitute n = and m = 1 into Eq. (11.3) for H atom emission wavelengths, we predict the same wavelength. To find a general relation between the integers that characterize H emission and those that give the same wavelength emitted by He +, we note that the constant in the He + expression,.794 nm, is one-fourth as large as the 35
constant in the H atom expression, Eq. (11.3), 91.176 nm. Thus we want integers n He, m He, n H, and m H such that 4 n He mhe n He mhe = n H mh n H mh. 11.3 The laser photon energy is E = hc/λ = 1.867 10 19 J photon 1 for λ = 1064 nm = 1.064 10 6 m. Thus, a 0.5 J pulse contains 0.5 J 1.867 10 19 J photon 1 =.678 1018 photons. The average laser power, at a 10 Hz (i.e., 10 pulses s 1 ) repetition rate, is (10 pulses s 1 ) (1.867 10 19 J photon 1 ) (.678 10 18 photon pulse 1 ) = 5 J s 1 = 5 W. To emit a mole of photons, we need (6.0 10 3 photons)/(.679 10 18 photons per pulse) =.49 10 5 pulses. At 10 pulses per second, the laser must operate for.49 10 5 pulses/10 pulses s 1 =.49 10 4 s = 6.5 hr. 11.4 The photon energy is E = 604.6 kev = 6.046 10 5 ev = 9.687 10 14 J, and since E = hν = hc/λ, the photon frequency is ν = E h = and the wavelength is 9.687 10 14 J 6.66 10 34 J Hz 1 = 1.46 100 Hz, λ = c ν =.998 108 m s 1 1.46 10 0 Hz =.050 10 1 m =.050 pm, which is at the boundary between the X-ray region and the γ ray region, just off the high-frequency, low-wavelength edge of the figure on page 350 in the text. 36
11.5 If the excited energy levels of Ca + have energies E 1 = 5.004 17 10 19 J and E = 5.048 44 10 19 J above the ground state, emission from these levels to the ground state leads to photons of wavelengths λ 1 and λ where λ 1 = hc E 1 = 3.969 58 10 7 m = 396.958 nm λ = hc E = 3.934 77 10 7 m = 393.477 nm. These wavelengths are at the extreme blue end of the visible spectrum. (Not everyone can see colors this blue. ) Emission between these two levels, however, leads to a very different wavelength: λ = hc hc = E E 1 4.47 10 1 J = 4.487 10 5 m = 44.87 µm. This is in the infrared (IR) region of the spectrum. 11.6 We can follow the logic of Example 11. here. The C energy level scheme must look qualitatively as shown below, with the middle energy state closer to the ground state than to the highest state: Larger E, Smaller λ{ Smaller E, Larger λ Highest state Middle state Ground state Since the energy, E, associated with a transition and the emission wavelength for the transition are reciprocally related: E = hc/λ, we can deduce that the larger emission wavelength, 0.6097 nm, is associated with the transition from the middle state to the ground state while the smaller wavelength, 0.3690 nm, corresponds to a transition from the highest to the middle state. If we assign the ground state an energy E 0 = 0, the energy of the middle state, E 1, is given by E 1 0 = E 1 E 0 = E 1 = hc λ = hc 0.6097 nm = 3.58 10 J. The energy difference between the highest and middle states, E = E E 1, is 37
E 1 = E E 1 = hc λ = hc 0.3690 nm = 5.383 10 J. Thus, the energy of the highest state is E = E 1 + E 1 = 5.383 10 J + 3.58 10 J = 8.641 10 J above the ground state. 11.7 The photoelectrons have an energy given by E = h(ν ν 0 ) = hν hν 0 where ν 0 is the threshold light frequency and hν 0 = φ is the work function. Thus, a plot of the electron energy versus the light frequency should be a straight line of slope h and intercept φ. We convert the wavelengths given in the problem to corresponding frequencies (ν = c/λ), the electron energies from ev units to joule units, and construct the graph below: 1.0 E/10 19 J 0 1.0.0 3.0 0 1.0.0 3.0 4.0 5.0 6.0 ν/10 14 s 1 The straight line shown above is the result of a least-squares fit that gives a slope 6.63 10 34 J s = h and an intercept.90 10 19 J = φ. In electron volt units, φ = 1.81 ev. 11.8 The de Broglie wavelength is given by Eq. (11.6), λ = h/mv. The molar mass of C 60 is 70 g mol 1, or 1.0 10 4 kg molecule 1. This is m, and if v = 130 m s 1, we find λ = 4.5 10 1 m = 0.045 Å, much smaller than the size of the molecule itself. For He at its boiling point with v = 160 m s 1, the de Broglie wavelength is considerably larger due to the smaller mass of He: m = 4 g mol 1 = 6.65 10 7 kg atom 1. We find this He atom s de Broglie wavelength is λ = 6.3 10 10 m = 6.3 Å, which is larger than the typical size of the atom. Thus, we could locate and treat C 60 molecules much as if they were classical particles of a definite size, but He atoms of this de Broglie wavelength must be considered as quantum mechanical waves. 38
11.9 We want to make the de Broglie wavelength λ = 0.3 µm = 3 10 7 m, and for an atomic beam of He with a speed v = 5RT/M, we can write λ = 3 10 7 m = h mv = h m 5RT/M. With m = 6.646 10 7 kg, the mass of one He atom, and M = 4.003 10 3 kg mol 1, the molar mass of He, we can solve for T, finding T = 1.06 10 5 K, which tells us that the experiment with He is not possible using ordinary atomic beam methods (or any other method currently known). For electrons, it is helpful to derive an expression between the electron kinetic energy in ev and its velocity. Since the kinetic energy is m e v / with m e, the electron mass, equal to 9.11 10 31 kg, and since 1 J = 6.4 10 18 ev, we can write E/eV = E/J 6.4 10 18 = (m e /kg) (v/m s 1 ) 6.4 10 18 or, solving for v, we have v = (5.93 10 5 m s 1 ) E/eV. Thus, the de Broglie wavelength expression is λ = h mv = h m e (5.93 10 5 m s 1 ) E/eV. Solving for E with λ = 0.3 µm gives E = 1.67 10 5 ev. This energy is below the limit of current technology for electron beams. Electron beams with energies below roughly 1 ev are very difficult to produce with useful intensity due to the inherent repulsion between electrons. At very small energies, the electrons are moving so slowly that they have plenty of time to repel each other and render any beam of them too diffuse to be useful. 11.10 We want to find a neutron temperature T such that λ = 1 Å = 1 10 10 m where T and λ are related through λ = h mv = h. m n 3k B T/m n With m n = 1.675 10 7 kg and k B = R/N A, we find T = 634 K. Colder neutrons would move slower and have larger de Broglie wavelengths. 39
11.11 From the geometry of the atomic packing, we can find d from some simple trigonometry: 60 d = s sin 60 =.15 Å s =.48 Å We can also find the de Broglie wavelength of an electron of kinetic energy E/eV from the expression derived above in Problem 11.9: h λ/m = m e (5.93 10 5 m s 1 ) E/eV = 1.7 10 9 E/eV or λ/å = 150.4/ E/eV. Using this expression, we find that electrons with E = 54 ev have λ = 1.67 Å while those with E = 65 ev have λ = 1.5 Å. These values correspond to diffraction spacings d given by d = λ sin θ = 1.67 Å sin 50 =.18 Å 1.5 Å sin 44 =.19 Å which are in good agreement with each other and with the.15 Å value we deduced from the atom-to-atom spacing and packing geometry. SECTION 11. 11.1 For each possible wavefunction, we are seeking a function that is finite at x = ±, has no discontinuities in its values (but perhaps in its first derivative), and, when squared and integrated over all space, leads to a finite value (for normalization requirements). 40
(a) The function ψ(x) = Ne a x satisfies all these requirements as long as a is a real number > 0. This function has a discontinuous first derivative at x = 0, but this is acceptable. We normalize it as follows, taking advantage of the symmetry of ψ about x = 0 (i.e., the integral from to 0 equals that from 0 to ): 1 = Ψ*Ψ dx = N e ax dx 0 = N 1 a or N = a 1/. (b) The function ψ(x) = N x is not acceptable, since this function diverges to infinity at x = ±. (c) The function ψ(x) = N cos(ax) e x / is acceptable as long as a is a real number constant. We normalize it as follows, again taking advantage of the symmetry of ψ about x = 0: 1 = Ψ*Ψ dx = N cos (ax)e x dx 0 At this point, we are faced with what is very likely an integral you have never seen before. It is a known integral, however, and it can be found in a good integral table:. cos (ax)e x dx = 0 π 4 1 + e a. If you did not find the value of this integral, do not worry. The important step is recognizing how to write the normalization integral in the first place. Given the value for this integral, however, we can go on to find N: N = π 1 + e a. (d) The function ψ(x) = N e +a x is only acceptable if a is real and < 0, in which case this function is identical to that in part (a). 41
11.13 We have the wavefunction ψ(x) = 1/ sin(πx), valid over the region 0 x 1. We note that the wavefunction is normalized so that the probability that the system is between points x = a and x = b is given by the integral b Ψ* Ψ dx a b = sin (πx) dx a Before we tackle this integral for the three cases mentioned in the problem, it is very helpful to look at a graph of Ψ to explore its symmetry:.0 1.5. Ψ 1.0 0.5 0 0 1/4 1/ 3/4 x 1 Since it is symmetric about x = 1/, we see that the probability of being anywhere between x = 0 and 1/ equals that for the interval 1/ x 1. Moreover, since these intervals span the whole region of x and since the probability of being somewhere between 0 and 1 is 1, the probability of being in the interval 0 x 1/ must be 0.5. For part (b), we see that the probability of being in the interval 0 x 1/4 or 3/4 x 1 must equal twice the probability of being between 0 and 1/4 alone. Here we must evaluate the probability integral above with a = 0 and b = 1/4. We find 1/4 Ψ* Ψ dx 0 1/4 = sin (πx) dx 0 = 1 8 1 4π = 9.08 10. Thus, the probability of being in the interval 0 x 1/4 or 3/4 x 1 is 9.08 10 = 0.18. Part (c) asks for the probability of being anywhere in the interval 1/4 x 3/4. We note that this probability plus that from part (b) must equal 1, since these intervals together cover the entire allowed region of 4
the system. Thus, the probability is 1 0.18 = 0.818. Finally, we note from the plot of ψ above that the system is most likely to be found at x = 1/, the point at which ψ is a maximum. 11.14 A linear operator O must satisfy these two criteria: O(ψ 1 + ψ + ) = Oψ 1 + Oψ + and O(cψ) = coψ for allowable wavefunctions ψ 1, etc., and any constant c. The operator take the square root fails both tests: ψ 1 + ψ + ψ 1 + ψ + and cψ c ψ, as does take the reciprocal : 1 ψ 1 + ψ + 1 ψ 1 + 1 ψ + and 1 cψ c 1 ψ, and square the function : ψ 1 + ψ + ψ 1 + ψ + and cψ cψ. In contrast, the operator integrate over all space is linear, because ψ 1 + ψ + dτ = ψ 1 dτ + ψ dτ + and cψ dτ = c ψ dτ. Likewise, multiply by i and take the first derivative is a linear operator, because d dx i ψ 1 + ψ + = d dx i ψ 1 + d dx i ψ + and d dx i cψ = c d dx i ψ. The operator add a constant to the function is not linear, as we can see if we let k represent the constant: k + ψ 1 + ψ + ψ 1 + k + ψ + k + and k + cψ c k + ψ. 11.15 Recall that the definition of a Hermitian operator A is ψ 1 * A ψ dτ = ψ Aψ 1 * dτ = ψ A * ψ 1 * dτ. 43
(a) If A and B are Hermitian, then C = AB is also Hermitian only if A and B commute, as we can show below: ψ 1 * A B ψ dτ = ψ A * B * ψ 1 * dτ (if C is Hermitian) = ψ A * φ * dτ (defining φ = Bψ 1 ) = φ * A ψ dτ (since A is Hermitian) = B * ψ * 1 A ψ dτ (by definition of φ) = B * ψ * 1 χ dτ (defining χ = Aψ ) = χb * ψ* 1 dτ (rearranging the integral) = ψ * 1 B χ dτ (since B is Hermitian) = ψ * 1 B A ψ dτ (by definition of χ). Thus, only if A B = BA (i.e., the operators commute) is the product of two Hermitian operators also Hermitian. (b) The sum of two Hermitian operators is Hermitian, because ψ 1 * A + B ψ dτ = ψ 1 * A ψ dτ + ψ 1 * B ψ dτ, and ψ A * + B * ψ 1 * dτ = ψ A * ψ 1 * dτ + ψ B * ψ 1 * dτ, but, since A and B are Hermitian, ψ A * ψ 1 * dτ = ψ1 * A ψ dτ and ψ B * ψ 1 * dτ = ψ1 * B ψ dτ. 44
Thus ψ 1 * A + B ψ dτ = ψ A * + B * ψ 1 * dτ. (c) If we repeat the line of reasoning from part (b) for the operator E = A + ib, we find that E is not Hermitian, because ψ 1 * A + ib ψ dτ = ψ 1 * A ψ dτ + i ψ 1 * B ψ dτ, but ψ A + ib ψ 1 * dτ = ψ A * ib * ψ 1 * dτ= ψ A * ψ 1 * dτ i ψ B * ψ 1 * dτ. We again have, since A and B are Hermitian, ψ A * ψ 1 * dτ = ψ1 * A ψ dτ and ψ B * ψ 1 * dτ = ψ1 * B ψ dτ, but here, these equalities prove that E is not Hermitian. 11.16 Substituting the definition R(r) = e αr into the differential equation gives de αr dr + r d e αr dr = r e αr. a a 0 0 Expanding the derivatives on the left gives de αr dr + r d e αr dr = αe αr + α r e αr. This shows that R(r) = e αr satisfies the differential equation if αe αr + α r e αr = r e αr or α = 1. a a 0 a 0 0 45
11.17 First, from the form of the Hamiltonian, we can tell that the potential energy of this system must be everywhere zero (or, stated more precisely, the potential energy is everywhere constant with the constant value chosen as zero), because the Hamiltonian has only one term and that term represents the kinetic energy of a particle of mass m moving along the x direction. The total energy of this state must be the eigenvalue found when H operates on ψ(x): Hψ(x) = Eψ(x) = m d dx L 1/ sin 4πx L = m (4π) L L 1/ sin 4πx L or, recognizing that the final expression is Eψ(x), we identify the energy as E = m (4π) L = 8 π ml. 11.18 Here, we use Eq. (11.13): i φ(t) t = Eφ(t) = i t eiω 0 t/4 = ω 0 4 e iω 0 t/4 = ω 0 4 φ(t), or E = ω 0 /4. Note that we could also use the general Eq. (11.14), φ(t) = C e iet/, to identify E in our particular φ(t). 11.19 First, we solve the quantum-mechanical part of this problem. We find the angular momentum (the eigenvalue J of the angular momentum operator J) for our particular wavefunction: Jψ(θ) = JΨ(θ) = i d dθ e i15θ π e i15θ = 15 π or J = 15. Classically, J = mvr, and solving for v with R = 10 m and m = 10 3 kg gives v = 15 mr = 1.58 10 8 m s 1. The particle is moving on a circle of circumference πr = 6.8 10 m. Thus, the time the particle needs to make one revolution is 46
t = πr v = 6.8 10 m 1.58 10 8 m s 1 = 3.97 106 s. This is clearly a long time! A handy number to keep in your head is the number of seconds in a year, which is exactly 31 536 000, but quite closely 10 π million, which is easier to remember and good enough for most estimates. Thus, 10 9 yr is about π 10 16 s. In a billion years, (1 Gyr), this particle has only just begun its first revolution about the circle: # of revolutions billion years π 10 16 s Gyr 1 3.97 10 6 s rev 1 = 7.9 10 11 rev Gyr 1. The moral here is that macroscopic objects moving with microscopic angular momenta are hardly moving at all. Put another way, a macroscopic angular momentum must correspond to a truly huge microscopic angular momentum quantum number. For example, if this particle revolves once per second, then it has a velocity v = 6.8 10 m s 1 and an angular momentum J = mvr = 6.8 10 7 kg m s 1. In units of, this is the very large number J = 6.8 10 7 kg m s 1 1.05 10 34 J s 6 10 7. The wavefunction is thus ψ(θ) = e ijθ/ π = e i(6 107 )θ π An angular momentum on the order of 15 is not unusual for a microscopic mass, and an angular momentum on the order of 10 7 is not unusual for a macroscopic mass. 11.0 The eigenvalue expressions here are Since A is Hermitian, we can write A ψ i = a i ψ i and Aψ j = a j ψ j.. 47
ψ i * A ψj dτ = ψ j A ψ i * dτ. Next, we invoke the eigenvalue expressions and take the eigenvalues (which are just constants) out of each integral: ψ i * A ψj dτ = a j ψ i * ψj dτ = ψ j A ψ i * dτ = ai * ψ j ψ i * dτ. Since the eigenvalues of Hermitian operators are real numbers, a i * = a i. Taking advantage of this and rearranging the final integral gives us a j ψ i * ψj dτ = a i ψ i * ψj dτ or a j a i ψ i * ψj dτ = 0. Since the eigenvalues are different, (a j a i ) 0. Thus, the integral must be zero and the wavefunctions are orthogonal. 11.1 For a system of only two states, Eq. (11.18) for the wavefunction of a general state of the system is ψ = c 1 ψ 1 + c ψ, and Eq. (11.19) gives us a relationship for the coefficients c 1 and c : c 1 + c = 1 where we have assumed real number coefficients. This equation has the form of the trigonometric identity cos γ + sin γ = 1, and thus we can associate c 1 = cos γ and c = sin γ. The general state of the system is defined in terms of the single parameter γ, and in general, a system of n eigenstates can always be described by n 1 parameters at most. 11. Here, we take advantage of the orthonormality of the complete set of wavefunctions ψ n that go into Eq. (11.18). We multiply Eq. (11.18) from the left by ψ n * and integrate over all space: 48
ψ n * ψ dτ = ψn * c n ψ n n dτ = ψ n * c0 ψ 0 + c 1 ψ 1 + + c n ψ n + dτ = c 0 ψ n * ψ0 dτ + c 1 ψ n * ψ1 dτ + + c n ψ n * ψn dτ + = c n where all the integrals in the third line vanish due to orthogonality except the n th one, which equals 1 by normalization. The specific case in the problem has ψ = Nq π 1/4 e q / = c 0 e q/ π 1/4 + c q 1 π 1/4 e q /. We first find c 0 and c without evaluating integrals. Canceling common factors in the expressions above for ψ gives Nq = c 0 c + c q. Since this must be true for all q, it must be true that or N = c and c 0 c = 0 c 0 = c and N = c 0. Substituting the first expression into the general requirement that c 0 + c = 1 gives c 1 = c 0 + c = + c or c = 3 49
so that c 0 = 1/ 3 and N = / 3. We can check this value for N simply by evaluating the normalization integral for ψ: 1 = ψ dτ = N π q 4 e q dq. We can derive the definite integral here from the one given in the problem. First, we find α e αq dq = q e αq dq = α π α = 1 π α 3/, followed by α q e αq dq = q 4 e αq dq = α 1 π α 3/ = 3 4 π α 5/, so that, setting α = 1, q 4 e q dq = 3 π 4. Thus, the normalization expression is 1 = ψ dτ = N π q 4 e q dq = N π 3 π 4 or N = 3 as we found before. We can also check the values for c 0 and c using the expression we proved at the start of this problem and the integrals derived above. We write 50
c 0 = = ψ 0 ψ dq = 3π 3π q e q dq = e q / q e q / dq 3π π = 1 3 and c = = ψ ψ dq = 6π 6π q 4 q e q dq = q 1 e q / q e q / dq 6π 6 π 4 π = 3, in agreement with our earlier results. SECTION 11.3 11.3 We wish to show that ( A) = <A > <A> = <(A <A>) >. To do so, we expand the last expression (which is the expression for ( A) in Eq. (11.)), first expanding the square, then averaging each term: <(A <A>) > = <A A<A> + <A> > = <A > <A> + <A> = <A > <A>. Note that the average of a sum: <a + b + c > is the sum of averages of individual terms: <a> + <b> + <c> + and that, since an average is just a number, averaging an average gives just the average: <<a>> = <a>. 11.4 We have Aψ i = a i ψ i and Bψ i = b i ψ i and we wish to show that A and B commute, i.e., that [A, B] = 0. Since any combination of operators is itself an operator, we can treat the commutator as an operator. Thus, we operate on ψ i with [A, B] and see if it leads to zero: 51
[A,B]ψ i = A B BA ψ i (by definition of the commutator) = AB ψ i BA ψ i (expanding the product) = Ab i ψ i Ba i ψ i (invoking the eigenvalue expressions) = b i A ψ i a i B ψ i (because A and B are Hermitian) = b i a i ψ i a i b i ψ i (again invoking eigenvalue expressions) = b i a i a i b i ψ i (factoring ψ i from the expression) = 0. Since the result is zero, the operators must commute. 11.5 If we define p x = p x, i.e., the operator for the x component of momentum is simply multiply by that component, we seek the corresponding x operator that still satisfies the commutation relation [x, p x ] = /i. Comparison with the traditional definitions of these operators suggests we try x = ( /i)( / p x ). This choice is correct, as we can see if we evaluate the result of operating with the commutator on an arbitrary wavefunction ψ: [x,p x ]ψ = xp x ψ p x x ψ = p i p x ψ p x ψ x i p x = i ψ + i p ψ x p x i p ψ x p x = i ψ. 11.6 For energy E 0, the wavefunction in region I is zero (choice (b)), because the potential energy is infinite in region I. It has no nodes (choice (a)) and does not oscillate (choice (c)) in region II, because E 0 is the lowest possible energy. It also has no nodes and does not oscillate in either region III or region IV, because E 0 is less than the potential energy in these regions and thus the system tunnels in these regions. Tunneling wavefunctions never have nodes or oscillate. For energy E 1, the wavefunction is still zero in region I. It oscillates with least wavelength for this energy in region II (choice (e)), because it has the greatest kinetic energy in this region. In region III, as for energy E 0, it has no nodes and does not oscillate, because it must tunnel in this region. In region IV, the wavefunction oscillates with greatest wavelength (choice d), because it has the smallest (positive) kinetic energy in this region. For energy E, the 5
wavefunction is still zero in region I (as it is for all energies). It oscillates with least wavelength in region II, with greatest wavelength in region III, and with an intermediate wavelength (choice (f)) in region IV. These choices again follow from the relative kinetic energy of the system in each of these regions and the general relationship that increasing kinetic energy decreases a wavefunction s wavelength. 11.7 We can divide the possible total energies into three categories, based on the energy zero choice made in the representation of the potential energy shown in the problem s figure. For all negative total energies, the system wavefunction is localized in the potential well region, and thus the energies are discrete. For all total energies > U, the motion of the system cannot reach r values close to zero, due to the steep potential energy wall, but it can extend all the way to r =. Thus, for these energies, the energy levels are continuous and the wavefunctions are delocalized out to r =. For total energies greater than zero but less than U, the system encounters a deep potential well at small r, but a barrier that decreases to zero as r. For these energies, tunneling through the barrier causes the wavefunction to be delocalized (compare to energy E 1 in the previous problem) and the energies to be continuous. GENERAL PROBLEMS 11.8 We use the relation λ = hc/e to connect an energy E to a wavelength λ. For a 3 ev chemical bond energy (which, with 1 ev 1.6 10 19 J, is about 4.8 10 19 J), λ 400 nm, at the border of the visible and UV spectral regions. For a 10 ev (1.6 10 18 J) ionization energy, λ 100 nm, in the so-called vacuum UV spectral region, since air absorbs wavelengths in this region. A 1.5 ev (.4 10 19 J) electron affinity corresponds to λ 800 nm, in the near IR spectral region, so-called because it is at that end of the infrared region that is near the visible. In contrast, the 1/40 ev (4.0 10 1 J) energy characteristic of room temperature corresponds to λ 50 000 nm = 50 µm, in the far IR region (i.e., at the opposite side of the IR from the region called near IR ). 11.9 Direct substitution of v = 0, 1,, and 10 into the bound state energy eigenvalue expression gives E 0 = 3/7, E 1 = 7/11, E = 11/15, and E 10 = 43/47. The classical turning points are those values of x for which E v = V(x): E v = V(x) = 1 1 x = 1+x 1 + x or E v 1 x + E v x + E v = 0 53
where the second expression is a convenient quadratic expression for the x coordinates of the classical turning points. Armed with these energies and this expression, we can construct the diagram below: V(x) 1.0 0.8 0.6 0.4 E 10 E E 1 E 0 0. 0 0 5 10 15 x 0 (The outer turning point for the level E 10 is slightly off the scale of this graph at x = 1.989.) If we ask for that v value that gives E v = 1, we find v =, which tells us that there are an infinite number of bound states in the Kratzer potential. 11.30 If ξ = x/x and if the de Broglie wavelength is λ = h/p where p = m(e V) with V = Eξ and E = /mx, some simple substitutions give an expression for λ(ξ): λ(ξ) = πx 1 ξ 1/. Thus, our trial wavefunction is y(ξ) = cos πx λ(ξ) = cos πx πx/(1 ξ = cos ξ(1 ξ ) 1/. ) 1/ (Note that the argument to the cosine function is a purely imaginary number for ξ > 1, but the cosine of a purely imaginary number is the hyperbolic cosine of the imaginary part of the number, which is purely real.) If we plot our trial wavefunction along with the true wavefunction ψ(ξ) = e ξ, we can see that our theory is really lousy. The trial wavefunction diverges at ±, as the graph at the top of the next page shows. 54
.0 1.5 y(ξ) 1.0 0.5 0 ψ(ξ) 1 0 1 ξ 11.31 The integral we seek here is the integral of the square of the wavefunction from 7 to + 7: 7 7 q 3 3q e q / 3 π dq. The figure below is the graph of ψ from Example 11.5 redrawn with a finer grid to aid a graphical integration. Each large grid rectangle, such as the shaded one, is 0. 0.04 = 0.008 area units large. 0.36 ψ 0.4 0.1 7 1 0 1 q 7 55
A rough count shows that there are about 114 or so of these area units under the curve for a numerical integral estimate of 114 0.008 = 0.91. A numerical integration on a computer gives the answer 0.9145 quoted in the problem. 11.3 We start with the definition of ( A) from Eq. (11.) and the expansion of ψ as written in the problem: ( A) = <(A <A>) > = ψ * (A <A>) ψ dτ = c * i φ * i i (A <A>) c j φ j j dτ. Next, we consider the effect of (A <A>) operating on the expansion of ψ, remembering that <A> is just a number: (A <A>) c j φ j j = (A <A>) (A <A>) c j φ j j = (A <A>) c j a j φ j j = c j a j φ j j <A> c j φ j j <A> c j a j φ j + <A> c j φ j. j j When we substitute this into the integral above and expand the sums, we find that, since the functions φ i are orthogonal, only those terms for which i = j survive. Integration over the others gives zero. Thus we have, recognizing that the functions φ i are also normalized: ( A) = i c i * φ i * c i a i φ i <A>c i a i φ i + <A> c i φ i dτ = φ * i φ i dτ c * i c i a i <A>ci * c i a i + <A> c * i c i i = c i a i <A>ai + <A> i = c i a i <A>. i 56
11.33 We manipulate the integral suggested in the problem as follows: <p> = = i φ * i d dx φ dx φ(x = ) φ dφ φ(x = ) = i = i φ dφ dx dx φ φ(x = ) φ(x = ) where the final result follows from the requirement that the wavefunction vanish at ±: φ(x = ±) = 0. (The complex conjugate of φ equals φ, since we assumed φ was real; thus, the complex conjugate sign is not needed after the first step.) 11.34 Using the definition of an expectation value, Eq. (11.1), we have = 0 <x > = ψ x dx 1 = X π x e x /X dx. We can evaluate the integral using the general expression derived in Problem 11.: <x 1 > = X π x e x /X 1 dx = X π π 1 X 3/ = X. On the other hand, the Virial Theorem for the harmonic oscillator tells us <V> = E/ where <V> is the expectation value of the potential energy operator V = E(x/X). Thus, we can write <V> = E = <E(x /X )> = E <x > X or <x > = X, in agreement with the previous calculation. 57