Chapter 1: From Classical to Quantum Mechanics Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q1.1) How did Planck conclude that the discrepancy between experiments and classical theory for blackbody radiation was at high and not low frequencies? The experimental results and the classical theory agree in the limit of low frequencies and diverge at high frequencies.
Q1.2) The inability of classical theory to explain the spectral density distribution of a blackbody was called the ultraviolet catastrophe. Why is this name appropriate? The divergence between the classical theory and the experimental results is very pronounced in the UV region. In particular, because the classical theory predicts that the spectral density increases as ν 2 , the total energy radiated by a blackbody is predicted to be infinite. This prediction was clearly wrong, and the lack of agreement with the experimental results was viewed by scientists of the time as a catastrophe.
Q1.3) Why does the analysis of the photoelectric effect based on classical physics predict that the kinetic energy of electrons will increase with increasing light intensity? In the classical theory, more light intensity leads to more energy absorption by the solid and the electrons. At equilibrium, as much energy must leave the surface as is absorbed. Therefore, the electrons will have a greater energy. Classical physicists thought that the greater energy would manifest as a higher speed. In fact, it appears as a larger number of electrons leaving the surface with a frequency determined by the photon energy and the work function of the surface.
Q1.4) What did Einstein postulate to explain that the kinetic energy of the emitted electrons in the photoelectric effect depends on the frequency? How does this postulate differ from the predictions of classical physics? Einstein postulated that the energy of light depends on the frequency, whereas in classical theory, the energy depends only on the intensity and is independent of the frequency. He further postulated that the energy of light could only be an integral multiple of hv.
Q1.5) Which of the experimental results for the photoelectric effect suggests that light can display particle-like behavior?
1-1
Chapter 1/From Classical to Quantum Mechanics The fact that at very low intensity the incident light is still able to eject electrons, if its frequency is above the threshold frequency, suggests that the energy of the light can be concentrated in a region of atomic dimensions at the surface of the solid. This suggests particle-like behavior.
Q1.6) In the diffraction of electrons by crystals, the volume sampled by the diffracting electrons is on the order of 3 to 10 atomic layers. If He atoms are incident on the surface, only the topmost atomic layer is sampled. Can you explain this difference? Whereas electrons can penetrate through the topmost layer into the solid, He atoms are too large to allow penetration at thermal energies. Therefore electrons sample a number of atomic layers below the surface, and He atoms are only sensitive to the outermost layer.
Q1.7) In the double-slit experiment, researchers found that an equal amount of energy passes through each slit. Does this result allow you to distinguish between purely particle-like or purely wave-like behavior? No. This result would be expected for both waves and particles. In the case of particles, the same number would pass through each slit for a large number of incident particles. For a wave, the fraction of the intensity of the wave that passes through each slit would be the same.
Q1.8) Is the intensity observed from the diffraction experiment depicted in Figure 1.6 the same for the angles shown in parts (b) and (c)? Yes. For all minima in a graph of intensity versus frequency, the intensity is zero as seen in Figure 1.5.
Q1.9) What feature of the distribution depicted as Case 1 in Figure 1.7 tells you that the broad distribution arises from diffraction? The intensity goes through a minimum and increases again on both sides of the maximum. This can only occur through wave interference.
Q1.10) Why were investigations at the atomic and subatomic levels required to detect the wave nature of particles? The wavelength of particles with greater mass is so short that particle diffraction could not be observed experimentally. Therefore, the wave-particle duality present even in heavier particles was not observed.
1-2
Chapter 1/From Classical to Quantum Mechanics
Problems P1.1) The distribution in wavelengths of the light emitted from a radiating blackbody is a sensitive function of the temperature. This dependence is used to measure the temperature of hot objects, without making physical contact with those objects, in a technique called optical pyrometry. In the limit (hc λ k T ) >> 1, the maximum in a plot of ρ ( λ , T ) versus λ is given by λmax = hc / 5kT . At what wavelength does the maximum in
ρ ( λ ,T ) occur for T = 450, 1500, and 4500 K?
hc . 5k T
According to Example Problem 3.1, λmax =
6.626 × 10−34 J s × 2.998 ×108 m s −1 = 6.40 × 10−6 m at 450 K. λmax for 1500 K and 5 × 1.381 × 10−23 J K −1 × 450 K 4500 K is 1.92 × 10–6 m and 6.39 × 10–7 m, respectively.
λmax =
P1.2) For a monatomic gas, one measure of the “average speed” of the atoms is the root 3kT , in which m is the molecular mass and k is the m Boltzmann constant. Using this formula, calculate the de Broglie wavelength for He and Ar atoms at 100 and at 500 K.
mean square speed, v rms = v 2
λ=
h = mv rms
h = 3k T m
1
2
=
6.626 × 10−34 J s 3 × 1.381 × 10−23 J K −1 × 100 K × 4.003amu × 1.661 × 10−27 kg amu −1
= 1.26 × 10−10 m for He at 100 K. λ =5.65 × 10–11 m for He at 500 K. For Ar, λ = 4.00 × 10–11 m and 1.79 × 10–11 m at 100 K and 500 K, respectively.
3kT , calculate the gas m temperatures of He and Ar for which λ = 0.20 nm, a typical value needed to resolve diffraction from the surface of a metal crystal. On the basis of your result, explain why Ar atomic beams are not suitable for atomic diffraction experiments.
P1.3) Using the root mean square speed, v rms = v 2
1
2
=
For He, h2 T= 3k m λ 2
=
( 6.626 ×10 3 × 1.381 ×10
−23
−34
J s)
−1
J K × 4.003amu × 1.661 × 10
For Ar, T = 4.0 K. 1-3
2
−27
kg amu × ( 0.20 × 10 m ) −1
−9
2
= 40 K
Chapter 1/From Classical to Quantum Mechanics The argon temperature is well below its liquifaction temperature at 4 K. It will not be possible to make an atomic beam of Ar atoms with this wavelength with conventional means.
P1.4) Electrons have been used to determine molecular structure by diffraction. Calculate the speed of an electron for which the wavelength is equal to a typical bond length, namely, 0.150 nm. p h 6.626 ×10−34 J s v= = = = 4.85 × 106 m s −1 −31 −9 m mλ 9.109 ×10 kg × 0.150 ×10 m
P1.5) Calculate the speed that a gas-phase oxygen molecule would have if it had the
same energy as an infrared photon (λ = 104 nm), a visible photon (λ = 500 nm), an ultraviolet photon (λ = 100 nm), and an X-ray photon (λ = 0.1 nm). What temperature would the gas have if it had the same energy as each of these photons? Use the root mean 1 3kT square speed, v rms = v 2 2 = , for this calculation. m v=
2E = m
2hc = mλ
2 × 6.626 × 10−34 Js × 2.998 × 108 m s −1 32.0 amu × 1.661 × 10
−27
kg ( amu ) × 10000 × 10 m −1
−9
= 864 m s −1 for
4
λ = 10 nm.The results for 500 nm, 100 nm and 0.1 nm are 3.87 × 103 m s–1, 8.65 × 103 m
s–1, and 2.73 × 105 m s–1.
We calculate the temperature using the formula 2 32.0 kg mol−1 × ( 864 m s −1 ) M v rms = = 958 K T= 3R 3× 8.314 J mol−1 K −1 λ = 104 nm. The results for 500 nm, 100 nm and 0.1 nm are 1.92 × 104 K, 9.60 × 104 K, and 9.56 × 107 K. 2
P1.6) Pulsed lasers are powerful sources of nearly monochromatic radiation. Lasers that emit photons in a pulse of 10-ns duration with a total energy in the pulse of 0.10 J at 1000 nm are commercially available. a) What is the average power (energy per unit time) in units of watts (1 W = 1 J/s) associated with such a pulse? b) How many 1000-nm photons are emitted in such a pulse?
∆E 0.10 J = = 1.0 × 107 J s −1 −8 ∆ t 1.0 × 10 s E E 0.10 J b) N = pulse = pulse = = 5.0 × 1017 −1 8 c 2.998 × 10 m s E photon h 6.626 × 10−34 J s −1 × λ 1.000 × 106 m
a) P =
1-4
Chapter 1/From Classical to Quantum Mechanics
P1.7) Assume that water absorbs light of wavelength 3.00 × 10–6 m with 100% efficiency. How many photons are required to heat 1.00 g of water by 1.00 K? The heat capacity of water is 75.3 J mol–1 K–1. hc = n C p ,m ∆T E = Nhν = N
λ
1.00g 75.3J K −1 mol−1 ×1.00 K × 3.00 ×10−6 m m C p , m ∆T λ = = 6.31×1019 N= −1 −34 −1 8 18.02g mol 6.626 ×10 J s × 2.998 ×10 m s M hc
P1.8) A 1000-W gas discharge lamp emits 3.00 W of ultraviolet radiation in a narrow range centered near 280 nm. How many photons of this wavelength are emitted per second? E 3.00 W ×1J s −1W −1 3.00 W ×1J s −1W −1 n′ = total = = = 4.23 ×1018 s −1 hc 6.626 ×10−34 J s × 2.998 ×108 m s −1 E photon λ 280 ×10−9 m
P1.9) A newly developed substance that emits 225 W of photons with a wavelength of 225 nm is mounted in a small rocket such that all of the radiation is released in the same direction. Because momentum is conserved, the rocket will be accelerated in the opposite direction. If the total mass of the rocket is 5.25 kg, how fast will it be traveling at the end of 365 days in the absence of frictional forces? The number of photons is given by 1 J s −1 1 J s −1 3.00 W × 225 W × E W = W n′ = total = = 2.547 ×1020 s −1 −34 8 −1 hc 6.626 ×10 J s × 2.998 ×10 m s E photon λ 225 ×10−9 m The momentum of one photon is h 6.626 × 10−34 J s p= = = 2.945 × 10−27 kg m s −1 λ 225 × 10−9 m The force is given by the rate of change of momentum. d ( n′p ) F= = 2.945 ×10−27 kg m s −1 × 2.547 ×1020 s −1 = 7.501×10−7 kg m s −2 dt The final speed is given by F 7.501×10−7 kg m s −2 86400s × 365days × v = v0 + at = t = = 4.51m s −1 m 5.25 kg day
P1.10) What speed does a H2 molecule have if it has the same momentum as a photon of wavelength 280 nm?
1-5
Chapter 1/From Classical to Quantum Mechanics p=
h
λ
vH2 =
= mH 2 v H 2 h mH 2 λ
=
6.626 ×10−34 J s 2.016 amu ×1.661×10−27 kg ( amu ) × 280 ×10−9 m −1
= 0.707m s −1
P1.11) The following data were observed in an experiment of the photoelectric effect from potassium: 10–19 Kinetic Energy (J) Wavelength (nm)
4.49
3.09
1.89
1.34
7.00
3.11
250
300
350
400
450
500
Graphically evaluate these data to obtain values for the work function and Planck’s constant. -19
4·10
-19
3·10 E (J)
-19
2·10
-19
1·10
6
6
2.5·10
3·10
6
3.5·10
6
4·10
1
The best-fit line is given by E = −3.97362 ×10−19 + 2.11171×10−25
1
λ
.
2.11171×10−25 J m ≈ 7.0 ×10−34 J s. The work function is 2.998 ×108 m s −1 hc − E where given by the intercept of the line with the x axis at y = 0. φ = Because the slope is hc, h =
λ0
λ0=
−25
2.1171×10 J m =5.30×10−7 m. This gives φ ≈ 4.0 ×10−19 J or 2.5eV. −19 3.97362×10 J
P1.12) Show that the energy density radiated by a blackbody Etotal ( T )
∞
∞
8π hν 3 1 dν depends on the temperature as T 4. hv / kT 3 −1 V c e 0 0 (Hint: Make the substitution of variables x = hν / kT . ) The definite integral = ∫ ρ (ν , T ) dν = ∫
∞
x3 π4 = ∫0 e x − 1 dx 15 . Using your result, calculate the energy density radiated by a blackbody at 800 and 4000 K. 1-6
Chapter 1/From Classical to Quantum Mechanics
∞
Etotal 8π hν 3 1 h =∫ ; dx = dν dν . Let x = hν hυ 3 kT kT V c 0 e kT − 1 ∞
8π hν 3 1 8π k 4T 4 d ν = ∫0 c3 ehυ kT − 1 h3c3
At 800 K, At 4000,
∞
x3 8π 5 k 4T 4 dx = ∫0 e x − 1 15 h 3 c 3
Etotal 8π 5 k 4T 4 8π 5 (1.381×10−23J K −1 ) 4 × (800 K) 4 = = = 3.10 ×10−4 J m −3 3 3 3 3 V 15 h c 15 × ( 6.626×10−34 J s ) ( 2.998 ×108 m s −1 ) Etotal 8π 5 × (1.381×10−23 J K −1 ) 4 × (4000 K) 4 = = 0.194 J m −3 3 3 34 8 1 − − V 15 × ( 6.626 ×10 J s ) × ( 2.998 ×10 m s )
P1.13) The power per unit area emitted by a blackbody is given by P = σ T 4 with σ = 5.67 × 10−8 W m −2 K −4 . Calculate the energy radiated by a spherical blackbody of radius 0.500 m at 1000 K per second. What would the radius of a blackbody at 2500 K be if it emitted the same energy as the spherical blackbody of radius 0.500 m at 1000 K? 2 4 E = Aσ T 4 =4π ( 0.500 m ) × 5.67 ×10 −8 J s −1 m −2 K −4 × (1000 K ) = 1.78 ×105 J s −1
Because the total energy radiated by the spheres must be equal, 4π r12σ T14 =4π r22σ T24 r2 =
r12T14 = T24
( 0.500 m ) (1000 K ) 4 ( 2500 K ) 2
4
= 0.0800 m
P1.14) In our discussion of blackbody radiation, the average energy of an oscillator E osc =
e
hν
hν kT
−1
was approximated as E osc =
Calculate the relative error =
(
hν
1 + hν
E − Eapprox
−1 kT )
= kT for hν
kT
<< 1 .
in making this approximation for E ν = 4 × 1012 s–1 at temperatures of 6000, 2000, and 500 K. Can you predict what the sign of the relative error will be without a detailed calculation?
1-7
Chapter 1/From Classical to Quantum Mechanics
Relative Error =
E − Eapprox E
hν − kT exp hν − 1 kT = hν exp hν − 1 kT
−34 12 −1 6.626 × 10 J s × 4.00 ×10 s hν − 1.381×10−23 J K −1 × 6000 K − kT −34 12 −1 6.626 ×10 J s × 4.00 ×10 s exp hν − 1 exp − 1 −23 −1 kT 1.381×10 J K × 6000 K = − 34 12 − 1 hν 6.626 ×10 J s × 4.00 ×10 s − 1 −34 12 −1 exp hν − 1 6.626 ×10 J s × 4.00 ×10 s exp −23 −1 kT 1.381×10 J K × 6000 K E − Eapprox
= −0.0162 for T = 6000 K. E The results for 2000 K, and 500 K are –0.0496 and –0.219. E − Eapprox Because Eapprox = kT > E, the relative error = is always a negative number. E Relative Error =
P1.15) The power (energy per unit time) radiated by black body per unit area of surface expressed in units of W m–2 is given by P = σ T 4 with σ = 5.67 × 10−8 W m −2 K −4 . The radius of the sun is 7.00 × 105 km and the surface temperature is 6000 K. Calculate the total energy radiated per second by the sun. Assume ideal blackbody behavior. E = PA = σ T 4 × 4π r 2 = 5.67 × 10−8 W m −2 K −4 × ( 6000 K ) × 4π × ( 7.00 × 108 m ) 4
2
= 4.52 × 1026 W
P1.16) A more accurate expression for E osc would be obtained by including additional terms in the Taylor-Mclaurin series. The Taylor-Mclaurin series expansion of f(x) in the vicinity of x0 is given by (see Math Supplement) 1 d 2 f ( x) 1 d 3 f ( x) d f ( x) 2 3 − + − + f ( x) = f ( x0 ) + ( x x ) ( x x ) ( x − x0 ) + ... 0 0 2 3 d x 2! d x 3! d x x = x0 x= x x= x 0
in powers of hν kT out E osc − kT in the vicinity of hν kT = 0. Calculate the relative error, , if you E osc
Use this formalism to better approximate E osc by expanding e to ( hν kT )
3
0
hν kT
1-8
Chapter 1/From Classical to Quantum Mechanics had not included the additional terms for ν = 1.00 × 1012 s–1 at temperatures of 800, 500, and 250 K. 2 3 hν hν 1 hν 1 hν kT The Taylor series expansion of e is 1 + + + + ... . Therefore kT 2 kT 6 kT
(
including terms up to hν
hν
E osc =
2
, kT ) 3
3
hν 1 hν 1 hν + + kT 2 kT 6 kT Eosc − kT hν = − kT hν 1 hν 2 1 hν 3 Eosc + + kT 2 kT 6 kT
hν 2
hν 1 hν 1 hν + + kT 2 kT 6 kT
3
6.626 ×10−34 J s ×1.00 ×1012s −1 1 6.626 ×10−34 J s ×1.00 ×1012s −1 2 + −23 −1 −23 −1 1.381 10 J K 800 K 2 1.381 10 J K 800 K × × × × Eosc − kT −23 −1 = 6.626 ×10−34 J s ×1.00 ×1012s −1 − 1.381×10 J K × 800 K × −34 12 −1 3 Eosc 1 6.626 ×10 J s ×1.00 ×10 s + 6 1.381×10−23 J K −1 × 800 K 2 − 34 12 − 1 − 34 12 − 1 6.626 ×10 J s ×1.00 ×10 s 1 6.626 ×10 J s ×1.00 ×10 s + −23 −1 1.381 10 J K 800 K 2 1.381×10−23 J K −1 × 800 K × × −34 12 −1 6.626 ×10 J s ×1.00 ×10 s 3 1 6.626 ×10−34 J s ×1.00 ×1012s −1 + 6 1.381×10−23 J K −1 × 800 K
Eosc − kT = −0.0306 for 800 K. The corresponding values for 500 K and 250 K are Eosc –0.0495 and –0.102.
P1.17) The observed lines in the emission spectrum of atomic hydrogen are given by 1 1 − 2 cm −1 , n > n1 . In the notation favored by spectroscopists, 2 n1 n
ν ( cm −1 ) = RH ( cm −1 )
E and RH = 109, 677 cm −1. The Lyman, Balmer, and Paschen series refers to λ hc n1 = 1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of v and E in each of these series? 1 → 0. Therefore The highest value for ν corresponds to n 1 ν = RH 2 cm −1 = 109, 677 cm −1 or Emax = 2.18 × 10–18 J for the Lyman series. 1 1 ν = RH 2 cm −1 = 27419 cm −1 or Emax = 5.45 × 10–19 J for the Balmer series, and 2
ν =
1
=
1-9
Chapter 1/From Classical to Quantum Mechanics
1
ν = RH 2 cm −1 = 12186 cm −1 or Emax = 2.42 × 10–19 J for the Paschen series 3
P1.18) A beam of electrons with a speed of 3.50 × 104 m/s is incident on a slit of width
200 nm. The distance to the detector plane is chosen such that the distance between the central maximum of the diffraction pattern and the first diffraction minimum is 0.500 cm. How far is the detector plane from the slit? The diffraction minima satisfy the condition sin θ =
λ
nλ , n = ± 1, ± 2,... and the first a
minimum is at sin θ = ± . We choose the plus sign (the minus sign gives the distance a from the slit in the opposite direction) giving λ h 6.626 ×10−34 J s = = 0.10392 sin θ = = a mva 9.109 ×10−31kg × 3.50 ×104 m s −1 × 200 ×10−9 m θ = 5.97 degrees The distance d from the screen and the position of the first minimum s are related by s 0.500 cm = = 4.78cm. d= tan θ 0.1045
P1.19) If an electron passes through an electrical potential difference of 1 V, it has an energy of 1 electron-volt. What potential difference must it pass through in order to have a wavelength of 0.100 nm? 2
h 1 1 h2 = E = me v 2 = me × 2 2 2me λ 2 me λ =
( 6.626 ×10
−34
J s)
2
2 × 9.109 ×10−31kg × (10−10 m )
2
= 2.41×10−17 J ×
1eV = 150.4 eV 1.602 ×10−19 J
The electron must pass through an electrical potential of 150.4 V.
P1.20) What is the maximum number of electrons that can be emitted if a potassium
surface of work function 2.40 eV absorbs 3.25 × 10–3 J of radiation at a wavelength of 300 nm? What is the kinetic energy and velocity of the electrons emitted? c 6.626 ×10−34 J s × 2.998 ×108 m s −1 1.602 ×10−19 J E = h −φ = − 2.40 eV × = 2.77 ×10−19 J −9 λ 300 ×10 m eV v=
2E = m
2 × 2.77 × 10−19 J = 7.80 × 105 m s −1 −31 9.109 × 10 kg
Etotal Etotal 3.25 × 10−3 J = = = 4.91× 1015 n= −34 8 −1 −9 E photon hc λ ( 6.626 × 10 J s × 2.998 × 10 m s ) 300 ×10 m electrons. 1-10
Chapter 1/From Classical to Quantum Mechanics
P1.21) The work function of platinum is 5.65 eV. What is the minimum frequency of light required to observe the photoelectric effect on Pt? If light with a 150-nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons? a) For electrons to be emitted, the photon energy must be greater than the work function of the surface. 1.602×10−19 J E = hν ≥ 5.65 eV × = 9.05 ×10−19 J eV −19 E 9.05 × 10 J ν≥ ≥ ≥ 1.37 × 1015 s −1 −34 h 6.626 × 10 J s b) The outgoing electron must first surmount the barrier arising from the work function, so not all the photon energy is converted to kinetic energy. hc Ee = hν − φ = −φ
λ
=
6.626 ×10 J s × 2.998 ×108 m s −1 − 9.05 × 10−19 J = 4.19 ×10−19 J −9 150 × 10 m
v=
−34
2 Ee 2 × 4.19 × 10−19 J = = 9.59 ×105 m s −1 −31 9.11× 10 kg me
P1.22) X-rays can be generated by accelerating electrons in a vacuum and letting them impact on atoms in a metal surface. If the 1000-eV kinetic energy of the electrons is completely converted to the photon energy, what is the wavelength of the X-rays produced? If the electron current is 1.50 × 10–5 A, how many photons are produced per second? hc 6.626 ×10−34 J s × 2.998 ×108 m s −1 λ= = = 1.24 nm 1.602 ×10−19 J E 1000 eV × eV current 1.50 ×10−5Cs −1 n= = = 9.36 ×1013 s −1. −19 charge per electron 1.602 ×10 C
P1.23) When a molecule absorbs a photon, both the energy and momentum are conserved. If a H2 molecule at 300 K absorbs an ultraviolet photon of wavelength 100 nm, what is the change in its velocity ∆v ? Given that its average speed is v rms = 3kT / m , what is ∆v / v rms ? Because momentum is conserved, p photon
6.626 ×10−34 J s = = = 6.626 ×10−27 kg m s −1 −9 λ 100 ×10 m h
1-11
Chapter 1/From Classical to Quantum Mechanics All this momentum is transferred to the H2 molecule ∆pH 2 = 6.626 ×10−27 kg m s −1 = m∆v ∆v H 2 =
6.626 ×10−27 kg m s −1 = 1.98 m s −1 −27 2.016 amu × 1.661×10 kg/amu
∆v 1.98 m s −1 = = v 3RT M
1.98 m s −1 −1
−1
3 × 8.314 J mol K × 298 K 2.016 ×10−3 kg mol−1
1-12
= 1.03 ×10−3
Chapter 2: The Schrödinger Equation Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q2.1) By discussing the diffraction of a beam of particles by a single slit, justify the statement that there is no sharp boundary between particle-like and wave-like behavior. For a low particle energy, corresponding to a long wavelength, the diffraction pattern is clearly resolved. As the energy increases, the wavelength decreases, the diffraction peak moves closer to the central peak and its intensity decreases. The diffraction peak does not disappear with increasing energy; it simply becomes difficult to observe.
Q2.2) Why does a quantum mechanical system with discrete energy levels behave as if
it has a continuous energy spectrum if the energy difference between energy levels ∆ E satisfies the relationship ∆ E << kT ?
If the difference in energy between levels becomes small compared to kT, the levels become “smeared out” and overlap. When this happens, the levels can no longer be distinguished, and from the viewpoint of the observer, the system has a continuous energy spectrum.
Q2.3) Why can we conclude that the wave function ψ ( x, t ) = ψ ( x)e −i ( E h )t represents a standing wave? It represents a standing wave because it can be written as the product of a function that depends only on time with a function that depends only on the spatial coordinate. Therefore the nodes do not move with time.
Q2.4) Why is it true for any quantum mechanical problem that the set of wave functions is larger than the set of eigenfunctions? The set of all wave functions must satisfy the boundary conditions as well as satisfy the conditions that allow us to interpret the square of the magnitude of the wave function in terms of probability. The set of eigenfunctions must satisfy an additional condition given by the eigenvalue equation. Because some wave functions won’t satisfy the eigenvalue equation, the set of wave functions is larger than the set of eigenfunctions.
Q2.5) Is it correct to say that because the de Broglie wavelength of a H2 molecule at 300 K is on the order of atomic dimensions that all properties of H2 are quantized? 2-1
Chapter 2/The Schrödinger Equation
No. The system must be treated using quantum mechanics if the characteristic size is comparable to the wavelength. For a H2 molecule moving in a one liter container, the wavelength is much smaller than the characteristic dimension of the container. In this case, the translational energy levels are so closely spaced that they can be described classically. Therefore, the pressure exerted by the H2 molecules on the walls of the container can be described classically.
Q2.6) In Figure 2.6 the extent to which the approximate and true functions agree was judged visually. How could you quantify the quality of the fit? You could use a function like
∫ ⎡⎣ f ( x ) − g ( x )⎤⎦ dx , where f(x) and g(x) are the 2
functions to be compared, as a quantitative measure of how alike the functions are. The value of the integral goes to zero as the two functions become identical.
Q2.7) If ψ ( x, t ) = A sin(kx − ωt ) describes a wave traveling in the plus x direction, how would you describe a wave moving in the minus x direction? Consider the nodes in the function ψ ( x, t ) = A sin(kx + ωt ). The wave amplitude ⎛x t⎞ is zero for 2π ⎜ + ⎟ = nπ where n is an integer. Solving for x, we obtain the ⎝λ T ⎠ ⎛n t ⎞ location of the nodes. x = λ ⎜ − ⎟ . We see that x decreases as t increases, ⎝2 T ⎠ showing that the wave is moving in the direction of negative x.
Q2.8) A traveling wave with arbitrary phase φ can be written as
ψ ( x, t ) = A sin(kx − ωt + φ ). What are the units of φ? Show that φ could be used to
represent a shift in the origin of time or distance. The units of φ are radians. We can rewrite this equation either as ⎡ φ⎤ ⎡ φ⎤ ψ ( x, t ) = A sin(k ⎢ x + ⎥ − ωt ) or ψ ( x, t ) = A sin(kx − ω ⎢t − ⎥ ) . The first of k⎦ ⎣ ω⎦ ⎣ these equations suggests a shift in x, and the second suggests a shift in t.
Q2.9) One source emits spherical waves and another emits plane waves. For which source does the intensity measured by a detector of fixed size fall off more rapidly with distance? Why? The intensity of the spherical source falls off more rapidly. The total intensity is the same for all spheres centered at the source. Larger spheres correspond to larger distances, and for a given detector area, the fraction of the intensity in the A area A is , where r is the radius of the sphere, which is equal to the distance 4π r 2 2-2
Chapter 2/The Schrödinger Equation from the source to the detector. Therefore the measured intensity for the spherical source decreases as 1/r2. There is no decrease in intensity for the source emitting plane waves. A well collimated light source, such as a laser, is an approximation to a plane wave source.
Q2.10) Distinguish between the following terms applied to a set of functions: orthogonal, normalized, and complete. Two functions φi(x) and φj(x) are orthogonal if ∫ φi* ( x ) φ j ( x ) dx = 0 , normalized if ∫ φi* ( x ) φi ( x ) dx = ∫ φ *j ( x ) φ j ( x ) dx = 1 , and complete if any well behaved ∞
function ψ(x) can be expanded in terms of the φi(x), ψ ( x ) = ∑ bmφm ( x ) . m =1
Problems P2.1) Assume that a system has a very large number of energy levels given by the formula ε = ε 0l 2 with ε 0 = 2.34 ×10−22 J, where l takes on the integral values 1, 2, 3, … . Assume further that the degeneracy of a level is given by gl = 2l . Calculate the ratios n5 / n1 and n10 / n1 for T = 100 K and T = 650 K, respectively. ⎡ − ( 52 ε 0 − ε 0 ) ⎤ ⎡ − ( ε 5 − ε1 ) ⎤ 2 × 5 n5 g5 ⎥ = exp ⎢ exp ⎢ ⎥= n1 g1 kT ⎢⎣ ⎥⎦ ⎣ kT ⎦ 2 ×1
⎡ −2.34×10−22 J × ( 25 − 1) ⎤ n5 10 (100 K) = exp ⎢ ⎥ = 0.086 −23 −1 × × 2 1.381 10 J K 100 K n1 ⎣ ⎦ ⎡ −2.34×10−22 J × ( 25 − 1) ⎤ n5 10 (650 K) = exp ⎢ ⎥ = 2.67 −23 −1 2 n1 ⎣1.381×10 J K × 650 K ⎦ ⎡ −2.34 ×10−22 J × (100 − 1) ⎤ 20 n10 −7 exp ⎢ (100 K ) = ⎥ = 5.2 ×10 −23 −1 2 n1 ⎣1.381×10 J K × 100 K ⎦ ⎡ −2.34 ×10−22 J × (100 − 1) ⎤ n10 20 (650 K) = exp ⎢ ⎥ = 0.757 −23 −1 n1 2 ⎣1.381×10 J K × 650 K ⎦
P2.2) Consider a two-level system with ε1 = 3.10 × 10−21J and ε 2 = 6.10 × 10−21J. If g2 = g1, what value of T is required to obtain n2 / n1 = 0.150? What value of T is required to obtain n2 / n1 = 0.999?
2-3
Chapter 2/The Schrödinger Equation
⎡ − ( ε 5 − ε1 ) ⎤ n2 g 2 = exp ⎢ ⎥ n1 g1 ⎣ kT ⎦ ⎛n ⎞ ⎛ g ⎞ (ε − ε ) ln ⎜ 2 ⎟ = ln ⎜ 2 ⎟ − 5 1 kT ⎝ n1 ⎠ ⎝ g1 ⎠ ⎡ ⎛ g2 ⎞ ⎛ n2 ⎞ ⎤ 1 k = ⎢ ln ⎜ ⎟ − ln ⎜ ⎟ ⎥ T ( ε 5 − ε1 ) ⎣ ⎝ g1 ⎠ ⎝ n1 ⎠ ⎦ T=
( ε 5 − ε1 )
⎡ ⎛g ⎞ ⎛ n ⎞⎤ k ⎢ ln ⎜ 2 ⎟ − ln ⎜ 2 ⎟ ⎥ ⎝ n1 ⎠ ⎦ ⎣ ⎝ g1 ⎠
3.00×10−21J = 115 K for n2 n1 = 0.150 T = 1.381×10−23 J K −1 × ⎡⎣ ln (1) − ln ( 0.150 ) ⎤⎦ for n2 n1 = 0.999 T =
3.00 × 10−21J = 2.17 × 105 K −23 −1 1.381 × 10 J K × ⎡⎣ ln (1) – ln ( 0.999) ⎤⎦
P2.3) To plot Ψ( x , t ) = A sin( kx − ωt ) as a function of one of the variables x and t, the Ψ ( x0 ,0) = –0.280, what is the Ψmax Ψ (0, t 0 ) = –0.309, what is the constant value of x0 in the upper panel of Figure 2.3? If Ψmax constant value of t0 in the lower panel of Figure 2.3? (Hint: The inverse sine function has two solutions within an interval of 2π. Make sure that you choose the correct one.) ⎛ x ⎞ t ⎛x t⎞ For the traveling wave Ψ ( x, t ) = A sin 2π ⎜ − ⎟ = A sin 2π ⎜ − −3 ⎟ ⎝λ T ⎠ ⎝ 1.46 1.00 x 10 ⎠ other variable needs to be set at a fixed value, x0 or t0. If
⎛ x0 ⎞ ⎛ x0 ⎞ Ψ ( x0 , 0) = sin 2π ⎜ ⎟ = −0.284; 2π ⎜ ⎟ Ψ max ⎝ 1.46 m ⎠ ⎝ 1.46 m ⎠ = −0.28 radians or π − ( −0.28 ) = 3.43 radians x = −6.4 ×10−2 m or 0.79 m
⎛ ⎞ ⎛ ⎞ Ψ (0, t0 ) −t 0 −t 0 = A sin 2π ⎜ = − 0.309; 2π ⎜ −3 ⎟ −3 ⎟ Ψ max ⎝ 1.00 x 10 s ⎠ ⎝ 1.00 x 10 s ⎠ = −0.31 radians or π − ( −0.31) = 3.45 radians t0 = 5.0 ×10−5s or − 5.5 ×10−4s To decide which of the two values best fits the data in Figure 2.2, it is necessary to plot the data.
2-4
Chapter 2/The Schrödinger Equation
It is seen that the data are fit by x0 = 0.79 m and t0= –5.5 × 10–4 s. Adding any positive or negative integral multiple of λ to x0 or T to t0 will give equally good agreement.
P2.4) A wave traveling in the z direction is described by the wave function
Ψ ( z, t ) = A1 x sin( kz − ωt + φ1 ) + A2 y sin( kz − ωt + φ2 ) where x and y are vectors of unit length along the x and y axes, respectively. Because the amplitude is perpendicular to the propagation direction, Ψ( z , t ) represents a transverse wave. a) What requirements must A1 and A2 satisfy for a plane polarized wave in the x-z plane? b) What requirements must A1 and A2 satisfy for a plane polarized wave in the y-z plane? c) What requirements must A1 and A2 and φ 1 and φ 2 satisfy for a plane polarized wave in a plane oriented at 45º to the x-z plane? d) What requirements must A1 and A2 and φ 1 and φ 2 satisfy for a circularly polarized wave? a) The amplitude along the x axis must oscillate, and the amplitude along the y axis must vanish. Therefore A1 ≠ 0 and A2 = 0. b) The amplitude along the y axis must oscillate, and the amplitude along the x axis must vanish. Therefore A1 = 0 and A2 ≠ 0.
2-5
Chapter 2/The Schrödinger Equation c) The amplitude along both the x and y axes must oscillate. Therefore A1 ≠ 0 and A2 ≠ 0 . Because they must oscillate in phase, φ1 = φ2 . d) The amplitude along both the x and y axes must oscillate with the same amplitude. Therefore A1 = A2 ≠ 0 . For circularly a polarized wave, the x and y components must be out of phase by π/2. Therefore φ1 =φ2 ±
π
2
. This can be seen by comparing the x and y
amplitudes for the positive sign.
π
Ψ ( z , t ) = A1 x sin(kz − ωt + φ1 ) + A1 y sin(kz − ωt + φ1 + ) 2 let kz + φ = kz ′
π
Ψ ( z , t ) = A1 x sin(kz ′ − ωt ) + A1 y sin(kz ′ − ωt + ) 2 π π⎤ ⎡ = A1 x sin( kz ′ − ωt ) + A1 y ⎢sin( kz ′ − ωt ) cos + cos( kz ′ − ωt ) sin ⎥ 2 2⎦ ⎣ = A1 x sin( kz ′ − ωt ) + A1 y cos(kz ′ − ωt ) The x and y amplitudes are π/2 out of phase and the sum of the squares of their amplitudes is a constant as required for a circle.
P2.5) Show that
b
a + ib ac + bd + i bc − ad = c + id c2 + d 2
g
a + ib ⎛ a + ib ⎞ ⎛ c − id ⎞ ac + bd + ibc − iad ac + bd + i ( bc − ad ) =⎜ = ⎟⎜ ⎟= c + id ⎝ c + id ⎠ ⎝ c − id ⎠ c2 + d 2 c2 + d 2
P2.6) Does the superposition ψ ( x , t ) = A sin( kx − ω t ) + 2 A sin( kx + ω t ) generate a standing wave? Answer this question by using trigonometric identities to combine the two terms.
ψ ( x, t ) = A sin( kx − ω t ) + 2 A sin( kx + ω t ) Using the identities sin (α ± β ) = sin α cos β ± cos α sin β
cos (α ± β ) = cos α cos β m sin α sin β the previous equation can be simplified to ψ ( x, t ) = A sin kx cos ω t − A cos kx sin ω t + 2 A sin kx cos ω t + 2 A cos kx sin ω t = 3A sin kx cos ω t + A cos kx sin ω t Because the wave function cannot be written as a single product of a function that is periodic in length with one that is periodic in time, the nodes will not be stationary. Therefore it is not a standing wave. 2-6
Chapter 2/The Schrödinger Equation
P2.7) Express the following complex numbers in the form reiθ . a) 2 − 4i
b) 6
c)
3+i 4i
d)
8+i 2 − 4i
⎛ Re z ⎞ In the notation reiθ , r = z = a 2 + b 2 and θ = cos −1 ⎜⎜ ⎟⎟ . ⎝ z ⎠ 1 ⎞ ⎛ a) 2 − 4i = 2 5 exp ⎜ i cos −1 ⎟ = 2 5 exp ( 0.352iπ ) 5⎠ ⎝ b) 6 = 6 exp ( i cos −1 1) = 6 exp ( 0 )
c)
3 + i 1 3i 10 1 ⎞ 10 ⎛ = − = exp ⎜ i cos −1 = exp ( 0.398iπ ) ⎟ 4i 4 4 4 4 10 ⎠ ⎝
d)
8+i 3 17i 13 6 ⎞ 13 ⎛ = + = exp ⎜ i cos −1 = exp ( 0.392iπ ) ⎟ 2 − 4i 5 10 2 2 5 13 ⎠ ⎝
P2.8) Express the following complex numbers in the form a + ib. a) 2e
iπ
b) 2 5 e
2
− iπ
2
c) eiπ
d)
3 2 iπ 4 e 5+ 3
To convert to the form a + ib, we use the equations Re z = z cos θ and Im z = z sin θ
π⎞ ⎛ + ⎜ 2 sin ⎟ i = 2i 2 ⎝ 2⎠ − iπ −π ⎛ −π + ⎜ 2 5 sin b) 2 5 e 2 = 2 5 cos 2 ⎝ 2 iπ c) e = cos π + ( sin π ) i = −1 a) 2e
iπ
2
= 2 cos
π
⎞ ⎟ i = −2 5i ⎠
3 2 iπ 4 3 2 π ⎛ 3 2 π⎞ 3 2 2 π ⎛ 3 2 2⎞ e = i cos + ⎜ sin ⎟ i = cos + ⎜ d) 5 + 3 4 ⎝ 5+ 3 4⎠ 4 ⎝ 5 + 3 2 ⎟⎠ 5+ 3 5+ 3 2 =
3 (1 + i ) 5+ 3
P2.9) Using the exponential representation of the sine and cosine functions 1 iθ 1 e + e − iθ ) and sin θ = ( eiθ − e − iθ ) , show that ( 2 2i 2 2 a) cos θ + sin θ = 1 d cos θ = − sin θ b) dθ cos θ =
b
g
2-7
Chapter 2/The Schrödinger Equation
FG H
c) sin θ +
IJ = cos θ 2K
π
a) 2
2
⎡1 ⎤ ⎡1 ⎤ cos2 θ + sin 2 θ = ⎢ ( eiθ + e − iθ ) ⎥ + ⎢ ( eiθ − e − iθ ) ⎥ ⎣2 ⎦ ⎣ 2i ⎦ 1 1 = ( 2 + e 2iθ + e −2iθ ) − ( −2 + e 2iθ + e −2iθ ) = 1 4 4 b) ⎡1 ⎤ d ⎢ ( eiθ + e − iθ )⎥ d ( cos θ ) 2 ⎦ = 1 ieiθ − ie − iθ = − 1 eiθ − e − iθ = − sin θ = ⎣ ( ) 2−( ) dθ dθ 2 c) ⎡ π⎤ ⎡ π⎤ π ⎞ 1 ⎛ i ⎣⎢θ + 2 ⎦⎥ − i ⎣⎢θ + 2 ⎦⎥ ⎞ 1 ⎛ i π2 iθ − i π2 − iθ ⎞ i iθ ⎛ − iθ sin ⎜ θ + ⎟ = ⎜ e −e ⎟ = ⎜ e e − e e ⎟ = ( e + e ) = cos θ ⎜ ⎟ 2 ⎠ 2i ⎝ ⎝ ⎠ 2i ⎠ 2i ⎝
P2.10) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue? ∂ a) sin θ cos φ ∂φ 2 1 d −x b) e 2 x dx sin θ d c) sin θ cosθ dθ
a) sin θ cos φ
∂ ∂φ
∂ sin θ cos φ = − sin θ sin φ . ∂φ 1 d − 1 x2 b) e 2 x dx 1 d − 12 x2 − 1 x2 = −e 2 e x dx sin θ d c) sin θ cos θ dθ sin θ d sin θ = sin θ cos θ dθ
Not an eigenfunction
Eigenfunction with eigenvalue –1
Eigenfunction with eigenvalue +1
2-8
Chapter 2/The Schrödinger Equation
P2.11) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue? d3 a) x 3 d x3 ∂ ∂ x +y b) x y ∂x ∂y 2 ∂ c) sin θ cos φ ∂θ 2 d d x3
a) x 3 d x3 =6 d x3
b) x y
x
Not an eigenfunction
x
∂ ∂ +y ∂x ∂y
∂ xy ∂ xy +y = 2 xy ∂x ∂y
c) sin θ cos φ
Eigenfunction with eigenvalue +2 ∂2 ∂θ 2
∂2 ( sin θ cos φ ) = − sin θ cos φ ∂θ 2
Eigenfunction with eigenvalue –1
P2.12) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue? 1 d ⎛ d ⎞ a) 3cos 2 θ − 1 ⎜ sin θ ⎟ sin θ d θ ⎝ dθ ⎠ b) e
−x
2
2
c) e −4iφ
d2 − x2 2 dx d2 dφ 2
1 d ⎛ d ⎞ ⎜ sin θ ⎟ sin θ d θ ⎝ dθ ⎠ d ( 3cos2 θ − 1) ⎞ 1 d ⎛ 1 d ⎜ sin θ ⎟= −6 cos θ sin 2 θ ) ( ⎜ ⎟ dθ sin θ d θ sin θ d θ ⎝ ⎠ 1 = ( 6 sin3 θ − 12 cos2 θ sin θ ) = 6 sin 2 θ − 12 cos2 θ sin θ = 6 − 18 cos2 θ = −6 ( 3cos2 θ − 1)
a) 3cos2 θ − 1
2-9
Chapter 2/The Schrödinger Equation Eigenfunction with eigenvalue –6. b) e
d2 − x2 2 dx
− 1 x2 2
− 1 x2
d 2e 2 − 1 x2 − 1 x2 − x 2 e 2 = −e 2 2 dx Eigenfunction with eigenvalue –1. d2 dφ 2
c) e −4iφ
d 2 e −4iφ = −16e −4iφ dφ 2 Eigenfunction with eigenvalue –16.
P2.13) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue? ∂2 a) e − i b 3 x + 2 y g ∂ x2 1 2 ∂ b) x 2 + y 2 x + y2 ) ( x ∂x c) sin θ cos θ sin θ
a) e
−i 3 x+2 y )
b)
d ⎞ ⎛ 2 ⎜ sin θ ⎟ + 6sin θ d θ ⎝ ⎠
∂2 ∂ x2
− i(3 x +2 y )
∂2e ( ∂ x2
d dθ
= −9 e
− i(3 x +2 y )
Eigenfunction with eigenvalue –9.
1 2 ∂ x + y2 ) ( x ∂x
x2 + y2
2 2 1 2 2 ∂ x + y (x + y ) ∂ x = x
c) sin θ cos θ
sin θ
d dθ
sin θ
d dθ
x2 + y 2
Eigenfunction with eigenvalue +1.
⎛ d ⎞ 2 ⎜ sin θ ⎟ + 6 sin θ θ d ⎝ ⎠
⎛ d sin θ cos θ ⎜ sin θ dθ ⎝
⎞ d 3 sin θ ⎡⎣1 − 2 sin 2 θ ⎤⎦ + 6 sin 3 θ cos θ ⎟ + 6 sin θ cos θ = sin θ dθ ⎠
(
= sin θ ( cos θ − 6 sin 2 θ cos θ ) + 6 sin 3 θ cos θ = sin θ cos θ
2-10
)
Chapter 2/The Schrödinger Equation Eigenfunction with eigenvalue +1.
P2.14) Which of the following wave functions are eigenfunctions of the operator d/dx? If they are eigenfunctions, what is the eigenvalue? a) a e −3 x + b e −3i x
a)
d ( a e −3 x + b e −3i x ) dx
b) sin 2 x
c) e −i x
d) cosa x
e) e − i x
2
= −3 a e −3 x − 3 i b e −3ix Not an eigenfunction
2
b)
d sin x = 2 sin x cos x dx
Not an eigenfunction
d e−i x = −i e − i x dx d cos a x d) = −a sin a x dx
c)
Eigenfunction with eigenvalue –i Not an eigenfunction
2
2 d e−i x = −2i x e − i x e) dx
Not an eigenfunction
P2.15) Which of the following wave functions are eigenfunctions of the operator d 2 dx 2 ? If they are eigenfunctions, what is the eigenvalue? a) a e −3 x + b e −3i x a)
d 2 ( a e −3 x + b e −3i x ) d x2
b) sin 2 x = 9 a e −3 x − 9 b e −3 i x
c) e −i x
e) e − i x
d) cosa x
Not an eigenfunction
b)
d 2 sin 2 x = −2 sin 2 x + 2 cos2 x 2 dx
c)
d 2 e−i x = −e −i x d x2
Eigenfunction with eigenvalue –1
d)
d 2 cos a x = − a 2 cos a x d x2
Eigenfunction with eigenvalue –a2
Not an eigenfunction
2
2 2 d 2 e−i x e) = −2i e − i x − 4 x 2 e − i x 2 dx
Not an eigenfunction
bg
$ $ f x , it is important P2.16) If two operators act on a wave function as indicated by AB to carry out the operations in succession with the first operation being that nearest to the
2-11
2
Chapter 2/The Schrödinger Equation
b g d b gi bg
b g d b gi
$ $ f x = A$ B$ f x and A$ 2 f x = A$ A$ f x . Evaluate the function. Mathematically, AB $ $ f x . The operators A$ and B$ are listed in the first following successive operations AB
bg
and second columns and f x is listed in the third column. 2 d d a) x 2 + ea x dx dx 2 ∂ ∂ b) (cos 3 y ) sin 2 2 ∂y ∂x ∂ ∂2 cos φ c) 2 sin θ ∂θ ∂φ
(
)
2 a x2 ⎤ ⎡ d ⎢d x + e ⎥ = d ⎡ 2 x + 2a x e a x 2 ⎤ = 2 + 4a 2 x 2 e a x 2 + 2a e a x 2 a) ⎦ ⎥ dx⎣ dx⎢ dx ⎣ ⎦ 2 ∂ 2 ⎡ ∂ ( cos 3 y sin x ) ⎤ ∂2 b) = ⎢ ⎥ [ 2 cos 3 y sin x cos x ] = −18 cos 3 y sin x cos x ∂ y2 ⎢ ∂x ∂ y2 ⎥ ⎣ ⎦ ⎡ 2 ⎛ cos φ ⎞ ⎤ ∂ ∂ ⎢ ⎜⎝ sin θ ⎟⎠ ⎥ ∂ ⎡ cos φ ⎤ cos φ cos θ ⎢ ⎥= c) ⎢− ⎥ = sin 2 θ ∂θ ⎢ ∂ φ 2 ⎥ ∂θ ⎣ sin θ ⎦ ⎢⎣ ⎥⎦
bg to carry out the operations in succession with the first operation being that nearest to the $ $ f b x g = A$ d B$ f b x gi and A$ f b x g = A$ d A$ f b x gi . Evaluate the function. Mathematically, AB $ $ f b x g . The operators A$ and B$ are listed in the first following successive operations AB two columns and f b x g is listed in the third column.
$ $ f x , it is important P2.17) If two operators act on a wave function as indicated by AB 2
a)
d dx
b) x c) y
∂ ∂x
x
x e−a x
2
d dx ∂ x ∂y
x e−a x
2
e
e
−a x2 + y2
j
Note that your answers to parts (a) and (b) are not identical. As we will learn in Chapter 7, the fact that switching the order of the operators x and d dx changes the outcome of $ $ f x is the basis for the Heisenberg uncertainty principle. the operation AB
bg
2-12
Chapter 2/The Schrödinger Equation
a)
(
)
(
)
2 2 2 d ⎡ x x e− a x ⎤ = 2 x e− a x − 2 a x3 e− a x ⎦ dx⎣
2 ⎤ 2 2 ⎡ d b) x ⎢ x e− a x ⎥ = x e− a x − 2 a x 3 e− a x ⎣d x ⎦ ⎡ ∂ e− a ( x2 + y 2 ) ⎤ ⎥ − a ( x2 + y2 ) ⎤ ∂ ⎢ ∂ ⎡ y x =y −2 a x y e = ⎢ ⎥ ⎢⎣ ⎥⎦ x ∂y ∂ c) ∂ x ⎢ ⎥ ⎣ ⎦ 2 2 −a ( x + y ) − a ( x2 + y 2 ) = −2 a y 2 e + 4 a2 x2 y2e
(
)
2 d2 − 4 x 2 on the function e − ax . What must the 2 dx value of a be to make this function an eigenfunction of the operator? 2 2 2 2 2 2 2 d 2 e − ax − 4 x 2 e − ax = −2ae − ax − 4 x 2 e − ax + 4a 2 x 2 e − ax = −2ae − ax + 4 ( a 2 − 1) x 2 e − ax 2 dx
P2.18) Find the result of operating with
2
For the function to be an eigenfunction of the operator, the terms containing x 2 e − ax must vanish. This is the case if a = ±1 .
P2.19) Find the result of operating with (1 r 2 ) ( d dr ) ( r 2 d dr ) + 2 r on the function Ae − br . What must the values of A and b be to make this function an eigenfunction of the operator? 1 d 2 dAe − br 2 Ae − br 1 d 2 Ae − br 2 − br + = − + r bAr e ( ) r 2 dr dr r r 2 dr r − br 1 2 Ae = 2 ( −2brAe − br + b2 r 2 Ae − br ) + r r − br 2 Ae = ( 2 − 2b ) + b2 Ae− br r Ae − br To be an eigenfunction of the operator, the terms in must vanish. This requires r that b = 1. There are no restrictions on the value of A. d2 d2 d2 + + on the function x 2 + y 2 + z 2 . dx 2 dy 2 dz 2 Is this function an eigenfunction of the operator? ⎛ d2 d2 d2 ⎞ 2 2 2 ⎜ 2 + 2 + 2 ⎟ ( x + y + z ) = 6 . Therefore, the function is not an eigenfunction of dy dz ⎠ ⎝ dx the operator.
P2.20) Find the result of operating with
2-13
Chapter 2/The Schrödinger Equation
P2.21) Show that the set of functions φn (θ ) = ei nθ , 0 ≤ θ ≤ 2π , are orthogonal if n and m are integers. To do so, you need to show that the integral 2π
∫ φ (θ ) φ (θ ) dθ = 0 for m ≠ n if n and m are integers. * m
n
0
2π
⎡ 1 i ( n − m )θ ⎤ = = = φ θ φ θ d θ e e d θ e d θ e ( ) ( ) ⎢ ⎥ n ∫0 ∫0 ∫0 ⎣ i (n − m) ⎦0 1 1 ⎡ e i ( n − m ) 2π − e 0 ⎤ = = ⎡ cos ( n − m ) 2π + sin ( n − m ) 2π − 1⎤⎦ ⎣ ⎦ i (n − m) i (n − m) ⎣
2π
2π
* m
− imθ
2π
inθ
i ( n − m )θ
Because n and m are integers, ( n − m ) is an integer and the arguments of the sine and cosine functions are integral multiples of 2π . 2π
1 ∫ φ (θ ) φ (θ ) dθ = i ( n − m ) [1 + 0 − 1] = 0 * m
n
0
P2.22) Show by carrying out the integration that sin ( mπ x/a ) and cos ( mπ x/a ) , where m is an integer, are orthogonal over the interval 0 ≤ x ≤ a . Would you get the same result if you used the interval 0 ≤ x ≤ 3a / 4? Explain your result. a
a ⎡ a ⎛ mπ x ⎞ ⎛ mπ x ⎞ 2 ⎛ mπ x ⎞ ⎤ 2 ∫0 cos ⎜⎝ a ⎟⎠ sin ⎜⎝ a ⎟⎠ d x = ⎢⎣ 2mπ sin ⎜⎝ a ⎟⎠ ⎥⎦0 = 2mπ ⎡⎣sin ( mπ ) − 0⎤⎦ = 0 a
3a 4
⎛ mπ x ⎞ ⎛ mπ x ⎞ ⎟ sin ⎜ ⎟d x = a ⎠ ⎝ a ⎠
∫ cos ⎜⎝ 0
3a 4
⎛ mπ x ⎞ ⎛ mπ x ⎞ ⎟ sin ⎜ ⎟d x a ⎠ ⎝ a ⎠
∫ cos ⎜⎝ 0
3a 4
a ⎡ 2 ⎛ 3mπ ⎞ ⎤ ⎡ a ⎛ mπ x ⎞ ⎤ sin 2 ⎜ sin ⎜ =⎢ ⎟⎥ = ⎟ − 0⎥ ≠ 0 2mπ ⎢⎣ ⎝ a ⎠⎦0 ⎝ 4 ⎠ ⎦ ⎣ 2mπ 3m = n where n is an integer. The length of the integration except for the special case 4 interval must be n periods (for n an integer) to make the integral zero.
P2.23) Normalize the set of functions φn (θ ) = ei nθ , 0 ≤ θ ≤ 2π . To do so, you need to multiply the functions by a so-called normalization constant N so that the integral
z
2π
NN
*
bg bg
φ *m θ φ n θ dθ = 1 for m = n
0 2π
2π
0
0
N N * ∫ e − i nθ ei nθ dθ = N N * ∫ dθ = 2π N N * = 1 This is satisfied for N = normalized functions are φn (θ ) =
1 i nθ e , 0 ≤ θ ≤ 2π . 2π
2-14
1 and the 2π
Chapter 2/The Schrödinger Equation
P2.24) In normalizing wave functions, the integration is over all space in which the wave function is defined. The following examples allow you to practice your skills in two- and three-dimensional integration. ⎛ nπ x ⎞ ⎛ mπ y ⎞ a) Normalize the wave function sin ⎜ ⎟ sin ⎜ ⎟ over the range ⎝ a ⎠ ⎝ b ⎠ 0 ≤ x ≤ a, 0 ≤ y ≤ b . The element of area in two-dimensional Cartesian coordinates is dx dy; n and m are integers and a and b are constants. −r
b) Normalize the wave function e a cosθ sin φ over the interval 0 ≤ r < ∞, 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2π . The volume element in three-dimensional spherical coordinates is r 2 sin θ dr dθ dφ . a) ⎛ nπ x ⎞ 2 ⎛ mπ y ⎞ 1 = N 2 ∫ ∫ sin 2 ⎜ ⎟ sin ⎜ ⎟d xd y ⎝ a ⎠ ⎝ b ⎠ 0 0 nπ x mπ y adu bdw Let u = ,w = ; then dx = and dy = a b nπ mπ nπ mπ a b 1= N2 sin 2 u sin 2 w d u d w ∫ ∫ nπ mπ 0 0 a b
mπ ⎤ ⎡ nπ 2 ⎤ a b ⎡ 2 u d u sin ⎢∫ ⎥ x ⎢ ∫ sin w d w ⎥ nπ mπ ⎣ 0 ⎦ ⎣0 ⎦ 1 1 sin 2ax Using the standard integral ∫ ( sin 2 ax ) dx = x − 2 4a a b ⎡⎛ mπ 1 1 1 ⎞ ⎤ ⎡ ⎛ nπ 1 ⎞⎤ 1= N2 − sin 2mπ − 0 − sin 0 ⎟ ⎥ × ⎢ ⎜ − sin 2nπ − 0 − sin 0 ⎟ ⎥ ⎜ ⎢ nπ mπ ⎣⎝ 2 4 4 4 ⎠ ⎦ ⎣⎝ 2 4 ⎠⎦
= N2
a b nπ mπ ab = N2 nπ mπ 2 2 4 2 N= ab = N2
2-15
Chapter 2/The Schrödinger Equation b) 2π
π
∞
1 = N 2 ∫ d φ ∫ dθ ∫ e 0
0
−2 r
2π
a
0
π
∞
cos2 θ sin 2 φ r 2 sin θ d r = N 2 ∫ sin 2 φ dφ ∫ cos2 θ sin θ dθ ∫ r 2 e 0
∞
Using the standard integral ∫ x n e − ax dx = 0
0
−2 r
a
dr
0
n! ( a > 0, n positive integer) a n +1
⎛ 2! a 3 ⎞ 1 π ⎛ 2π 1 ⎞ 1 1= N2 ⎜ − sin 2π − 0 − sin 0 ⎟ × ( cos3 0 − cos π ) × ⎜ 3 ⎟ = N 2 a 3 4 6 ⎝ 2 4 ⎠ 3 ⎝ 2 ⎠ 6 π a3
N=
P2.25) Show that the following pairs of wave functions are orthogonal over the indicated range. a) e
1 − α x2 2
and ( 2α x 2 − 1) e
1 − α x2 2
, −∞ ≤ x < ∞ where α is a constant that is greater than
zero. ⎛ r r ⎞ b) ⎜ 2 − ⎟ e − r / 2 a0 and e − r / 2 a0 cosθ over the interval 0 ≤ r < ∞, 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2π a0 a0 ⎠ ⎝
a) ∞
∫ ( 2α x
2
− 1) e
1 − α x2 2
e
1 − α x2 2
∞
dx=
−∞
∫ ( 2α x
2
− 1) e
−α x 2
−∞
∞
∞
∞
d x = 2α ∫ x e 2
−∞
−α x 2
∞
d x−
∫
2
e −α x d x
−∞
= 4α ∫ x 2 e −α x d x − 2 ∫ e −α x d x because the integrand is an even function of x. 2
0
2
0
∞
Using the standard integrals ∫ x 2 n e − ax dx = 2
0
∞
and
∫ 0
2 ⎛π ⎞ e − ax dx = ⎜ ⎟ ⎝ 4a ⎠
∞
∫ ( 2α x
−∞
2
− 1) e
1 − α x2 2
e
1
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1) π ( a > 0, n positive integer) 2n +1 a n a
2
1 − α x2 2
d x = 4α
π π − =0 2α α α 1
2
2-16
Chapter 2/The Schrödinger Equation
b) 2π
∫ 0
π ∞ ⎛ r ⎞ r − r / 2 a0 2 d φ ∫ cos θ sin θ dθ ∫ ⎜ 2 − ⎟ e − r / 2 a0 e r dr a0 ⎠ a0 0 0 ⎝
⎛π ⎞⎛∞⎛ r ⎞ r − r / 2 a0 2 ⎞ = 2π ⎜ ∫ cos θ sin θ dθ ⎟ ⎜ ∫ ⎜ 2 − ⎟ e − r / 2 a0 e r d r⎟ a0 ⎠ a0 ⎝0 ⎠⎝ 0 ⎝ ⎠ ⎛∞⎛ 1 r ⎞ r − r / 2 a0 2 ⎞ ⎛ 1 ⎞ = 2π ⎜ ∫ ⎜ 2 − ⎟ e − r / 2 a0 e r d r ⎟ ⎜ cos2 0 − cos2 π ⎟ = 0 2 a0 ⎠ a0 ⎠ ⎝0⎝ ⎠⎝ 2
⎛ nπ x ⎞ ⎛ mπ x ⎞ cos ⎜ ⎟ cos ⎜ ⎟ d x = 0, m ≠ n , the functions ⎝ d ⎠ ⎝ d ⎠ cos ( nπ x/d ) for n = 1, 2, 3, ... form an orthogonal set. What constant must these functions
P2.26) Because
∫
d
0
be multiplied by to form an orthonormal set? d
d ⎛ mπ x ⎞ ⎛ mπ x ⎞ ⎛ 2mπ x ⎞ ⎤ 2 ⎡x 1 = N ∫ cos ⎜ sin ⎜ ⎟ cos ⎜ ⎟d x = N ⎢ + ⎟⎥ 0 ⎝ d ⎠ ⎝ d ⎠ ⎝ d ⎠⎦0 ⎣ 2 4mπ 1 1 where we have used the standard integral ∫ ( cos2 ax ) dx = x + sin 2ax 2 4a d 0 d ⎡d ⎤ d 1= N2 ⎢ + sin ( 2mπ ) − − sin ( 0 )⎥ = N 2 = 1 2 4mπ ⎣ 2 4mπ ⎦ 2 d
2
N=
2 d
P2.27) Use a Fourier series expansion to express the function f ( x) = x, − b ≤ x ≤ b in ⎛ nπ x ⎞ ⎛ nπ x ⎞ the form f ( x ) = d 0 + ∑ cn sin ⎜ ⎟ + d n cos ⎜ ⎟ . Obtain d0 and the first five ⎝ b ⎠ ⎝ b ⎠ n =1 coefficients cn and dn. m
2-17
Chapter 2/The Schrödinger Equation b
b
d0 =
1 1 f ( x ) dx = x dx = 0 2 b −∫b 2 b −∫b
dn =
1 1 ⎛ nπ x ⎞ ⎛ nπ x ⎞ f ( x ) cos ⎜ ⎟ dx = ∫ x cos ⎜ ⎟ dx b −∫b b b ⎝ ⎠ ⎝ b ⎠ −b b
b
Using the standard integral ∫ x cos ax =
⎛ nπ ( −b ) ⎞ ⎛ b ⎞ ⎛ nπ ( −b ) ⎞ ⎤ ⎞ ⎛ nπ b ⎞ ⎛ b ⎞ ⎟−⎜ ⎟⎥ = 0 ⎟ b sin ⎜ ⎟−⎜ ⎟ cos ⎜ ⎟ ( −b ) sin ⎜ b b ⎠ ⎝ b ⎠ ⎝ nπ ⎠ ⎝ ⎠ ⎝ nπ ⎠ ⎝ ⎠ ⎦⎥
1 ⎡⎛ b ⎞ ⎛ nπ b ⎞ ⎛ b ⎢⎜ ⎟ cos ⎜ ⎟+⎜ b ⎣⎢⎝ nπ ⎠ ⎝ b ⎠ ⎝ nπ 2
dn =
cos ax x sin ax + a2 a 2
All the d n = 0 because f ( x ) is an odd function of x. 1 1 ⎛ nπ x ⎞ ⎛ nπ x ⎞ f ( x ) sin ⎜ ⎟ dx = ∫ x sin ⎜ ⎟ dx ∫ b −b b −b ⎝ b ⎠ ⎝ b ⎠ b
cn =
b
Using the standard integral ∫ x sin ax = − 1 ⎡⎛ b ⎞ ⎛ nπ x ⎞ ⎛ b cn = ⎢⎜ ⎟ sin ⎜ ⎟−⎜ b ⎣⎢⎝ nπ ⎠ ⎝ b ⎠ ⎝ nπ 2
b
⎞ ⎛ nπ x ⎞ ⎤ ⎟ x cos ⎜ ⎟⎥ ⎠ ⎝ b ⎠ ⎦⎥ − b ⎛ nπ ( −b ) ⎞ ⎛ b ⎞ ⎛ nπ ( −b ) ⎞ ⎤ ⎞ ⎛ nπ b ⎞ ⎛ b ⎞ ⎟+⎜ ⎟⎥ ⎟ b cos ⎜ ⎟−⎜ ⎟ sin ⎜ ⎟ ( −b ) cos ⎜ b b ⎠ ⎦⎥ ⎠ ⎝ b ⎠ ⎝ nπ ⎠ ⎝ ⎠ ⎝ nπ ⎠ ⎝
cn =
1 ⎡⎛ b ⎞ ⎛ nπ b ⎞ ⎛ b ⎢⎜ ⎟ sin ⎜ ⎟−⎜ b ⎣⎢⎝ nπ ⎠ ⎝ b ⎠ ⎝ nπ
cn =
2 ⎤ 2 ⎡⎛ b ⎞ b2 − sin n cos ( nπ ) ⎥ π ( ) ⎢⎜ ⎟ b ⎣⎢⎝ nπ ⎠ nπ ⎦⎥
2
x cos ax sin ax + a a2
2
2b n +1 ⎛ 2b ⎞ cn = ⎜ ( −1) ⎟ cos nπ = n n π π ⎝ ⎠ Therefore, d 0 = 0 and d1 − d 5 = 0. c1 =
2b
π
, c2 = −
b
π
, c3 =
2b b 2b , c4 = − , and c5 = 3π 2π 5π
P2.28) Carry out the following coordinate transformations. a) Express the point x = 3, y = 2, and z = 1 in spherical coordinates. π 3π b) Express the point r = 5, θ = , and φ = in Cartesian coordinates. 4 4 a) r =
x 2 + y 2 + z 2 = 32 + 22 + 1 = 14 z 1 θ = cos −1 = cos −1 = 1.30 radians 2 2 2 14 x +y +z 2 y φ = tan −1 = tan −1 = 0.588 radians x 3 π 3π b) x = r sin θ cos φ = 5sin cos = −2.5 4 4 π 3π y = r sin θ sin φ = 5sin sin = 2.5 4 4 π 5 z = r cos θ = 5cos = 4 2
2-18
Chapter 2/The Schrödinger Equation
P2.29) Operators can also be expressed as matrices and wave functions as column
⎛α β ⎞ ⎛a⎞ vectors. The operator matrix ⎜ ⎟ acts on the wave function ⎜ ⎟ according to the ⎝δ ε ⎠ ⎝b⎠ ⎛ α β ⎞⎛ a ⎞ ⎛ α a + β b ⎞ rule ⎜ ⎟⎜ ⎟ = ⎜ ⎟ . In words, the 2 × 2 matrix operator acting on the two⎝ δ ε ⎠⎝ b ⎠ ⎝ δ a + ε b ⎠ element column wave function generates another two-element column wave function. If the wave function generated by the operation is the original wave function multiplied by a constant, the wave function is an eigenfunction of the operator. What is the effect of the ⎛0 1⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎛ 1⎞ ⎛ −1 ⎞ operator ⎜ ⎟ on the column vectors ⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ and ⎜ ⎟ ? Are these wave ⎝1 0⎠ ⎝ 0 ⎠ ⎝ 1 ⎠ ⎝ 1⎠ ⎝1⎠ functions eigenfunctions of the operator? See the Math Supplement for a discussion of matrices. ⎛ 0 1⎞ ⎛ 1⎞ ⎛ 0⎞ ⎜ 1 0⎟ ⎜ 0⎟ = ⎜ 1⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ 0 1⎞ ⎛ 0⎞ ⎛ 1⎞ ⎜ 1 0⎟ ⎜ 1⎟ = ⎜ 0⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ 0 1 ⎞ ⎛ 1 ⎞ ⎛ 1⎞ ⎜ 1 0 ⎟ ⎜ 1 ⎟ = ⎜ 1⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ 0 1 ⎞ ⎛ −1 ⎞ ⎛ 1 ⎞ ⎛ −1 ⎞ ⎜ 1 0 ⎟ ⎜ 1 ⎟ = ⎜ −1⎟ = ( −1) ⎜ 1 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ 1⎞ ⎛ −1 ⎞ Only ⎜ ⎟ and ⎜ ⎟ are eigenfunctions with the eigenvalues 1 and –1, respectively. ⎝ 1⎠ ⎝1⎠
P2.30) Let
FG 1IJ and FG 0IJ represent the unit vectors along the x and y directions, H 0K H 1K
⎛ cos θ − sin θ ⎞ respectively. The operator ⎜ ⎟ effects a rotation in the x-y plane. Show that ⎝ sin θ cos θ ⎠ ⎛a⎞ ⎛1⎞ ⎛0⎞ the length of an arbitrary vector ⎜ ⎟ = a ⎜ ⎟ + b ⎜ ⎟ , which is defined as a 2 + b 2 , is ⎝b⎠ ⎝0⎠ ⎝1⎠ unchanged by this rotation. See the Math Supplement for a discussion of matrices.
⎛ cos θ − sin θ ⎞ ⎛ a ⎞ ⎛ a cos θ − b sin θ ⎞ ⎜ sin θ cos θ ⎟ ⎜ b ⎟ = ⎜ a sin θ + b cos θ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ The length of the vector is given by
( a cos θ − b sin θ )
2
+ ( a sin θ + b cos θ ) = ( a 2 cos2 θ + b2 sin 2 θ − 2ab sin θ cos θ 2
+ a 2 sin 2 θ + b2 cos2 θ + 2ab sin θ cos θ )1 2 = a 2 ( cos2 θ + sin 2 θ ) + b2 ( cos2 θ + sin 2 θ ) = a 2 + b 2 This result shows that the length of the vector is not changed.
2-19
Chapter 4: Using Quantum Mechanics on Simple Systems Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q4.1) Why are standing-wave solutions for the free particle not compatible with the classical result x = x0 + v0t ? Because the particle is moving, we must represent it by a traveling wave, for which the nodes move with time. This is not possible using a standing wave, because the nodes do not move with time.
Q4.2) Why is it not possible to normalize the free-particle wave functions over the whole range of motion of the particle? This is the case because
∫
L −L
L
ψ * ( x )ψ ( x ) dx = A+ A+ ∫ e − ik x e + ik x dx = A+ A+ 2 L diverges as −L
L→∞.
Q4.3) Show that for the particle in the box total energy eigenfunctions,
dψ ( x ) 2 nπ x sin a , ψ ( x ) is a continuous function at the edges of the box. Is a dx a continuous function of x at the edges of the box?
ψ n ( x) =
2 nπ x sin = 0 at x = 0 and x = a. Because the wave function is zero a a everywhere outside the box, it has the value zero at x = 0 and x = a. Therefore, ψ ( x ) is a
ψ n ( x) =
continuous function of x because there is no abrupt change in the function at x = 0 or x = a. dψ ( x ) 2 nπ d 2 nπ x nπ x = sin cos = = 1 at x = 0 dx dx a a a a a cos nπ = 1 for n zero or even and − 1 for n odd at x = a.
4-1
Chapter 4/Using Quantum Mechanics on Simple Systems dψ ( x ) = 0 everywhere outside the box because ψ(x) = 0 outside the box. dx Because
dψ ( x ) can be one when approaching x = a from inside the box, and is always dx
equal to zero outside of the box,
dψ ( x ) is not a continuous function at x = a. dx
Q4.4) Can the particles in a one-dimensional box, a square two-dimensional box, and a cubic three-dimensional box all have degenerate energy levels? The two and three dimensional boxes can have degenerate energy levels if the lengths along the x and y or x, y, and z directions are the same or if any of the quantities 2 nx2 n y nz2 are equal. The one dimensional box levels have only one state per level. , , or a 2 b2 c2
Q4.5) We set the potential energy in the particle in the box equal to zero and justified it by saying that there is no absolute scale for potential energy. Is this also true for kinetic energy? 1 2 mv , and the velocity is measured relative to the frame of reference. The 2 =2k 2 same is true if we express the kinetic energy as Ekinetic = , because the value of 2m p = =k depends on the frame of reference if the observer is moving relative to the source.
Yes. Ekinetic =
Q4.6) Why are traveling-wave solutions for the particle in the box not compatible with the boundary conditions? The nodes in traveling waves move with time. This is incompatible with the boundary conditions for the particle in the box.
Q4.7) Why is the zero point energy lower for a He atom in a box than for an electron? h2 varies inversely with the mass, and the mass of a He 8ma 2 atom is much greater than the mass of an electron. The zero point energy E =
Q4.8) Invoke wave-particle duality to address the following question: How does a particle get through a node in a wave function to get to the other side of the box? The question poses a problem only if we try to see the “particle” as a pure particle rather than as a wave-particle. The essence of particle-wave particle duality is that some properties are more easily addressed in a wave picture, and some properties are more 4-2
Chapter 4/Using Quantum Mechanics on Simple Systems easily addressed in a particle picture. Address this question by looking at the “particle” as a wave. We can easily make a guitar string vibrate over its whole length and still have nodes by holding a finger at the fifth fret. The wave-like properties of the “particle” allow it to have a nonzero probability of being found within any interval dx throughout the box, even if the amplitude of the wave function is zero at a number of special points.
Q4.9) What is the difference between probability and probability density? Probability is expressed as ψ * (τ )ψ (τ ) dτ and is the probability density ψ * (τ )ψ (τ ) integrated over the interval d τ .
Q4.10) Explain using words, rather than equations, why if V ( x, y, z ) ≠ Vx ( x) + Vy ( y ) + Vz ( z ) ,
the total energy eigenfunctions cannot be written in the form ψ ( x, y, z ) = X ( x) Y ( y ) Z ( z ) .
If the potential has the form V ( x, y, z ) = Vx ( x ) + Vy ( y ) + Vz ( z ) , then the total energy can be written in the form E = Ex + Ey + Ez. This allows the Schrödinger equation to be separated into separate equations for x, y and z and the wave functions to be written in the form ψ ( x, y, z ) = X ( x ) Y ( y ) Z ( z ) . However, if V ( x, y , z ) ≠ Vx ( x ) + V y ( y ) + Vz ( z ) , this separation into three separate Schrödinger equations is not possible, and ψ ( x, y , z ) cannot be factored into three terms, each of which depends only on one variable.
Problems P4.1) Show by examining the position of the nodes that Re A+ ei b kx −ω t g and Re A− ei b − kx −ω t g represent plane waves moving in the positive and negative x directions, respectively. The notation Re[ ] refers to the real part of the function in the brackets. i kx −ω t ) = cos ( kx − ω t ) . The nodes of this function occur at Re A+ e ( kx − ω t = ( 2n + 1) x = ( 2n + 1)
π 2
π
+ω t 2 Because the position of the nodes is increasing as t increases, this function represents a plane wave moving in the positive x direction. i − kx −ω t ) = cos ( −kx − ω t ) = cos ( kx + ω t ) . The nodes of this function occur at Re A− e (
kx + ω t = ( 2n + 1) x = ( 2n + 1)
π 2
π
−ω t 2 Because the position of the nodes is decreasing as t increases, this function represents a plane wave moving in the negative x direction.
4-3
Chapter 4/Using Quantum Mechanics on Simple Systems =2 k 2 P4.2) Show that the energy eigenvalues for the free particle, E = , are consistent with 2m 1 the classical result E = mv 2 . 2 2 1 p E = m v2 = 2 2m h From the de Brogliie relation, p =
λ
2
E=
2
2
1 h =k , showing consistency between the classical and quantum result. = 2m λ 2m
P4.3) Are the total energy eigenfunctions for the free particle in one dimension, ψ + ( x ) = A+ e
2 mE
+i
=2
x
and ψ − ( x ) = A− e
−i
2 mE
=2
x
, eigenfunctions of the one-dimensional linear
momentum operator? If so, what are the eigenvalues? 2 mE
+i x +i d 2mE 2 2 = − (i ) = A+ e = A+ e 2 dx = Eigenfunction with eigenvalue + = k
−i =
2 mE
x −i −i d 2mE 2 2 A− e = A− e = − (i ) = 2 dx = Eigenfunction with eigenvalue −= k
−i =
2 mE =2
2 mE =2
x
x
= = k A+ e + ikx
= − = k A− e − ikx
P4.4) Is the superposition wave function for the free particle
bg
ψ + x = A+ e
+i
2 mE x =2
−i
2 mE 2
x
+ A− e = an eigenfunction of the momentum operator? Is it an eigenfunction of the total energy operator? Explain your result. +i d A+ e −i = d x +i
2 mE =2
2 mE =2
x
x
+ A− e
−i
−i
2 mE =2
2 mE 2
x
+i 2mE 2 = − (i ) = A e + =2
2 mE =2
x
+ (i )
2
−i 2mE = A e − =2
2 mE
x
=2
x
= = k A+ e − = k A− e = This function is not an eigenfunction of the momentum operator, because the operation does not return the original function multiplied by a constant. 2 mE 2 mE 2 mE 2 mE 2 2 +i −i +i −i x x x x =2 d 2 2mE 2mE 2 = 2 = =2 =2 =2 =2 − + = − − A e A e i A e i A e ( ) ( ) + − + − 2m d x 2 2m = 2 2m = 2 2 mE 2 mE +i −i x x 2 =2 = E A+ e + A− e = This function is an eigenfunction of the total energy operator. Because the energy is
4-4
Chapter 4/Using Quantum Mechanics on Simple Systems proportional to p2, the difference in sign of the momentum of these two components does not affect the energy.
P4.5) Consider a particle in a one-dimensional box defined by
V ( x ) = 0, a > x > 0 and V ( x ) = ∞, x ≥ a, x ≤ 0. Explain why each of the following
unnormalized functions is or is not an acceptable wave function based on criteria such as being consistent with the boundary conditions, and with the association of ψ * x ψ x dx with probability. nπ x D a) A cos b) B x + x 2 c) C x 3 x − a d) nπ x a sin a nπ x is not an acceptable wave function because it does not satisfy the boundary a) A cos a condition that ψ ( 0 ) = 0.
bg bg
c
h
b g
b) B ( x + x 2 ) is not an acceptable wave function because it does not satisfy the boundary
condition that ψ ( a ) = 0.
c) C x 3 ( x − a ) is an acceptable wave function. It satisfies both boundary conditions and can be normalized. D d) is notan acceptable wave function. It goes to infinity at x = 0 and cannot be nπ x sin a normalized in the desired interval.
P4.6) Evaluate the normalization integral for the eigenfunctions of H for the particle in the 1 − cos 2 y nπ x 2 . box ψ n ( x ) = A sin using the trigonometric identity sin y = 2 a a nπ x 1 = ∫ A2 sin 2 dx a 0 let y =
nπ x a dy ; dx = a nπ
a 1= A nπ 2
nπ
a ∫0 sin y dy = A nπ 2
2
nπ
nπ
1 − cos 2 y sin 2 y 2 a y ∫0 2 dy = A nπ 2 − 4 0
A2 a A2 a A2 a nπ − ( sin 2nπ − sin 0 ) = 2 nπ 2 nπ 2 2 A= a
=
4-5
Chapter 4/Using Quantum Mechanics on Simple Systems
bg
P4.7) Use the eigenfunction ψ ( x ) = A′e + ikx +B′e −ikx rather than ψ x = A sin kx + B cos kx to apply the boundary conditions for the particle in the box. a) How do the boundary conditions restrict the acceptable choices for A′ and B′ and for k? b) Do these two functions give different probability densities if each is normalized? a)
ψ ( x ) = A′e + ik x +B ′e − ik x = A′ cos k x + B ′ cos ( − k x ) + i ( A′ sin k x + B ′ sin ( − k x ) ) = A′ cos k x + B ′ cos k x + i ( A′ sin k x − B ′ sin k x )
ψ ( 0 ) = A′ + B ′ = 0, giving A′ = − B ′. Therefore
ψ ( x ) = A′ cos k x − A′ cos k x + i ( A′ sin k x + A′ sin k x ) = 2iA′ sin k x ψ ( a ) = 2iA′ sin k a = 0
This leads to the same condition on k, namely k a = nπ . The amplitude is now 2 iA′ rather than A.
b) The wave function after the imposition of the boundary conditions is ψ ( x ) = A′ ( e+ ik x − e−ik x ) = 2iA′ sin k x. After the normalization condition is imposed,
2 . Therefore functions are indistinguishable, because both will give the same a probability densities when normalized.
( 2iA′)( −2iA′) =
P4.8) Calculate the probability that a particle in a one-dimensional box of length a is found between 0.31a and 0.35a when it is described by the following wave functions: 2 π x sin a) a a 2 3π x sin a a What would you expect for a classical particle? Compare your results in the two cases with the classical result. b)
a) Using the standard integral ∫ sin 2 ( by ) dy = 0.35 a
y 1 − sin ( 2by ) 2 4b 0.35 a
x 2 a π x 2π x P= sin 2 sin dx = − ∫ a 0.31a a a 0.31a 2 4π
2 0.35a a 0.31a a − + sin ( 0.70 π ) − sin ( 0.62 π ) a 2 4π 2 4π 1 = 0.04 + sin ( 0.62 π ) − sin ( 0.70 π ) = 0.059 2π =
4-6
Chapter 4/Using Quantum Mechanics on Simple Systems b) y 1 − sin ( 2by ) 2 4b 0.35 a 2 2 0.35a a 0.31a a 3π x P= sin 2 − sin ( 2.10 π ) − + sin (1.86 π ) dx = ∫ a 0.31a a 2 12π 2 12π a 1 = 0.04 + sin (1.86 π ) − sin ( 2.10 π ) = 0.0010 6π Because a classical particle is equally likely to be in any given interval, the probability will be 0.04 independent of the energy. In the ground state, the interval chosen is near the maximum of the wave function so that the quantum mechanical probability is greater than the classical probability. For the n = 3 state, the interval chosen is near a node of the wave function so that the quantum mechanical probability is much less than the classical probability. Using the standard integral ∫ sin 2 ( by ) dy =
P4.9) Are the eigenfunctions of H for the particle in the one-dimensional box also
eigenfunctions of the position operator x ? Calculate the average value of x for the case where n = 3. Explain your result by comparing it with what you would expect for a classical particle. Repeat your calculation for n = 5 and, from these two results, suggest an expression valid for all values of n. How does your result compare with the prediction based on classical physics? No, they are not eigenfunctions because multiplying the function by x does not return the function multiplied by a constant. For n = 3, a
a
2 3π x x = ∫ ψ ( x ) xψ ( x ) d x = ∫ x sin 2 d x a0 a 0 *
Using the standard integral ∫ x sin 2 ( b x ) d x =
x 2 cos ( 2bx ) x sin ( 2bx ) − − 4 8b2 4b
a
6π x 6π x x sin 2 cos 2 x a a x = − − 2 a4 3π x 3π x 4 8 a 0 a x =
a 1 1 2 a 2 cos ( 6π ) a sin ( 6π ) cos ( 0 ) 0 sin ( 0 ) 2 a 2 − + − −0+ − 0 = − = − 2 2 2 2 a4 72π 12π a 4 72π 72π 72π 12π 2
For n = 5,
4-7
Chapter 4/Using Quantum Mechanics on Simple Systems a
a
2 5π x x = ∫ ψ ( x ) xψ ( x ) d x = ∫ x sin 2 d x a0 a 0 *
Using the standard integral ∫ x sin 2 ( b x ) d x =
x 2 cos ( 2bx ) x sin ( 2bx ) − − 4 8b2 4b
a
10π x 10π x x sin 2 cos 2 x a a x = − − 2 a4 5π x 5π x 4 8 a 0 a
2 a 2 cos (10π ) a sin (10π ) cos ( 0 ) 0 sin ( 0 ) 2 a 2 1 1 − + − −0+ − 0 − = − 2 2 2 2 200π a4 20π 200π 20π a 4 200π 200π a = 2 a The general expression valid for all states is x = . Classical physics gives the same result 2 because the particle is equally likely to be at any position. The average of all these values is the midpoint of the box. x =
P4.10) Are the eigenfunctions of H for the particle in the one-dimensional box also
eigenfunctions of the momentum operator p x ? Calculate the average value of px for the case n = 3. Repeat your calculation for n = 5 and, from these two results, suggest an expression valid for all values of n. How does your result compare with the prediction based on classical physics?
For n = 3, a
a d −2i= 3π 3π x 3π x p = ∫ ψ ( x ) −i = ψ ( x) d x = sin cos d x ∫ d x a a 0 a a 0 cos2 ( bx ) Using the standard integral ∫ sin ( b x ) cos ( b x ) d x = 2b 2 2 1 −2i= 3π cos ( 3π ) cos ( 0 ) −2i= 3π 1 p = − − =0 = a a a a 2b 2b 2b 2b For n = 5, a a d −2i= 5π 5π x 3π x p = ∫ ψ * ( x ) −i = x d x ψ sin = ( ) cos d x ∫ d x a a a a 0 0 *
cos2 ( bx ) Using the standard integral ∫ sin ( b x ) cos ( b x ) d x = 2b 2 2 1 −2i= 5π cos ( 5π ) cos ( 0 ) −2i= 5π 1 p = − − =0 = a a a a 2b 2b 2b 2b
4-8
Chapter 4/Using Quantum Mechanics on Simple Systems This is the same result that would be obtained using classical physics. The classical particle is equally likely to be moving in the positive and negative x directions. Therefore the average of a large number of measurements of the momentum is zero for the classical particle moving in a constant potential.
P4.11) It is useful to consider the result for the energy eigenvalues for the one-dimensional h2n2 , n = 1,2,3,...as a function of n, m, and a. 8ma 2 a) By what factor do you need to change the box length to decrease the zero point energy by a factor of 400 for a fixed value of m? b) By what factor would you have to change n for fixed values of a and m to increase the energy by a factor of 400? c) By what factor would you have to increase a at constant n to have the zero point energies of an electron be equal to the zero point energy of a proton in the box?
box En =
( a ′) E a) n = , = 400; En ′ a2 2
b)
En n2 = , = 400; 2 En ′ ( n ′)
a′ = 20 a n = 20 n′
c)
1 1 = 2 me ae m p a 2p ae = ap
mp me
=
1.673 × 10−27 kg = 42.9 9.109 × 10−31 kg
2 nπ x mπ x sin + sin an a a a eigenfunction of the total energy operator for the particle in the box? 2 =2 d 2 2 nπ x mπ x ˆ sin sin Hψ ( x ) = − + 2m d x 2 a a a a
P4.12) Is the superposition wave function ψ ( x ) =
h2n2 = 8ma 2
2 2 2 nπ x h m sin + 2 a a 8ma
2 mπ x sin a a
Because the result is not the wave function multiplied by a constant, the superposition wave function is not an eigenfunction of the total energy operator.
4-9
Chapter 4/Using Quantum Mechanics on Simple Systems
x P4.13) The function ψ ( x ) = Ax 1 − is an acceptable wave function for the particle in the
a one dimensional infinite depth box of length a. Calculate the normalization constant A and the expectation values x and x 2 . 2
a
a x4 x x3 1 = A ∫ x 1 − dx = A2 ∫ 2 − 2 + x 2 dx a a a 0 0 2
a
a
x5 a3 a3 a3 x4 x3 a3 + = A2 − + = A2 1= A 2 − 2a 3 0 2 3 0 30 5a 5 2
A=
30 a3 a
2
a
30 x x = ∫ ψ ( x ) xψ ( x ) dx = 3 ∫ x x 1 − dx a 0 a 0 *
a
30 x 4 2 x 5 30 a 4 2a 4 a 4 30 a 4 a x6 = 3 − + 2 = 3 − + = 3 = 5a 6a 0 a 6 5 4 a 60 2 a 4 a
x 2 = ∫ ψ * ( x ) x 2 ψ ( x ) dx = 0
2
a
30 2 x x x 1 − dx 3 ∫ a 0 a
a
30 x 5 x 6 30 a 5 a 5 a 5 30 a 5 x7 2a 2 = 3 − + = − + = 3 = 3 7 a 105 a 5 3a 7a 2 0 a 3 5 7
P4.14) Derive an equation for the probability that a particle characterized by the quantum number n is in the first quarter ( 0 ≤ x ≤ approaches the classical limit as n → ∞.
a ) of an infinite depth box. Show that this probability 4
Using the standard integral ∫ sin 2 ( b y ) d y = 2 P= a =
0.25 a
∫ 0
y 1 − sin ( 2b y ) 2 4b 0.25 a
2 x a nπ x 2nπ x sin sin dx = − a 2 4nπ a a 0 2
2 a a a 1 1 nπ 0 nπ sin sin ( 0 ) = − sin − − + a 8 4nπ 2 2 4nπ 2 4 2nπ
1 . 4 This is the classical value, because the particle is equally likely to be found anywhere in the box. As n → ∞, the second term goes to zero, and the probability approaches
4-10
Chapter 4/Using Quantum Mechanics on Simple Systems
P4.15) What is the solution of the time-dependent Schrödinger equation Ψ ( x, t ) for the total
FG H
IJ K
2 4π x E sin in the particle in the box model? Write ω = a a = explicitly in terms of the parameters of the problem.
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energy eigenfunction ψ 4 x =
ψ ( x, t ) = ψ ( x ) e − iω t = ψ ( x ) e Because E =
ψ ( x, t ) =
−i
Et =
n2 h2 16 h 2 = , 8 m a2 8 m a2 4π h t
2 4π x − i m a 2 sin e a a
P4.16) For a particle in a two-dimensional box, the total energy eigenfunctions are
n πy n xπ x sin y . a b a) Obtain an expression for Enx ,ny in terms of nx, ny, a, and b by substituting this wave function
ψ n n ( x, y ) = N sin x y
into the two-dimensional analog of Equation (4.19). b) Contour plots of several eigenfunctions are shown here. The x and y directions of the box lie along the horizontal and vertical directions, respectively. The amplitude has been displayed as a gradation in colors. Regions of positive and negative amplitude are indicated. Identify the values of the quantum numbers nx and ny for plots a–f.
b
a
+
d
e -
+
c +
-
-
+
+
-
+
f
+ -
- - -
+ +
-
+
-
+
+
4-11
-
Chapter 4/Using Quantum Mechanics on Simple Systems a) −
2 =2 ∂ 2 4 n π x n yπ y ∂ 4 n xπ x n y π y + sin sin 2 2 sin x a sin a 2 2 2m ∂ x a a a ∂ y a 2
2 2 = 2 n xπ x 4 n xπ x n y π y = n y π x 4 n xπ x n y π y = + sin sin 2 2 sin sin 2m a a a a 2m a a a a
h 2 ( n x2 + n 2y ) 4 n xπ x n y π y = 2 sin sin 2 8m a a a a Because application of the total energy operator returns the wave function multiplied by a n πy n πx constant, ψ nx n y ( x, y ) = N sin x sin y is an eigenfunction of the total energy operator. a b h 2 ( n x2 + n 2y ) b) From the result of part (a), Enx ,n y = 8 m a2 a: nx = 1 ny = 1 b: nx = 2 ny = 3 c: nx = 3 ny = 1 d: nx = 2 ny = 2 e: nx = 1 ny = 5 f: nx = 2 ny = 1
P4.17) Normalize the total energy eigenfunction for the rectangular two-dimensional box, nxπ x n yπ y sin a b in the interval 0 ≤ x ≤ a, 0 ≤ y ≤ b.
ψ n ,n ( x, y ) = N sin x
y
b n π y n xπ x 2 y 1 = ∫ ∫ ψ ( x, y )ψ ( x, y ) d x d y = N ∫ sin d x sin d y ∫ a b 0 0 0 0 x sin ( 2α x ) Using the standard integral ∫ sin 2α x d x = − 2 4α a b n yπ y n π x N 2 ∫ sin 2 x d x ∫ sin 2 d y a b 0 0 a b
a
*
2
2
a b a a ab = N2 − sin n yπ − sin 0 ) = N 2 ( sin nxπ − sin 0 ) × − ( 4 2 4n xπ 2 4n yπ N=
4 and ψ ( x, y ) = ab
4 n π x n yπ y sin x sin ab a b
P4.18) Consider the contour plots of Problem P4.16. a) What are the most likely area or areas dxdy to find the particle for each of the eigenfunctions of H depicted in plots a–f? 4-12
Chapter 4/Using Quantum Mechanics on Simple Systems b) For the one-dimensional box, the nodes are points. What form do the nodes take for the two-dimensional box? Where are the nodes located in plots a–f? How many nodes are there in each contour plot? a) What is the most likely point to find the particle for each of the eigenfunctions of Hˆ depicted in plots a–e? The most likely location of the particle is the set of points at which the wave function has its largest positive or negative value as indicated by the gray scale. There are 1, 6, 3, 4, 5, and 2 equivalent points in plots a–e respectively. b) For the one-dimensional box, the nodes are points. What form do the nodes take for the two-dimensional box? Where are the nodes located in plots a–e? How many nodes are there in each contour plot? The nodes are lines that lie between values at which the absolute magnitude of the wave function reaches its maximum value. There are 0, 3, 2, 2, 4, and 1 nodes in plots a–e, respectively.
P4.19) Show by substitution into Equation (4.19) that the eigenfunctions of H for a box with lengths along the x, y, and z directions of a, b, and c, respectively, are n πx n πy nπz ψ nx ,ny ,nz ( x, y, z ) = N sin x sin y sin z a b c b) Obtain an expression for Enx ,ny ,nz in terms of nx, ny, nz and a, b, and c. a) −
=2 ∂ 2 2m ∂ x 2
2 2 n xπ x n y π y n z π z = ∂ n xπ x n y π y n zπ N sin sin sin − a b c 2m ∂ y 2 N sin a sin b sin c =2 ∂ 2 n π x n yπ y nzπ z − N sin x sin sin 2 2m ∂ z a b c 2
z
2 2 = 2 n xπ x n xπ x n y π y n z π z = n y π x n xπ x n y π y n zπ z + sin sin sin = N N sin sin sin 2m a a b c 2m b a b c 2 = 2 nzπ x n xπ x n y π y n z π z + sin N a sin b sin c 2m c
h2 = 8m
nx2 n 2y nz2 n xπ x n y π y n z π z 2 + 2 + 2 N sin sin sin b c a b c a
4-13
Chapter 4/Using Quantum Mechanics on Simple Systems Because application of the total energy operator returns the wave function multiplied by a n π x n yπ y nzπ z constant, ψ nx ,n y ,nz ( x, y , z ) = N sin x sin is an eigenfunction of the sin a b c total energy operator. 2 h 2 nx2 n y nz2 + + b) Based on the result from part (a), Enx ,n y ,nz = 8 m a 2 b2 c 2
P4.20) Normalize the total energy eigenfunctions for the three-dimensional box in the interval 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c. a b c
1 = ∫ ∫ ∫ ψ * ( x, y , z )ψ ( x, y , z ) d x d y d z 0 0 0
a b c n π y n π x nπ z 2 y = N 2 ∫ sin 2 x d x sin d y sin 2 z d z ∫ ∫ b a b 0 0 0 x sin ( 2α x ) Using the standard integral ∫ sin 2α x d x = − 2 4α a b c n yπ y n π x 2 n zπ z 1 = N 2 ∫ sin 2 x d x ∫ sin 2 d y ∫ sin d z b a b 0 0 0
c a b a a a abc 1= N2 − sin n yπ − sin 0 ) × − ( sin nxπ − sin 0 ) × − ( sin nzπ − sin 0 ) = N 2 ( 8 2 4 n xπ 2 4n yπ 2 4rπ N=
8 and ψ ( x, y ) = abc
8 n π x n y π y n zπ sin x sin sin abc a b c
z
P4.21) In discussing the Boltzmann distribution in Chapter 2, we used the symbols gi and gj to indicate the degeneracies of the energy levels i and j. By degeneracy, we mean the number of distinct quantum states (different quantum numbers) all of which have the same energy. 5h 2 a) Using your answer to Problem P4.16a, what is the degeneracy of the energy level for 8ma 2 the square two-dimensional box of edge length a? 9h 2 for b) Using your answer to Problem P4.19b, what is the degeneracy of the energy level 8 ma 2 a three-dimensional cubic box of edge length a? a) What is the degeneracy of the energy level
5h 2 for the square two-dimensional box? 8m a 2
The only pairs nx, ny that satisfy the equation n x2 + n 2y = 5 are 2, 1 and 1, 2. Therefore the degeneracy of this energy level is 2. b) What is the degeneracy of the energy level
9h 2 for a three-dimensional cubic box of edge 8 m a2
length a? 4-14
Chapter 4/Using Quantum Mechanics on Simple Systems
The only trios of nonzero numbers n, q, r that satisfy the equation n x2 + n 2y + nz2 = 9 are 2, 2, 1; 2, 1, 2; and 1, 2, 2. Therefore the degeneracy of this energy level is 3.
P4.22) This problem explores under what conditions the classical limit is reached for a macroscopic cubic box of edge length a. An argon atom of average translational energy 3/2 kT is confined in a cubic box of volume V = 0.500 m3 at 298 K. Use the result from Equation (4.25) for the dependence of the energy levels on a and on the quantum numbers nx, ny, and nz. a) What is the value of the “reduced quantum number” α = nx2 + n 2y + nz2 for T = 298 K? b) What is the energy separation between the levels a and a + 1? (Hint: Subtract Ea+1 from Ea before plugging in numbers. E − Eα c) Calculate the ratio α +1 and use your result to conclude whether a classical or kT quantum mechanical description is appropriate for the particle. h2 Enx ,ny , nz = n 2 + ny2 + nz2 ) 2 ( x 8ma
(n
2 x
α=
2 y
+n +n
(n
2 x
2 z
)= 2 y
8 m a 2 Enx ,ny , nz
+n +n
h2 2 z
)=
8 m a 2 Enx , ny ,nz h2 2
α=
8×39.95amu ×1.661x10−27 kg(amu)−1 × ( 0.500 m3 ) 3 ×1.5 × 1.381×10−23 J K −1 × 298 K
( 6.626 ×10
−34
J s)
2
α = 6.86 ×1010 b) What is the energy separation between the levels α and α + 1? (Hint: Subtract Eα+1 from Eα before plugging in numbers. h 2 ( 2 α + 1) h2 2 2 Eα +1 − Eα = α + 1 − α = [ ] 8 m a2 8 m a2
(
=
( 6.626 ×10
)
−34
Js ) ( 2 × 6.86 ×1010 +1) 2
8 × 39.95amu × 1.661×10−27 kg(amu) −1 × ( 0.500 m3 )
2
3
= 1.80 ×10−31J
Eα +1 − Eα and use your result to conclude whether a classical or quantum kT mechanical description is appropriate for the particle.
c) Calculate the ratio
Eα +1 − Eα 1.80 ×10−31J = = 4.44 ×10−7 1.361×10−23 J K −1 × 298 K kT 4-15
Chapter 4/Using Quantum Mechanics on Simple Systems Because ∆E<
P4.23) Generally, the quantization of translational motion is not significant for atoms because of their mass. However, this conclusion depends on the dimensions of the space to which they are confined. Zeolites are structures with small pores that we describe by a cube with edge length 1 nm. Calculate the energy of a H2 molecule with nx = ny = nz = 10. Compare this energy to kT at T = 300 K. Is a classical or a quantum description appropriate? h2 Enx ,ny , nz = n 2 + n y2 + nz2 ) 2 ( x 8ma =
( 6.626 ×10
−34
J s ) (102 + 102 + 102 ) 2
8×2.016 amu ×1.661x10 kg(amu) × (10 m ) −27
−1
−9
2
= 4.92 × 10−21 J
Using the results of P4.22, the ratio of the energy spacing between levels and kT determines if a classical or quantum description is appropriate. h 2 ( 2 α + 1) h2 2 2 Eα +1 − Eα = α + 1 − α = where α = ( nx2 + n y2 + nz2 ) [ ] 2 2 8ma 8ma
(
=
( 6.626 ×10
)
−34
Js )
2
(
)
300+1
8 × 2.016 amu ×1.661×10 kg(amu) × (10 m ) −27
−1
−9
2
= 3.00 ×10−22 J
Eα +1 − Eα 3.00 ×10−22 J = = 0.073 kT 1.361×10−23 J K −1 × 298 K Because this ratio is not much smaller than one, a quantum description is appropriate.
P4.24) Two wave functions are distinguishable if they lead to a different probability density. Which of the following wave functions are distinguishable from sin θ + i cos θ ? 2 2 b) i ( sin θ + i cos θ ) +i a) ( sin θ + i cos θ ) 2 2 2 2 +i c) − ( sin θ + i cos θ ) d) ( sin θ + i cos θ ) − 2 2 e) ( sin θ − i cos θ ) f) eiθ
Two wave functions ψ 1 and ψ 2 are indistinguishable if ψ 1*ψ 1 = ψ *2ψ 2 . For the wave function in
the problem, ( sin θ − i cos θ )( sin θ + i cos θ ) = ( sin 2 θ + cos 2 θ ) = 1 .
2 2 2 2 2 2 −i +i a) ( sin θ − i cos θ ) ( sin θ + i cos θ ) = ( sin θ + cos θ ) = 1 2 2 2 2 indistinguishable.
4-16
Chapter 4/Using Quantum Mechanics on Simple Systems b) −i ( sin θ − i cos θ ) i ( sin θ + i cos θ ) = ( sin 2 θ + cos2 θ ) = 1 indistinguishable
c) − ( sin θ − i cos θ ) − ( sin θ + i cos θ ) = ( sin 2 θ + cos2 θ ) = 1 indistinguishable 2 2 2 2 2 2 d) ( sin θ − i cos θ ) − −i +i ( sin θ + i cos θ ) − = ( sin θ + cos θ ) = 1 2 2 2 2 indistinguishable e) ( sin θ + i cos θ )( sin θ − i cos θ ) = ( sin 2 θ + cos2 θ ) = 1 indistinguishable f) e − iθ eiθ = 1 indistinguishable
P4.25) Suppose that the wave function for a system can be written as
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1 1 3+ 2 i φ 3 x and that φ 1 x , φ 2 x and φ 3 x are normalized 2 4 4 eigenfunctions of the operator E kinetic with eigenvalues E1, 3E1, and 7E1, respectively. a) Verify that ψ ( x ) is normalized.
ψ x = φ1 x + φ 2 x +
b) What are the possible values that you could obtain in measuring the kinetic energy on identically prepared systems? c) What is the probability of measuring each of these eigenvalues? d) What is the average value of Ekinetic that you would obtain from a large number of measurements? a) We first determine if the wave function is normalized. 3− 2 i 3+ 2 i * 1 * 1 * * ∫ψ ( x )ψ ( x ) dx = 4 ∫ φ1 ( x ) φ1 ( x ) dx + 16 ∫ φ 2 ( x ) φ2 ( x ) dx + 4 4 ∫ φ 3 ( x ) φ3 ( x ) dx 1 * 3+ 2i * 3− 2i * 1 φ1 ( x ) φ2 ( x ) dx + φ1 ( x ) φ3 ( x ) dx + φ 3 ( x ) φ1 ( x ) dx φ1 ( x ) ∫ ∫ ∫ 4 16 16 2 3+ 2i * 3− 2i * φ 2 ( x ) φ3 ( x ) dx + + ∫ ∫ φ 3 ( x ) φ2 ( x ) dx 8 8 +
All but the first three integrals are zero because the functions φ1 ( x ) , φ2 ( x ) , and φ3 ( x ) are orthogonal. The first three integrals have the value one, because the functions are normalized. Therefore,
1 1 3− 2i 3+ 2i 1 1 11 ∫ψ ( x )ψ ( x ) dx = 4 + 16 + 4 4 = 4 + 16 + 16 = 1 *
4-17
Chapter 4/Using Quantum Mechanics on Simple Systems b) The only possible values of the observable kinetic energy that you will measure are those corresponding to the finite number of terms in the superposition wave function. In this case, the only values that you will measure are E1, 3E1, and 7E1. c) For a normalized superposition wave function, the probability of observing a particular eigenvalue is equal to the square of the magnitude of the coefficient of that kinetic energy eigenfunction in the superposition wave function. These coefficients have been calculated above. The probabilities of observing E1, 3E1, and 7E1 are ¼, 1/16, and 11/16, respectively. d) The average value of the kinetic energy is given by 1 1 11 E = ∑ Pi Ei = E1 + 3E1 + 7 E1 = 5.25E1 4 16 16
4-18
Chapter 7: A Quantum Mechanical Model for the Vibration and Rotation of Molecules Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q7.1) What is the functional dependence of the total energy of the quantum harmonic oscillator on the position variable x? It is independent of x. Both the kinetic and potential energy depend on x, but the total energy is independent of x.
Q7.2) The two linearly independent total energy eigenfunctions for rotation in two
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1 imlφ 1 − imlφ e and Φ − φ = e . What is the difference in 2π 2π motion for these two solutions? Explain your answer. dimensions are Φ + φ =
From Section 7.4, the angular momentum operator for rotation in two dimensions ∂ is lˆ = −ih if we apply this operator to the two eigenfunctions ∂φ 1 imlφ 1 − imlφ Φ+ φ = e and Φ − φ = e , 2π 2π mh ∂ Φ + (φ ) = l eimlφ = ml hΦ + (φ ) and lˆΦ + (φ ) = −ih ∂φ 2π −ml h − imlφ ∂ Φ − (φ ) = lˆΦ − (φ ) = −ih e = − ml hΦ − (φ ) ∂φ 2π Therefore, the direction of rotation is opposite for these two eigenfunctions.
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Q7.3) Spatial quantization was discussed in Supplemental Section 7.8. Suppose that we have a gas consisting of atoms, each of which has a nonzero angular momentum. Are all of their angular momentum vectors aligned? No. Only the application of an eternal field that makes the energy depend on the component of the angular momentum will lead to spatial quantization. Even in this case, not all molecules will have the same component of angular momentum along the field direction. The number of molecules with a particular value for the component of angular momentum along the field direction will be determined by the Boltzmann distribution.
7-1
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
Q7.4) Why can the angular momentum vector lie on the z axis for two-dimensional rotation in the x-y plane but not for rotation in three-dimensional space? For rotation in two dimensions, the angular momentum vector has only one component, and the vector is perpendicular to the plane of rotation. For rotation in three dimensions, the angular momentum vector has three components. Because the corresponding angular momentum operators do not commute, all three components of the angular momentum can’t be known simultaneously. Therefore, the angular momentum vector can’t lie on the z axis, because this requires that the x and y components are known to be zero, and the z component is known through the value of L2.
Q7.5) The zero point energy of the particle in the box goes to zero as the length of the box approaches infinity. What is the appropriate analogue for the quantum harmonic oscillator? The force constant k goes to zero. This corresponds in a diatomic molecule to a very weak bond, such as for Ar2.
Q7.6) Explain in words why the amplitude of the total energy eigenfunctions for the quantum mechanical harmonic oscillator increases with |x| as shown in Figure 7.4. The velocity of the particle goes through zero at the classical turning point. Therefore, it spends more time near the turning point than in the same interval near x = 0. Therefore, the probability density must be highest at the classical turning point.
Q7.7) Why does the energy of a rotating molecule depend on l, but not on ml? The energy depends only on the frequency of rotation, and not on the orientation of the rotation axis. The quantum number l determines the speed, and ml determines the orientation of the rotational axis.
Q7.8) Are the real functions listed in Equations (7.34) and (7.35) eigenfunctions of lz? Justify your answer. No. They are formed by the combination of Yl ml and Yl − ml . Therefore, the functions are eigenfunctions of Hˆ and lˆ 2 , but not of lˆ . z
Q7.9) Why is only one quantum number needed to characterize the eigenfunctions for rotation in two dimensions, whereas two quantum numbers are needed to characterize the eigenfunctions for rotation in three dimensions? For rotation in two dimensions, the angular momentum vector is perpendicular to the plane of rotation, and therefore has only one component. Only one quantum 7-2
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules number is needed to describe this vector. By contrast, the angular momentum vector for 3-D rotation has three components. However, the commutation relations say that we can only know the magnitude of the vector and one of its components simultaneously. Two quantum numbers are needed to describe this vector.
Q7.10) What makes the z direction special so that we say that lˆ2 , H$ , and lˆz commute, whereas the individual angular momentum operators lˆz , lˆy and lˆx don’t commute? There is nothing special about the z direction. We choose the z direction because the angular momentum operator for this component has a simple form. We can rotate our coordinate system to make any direction that we choose to lie along the z axis. The important point is that we can know the magnitude of the angular momentum and only one of its components.
Problems P7.1) The force constant for a H19F molecule is 966 N m–1. a) Calculate the zero point vibrational energy for this molecule for a harmonic potential. b) Calculate the light frequency needed to excite this molecule from the ground state to the first excited state. a)
966 N m −1 k ⎛ 1⎞ 3 −34 × × × 1+ = 1.055 10 J s 1.0078 × 18.9984 µ ⎜⎝ 2 ⎟⎠ 2 ×1.66 x10−27 kg amu −1 1.0078+18.9984 E1 = 1.23 ×10−19 J E1 = h
E0 = h
k ⎛1⎞ 1 = E1 = 4.10 ×10−20 J µ ⎜⎝ 2 ⎟⎠ 3
E1 − E0 1.23 ×10−19 J − 4.10 ×10−20 J = 1.24 ×1014s −1 = b) ν = −34 h 6.626 ×10 J s
P7.2) By substituting in the Schrödinger equation for the harmonic oscillator, show that the ground-state vibrational wave function is an eigenfunction of the total energy operator. Determine the energy eigenvalue.
7-3
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules 2 h2 d ψ n ( x ) k x 2 ψ n ( x ) = Enψ n ( x ) − + 2µ d x 2 2 14 1 ⎧⎪ ⎛ α ⎞1 4 − 1α x2 ⎫⎪ ⎛ α ⎞ − 2α x 2 d ⎨α x ⎜ ⎟ e 2 ⎬ e 14 14 1 2 1 2 ⎟ 2 d ⎜ 2 2 2 − α x k x ⎛α ⎞ h h ⎪⎩ ⎝ π ⎠ ⎭⎪ + k x ⎛ α ⎞ e − 2α x ⎝π ⎠ 2 e − + = ⎜ ⎟ ⎜ ⎟ d x2 dx 2µ 2 ⎝π ⎠ 2µ 2 ⎝π ⎠ 2
h2 2µ
=
h2 = 2µ
14 1 ⎧⎪ ⎛ α ⎞1 4 − 1α x2 − α x2 ⎫ ⎪ k x2 2 2 ⎛α ⎞ 2 2 e x e α α − ⎨ ⎜ ⎟ ⎬+ ⎜ ⎟ 2 ⎝π ⎠ ⎪⎩ ⎝ π ⎠ ⎪⎭
⎛ α ⎞ − 2α x 2 ⎜ ⎟ e ⎝π ⎠ 14
1
14 14 14 1 1 1 ⎧⎪ ⎛ α ⎞1 4 − 1α x2 − α x2 ⎫ ⎪ h 2α 2 x 2 ⎛ α ⎞ − 2α x 2 h 2 ⎛ α ⎞ − 2α x2 2 2 ⎛α ⎞ 2 2 α⎜ ⎟ e −α x ⎜ ⎟ e = ⎨α ⎜ ⎟ e ⎬+ ⎜ ⎟ e 2µ ⎝ π ⎠ 2µ ⎝ π ⎠ ⎝π ⎠ ⎩⎪ ⎝ π ⎠ ⎭⎪
h2 = 2µ
kµ h2
⎛ α ⎞ − 2α x 2 h = ⎜ ⎟ e 2 ⎝π ⎠ 14
k ⎛α ⎞ µ ⎜⎝ π ⎟⎠
14
1
e
1 − α x2 2
= E1ψ 1 ( x ) with E1 =
h 2
k
µ
P7.3) Show by carrying out the appropriate integration that the total energy ⎛ α ⎞ − α x2 eigenfunctions for the harmonic oscillator ψ 0 ( x ) = ⎜ ⎟ e 2 and ⎝π ⎠ 14
1
b g FGH 4απ IJK c2α x − 1he are orthogonal over the interval −∞ < x < ∞ and that ψ b x g is normalized over the same interval. In evaluating integrals of this type, z f bxg d x = 0 if f(x) is an odd function of x and z f bxg d x = 2z f bxg d x if f(x) is an 14
ψ2 x =
1 − α x2 2
2
2
∞
∞
∞
−∞
−∞
0
even function of x.
We use the standard integrals ∞
∫
e
− ax 2
0
⎛π ⎞ dx = ⎜ ⎟ ⎝ 4a ⎠
1
∫
∞
0
14
∞
* 2
14
⎛ α2 ⎞ =⎜ 2⎟ ⎝ 4π ⎠
∫ ( 2α x ∞
−∞
⎛ α2 ⎞ ⎛ 1 = ⎜ 2 ⎟ ⎜ 2α 4α ⎝ 4π ⎠ ⎝ 14
2
1 • 3 • 5 • • • (2n − 1) π and 2n +1 a n a
2
⎛α ⎞ ∫−∞ψ ( x )ψ 0 ( x ) d x = ∫−∞ ⎜⎝ 4π ⎟⎠ ∞
x 2n e−a x d x =
2
− 1) e
( 2α x
2
− 1) e
1 − α x2 2
14
−α x 2
⎛ α2 ⎞ 2 = dx ⎜ 2⎟ ⎝ 4π ⎠
π 1 π⎞ − ⎟=0 α 2 α⎠
7-4
⎛α ⎞ ⎜ ⎟ ⎝π ⎠
∫ ( 2α x ∞
0
14
e 2
1 − α x2 2
dx
− 1) e −α x d x 2
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules ⎛α ⎞ ∫−∞ψ ( x )ψ 2 ( x ) d x = ∫−∞ ⎜⎝ 4π ⎟⎠ ∞
14
∞
* 2
⎛α ⎞ = 2⎜ ⎟ ⎝ 4π ⎠
12
∫ ( 4α ∞
2
0
( 2α x
2
− 1) e
⎛α ⎞ ⎜ ⎟ ⎝ 4π ⎠
14
1 − α x2 2
( 2α x
2
− 1) e
1 − α x2 2
dx
x 4 − 4α x 2 + 1) e −α x d x 2
12 ⎛α ⎞ ⎛ 2 3 = 2⎜ ⎟ ⎜ 4α 3 2 2α ⎝ 4π ⎠ ⎝
π 1 π 1 π ⎞ ⎛α ⎞ − 4α 2 + ⎟ = 2⎜ ⎟ α 2α α 2 α⎠ ⎝ 4π ⎠
12
π α
⎛3 ⎜ −1+ ⎝2
1⎞ ⎟ =1 2⎠
P7.4) Evaluate the average kinetic and potential energies, E kinetic and E potential , for the ground state (n = 0) of the harmonic oscillator by carrying out the appropriate integrations. ∞ 2 1 • 3 • 5 • • • (2n − 1) π We use the standard integrals ∫ x 2 n e − a x d x = and 0 2n +1 a n a ∞
∫ 0
e
− ax 2
⎛π ⎞ dx = ⎜ ⎟ ⎝ 4a ⎠
1
2
⎛1 ⎞ E potential = ∫ψ 0* ( x ) ⎜ k x 2 ⎟ψ 0 ( x ) dx ⎝2 ⎠ 12 ∞
1 ⎛α ⎞ = k⎜ ⎟ 2 ⎝π ⎠
⎛α ⎞ =k ⎜ ⎟ ⎝π ⎠
12
∫
2 −α x 2
xe
−∞
12∞
⎛α ⎞ dx = k ⎜ ⎟ ⎝π ⎠
2 −α x ∫ x e dx 2
0
1 π 1 h k =k = 4α α 4α 4 µ
⎛ h2 ∂ 2 ⎞ Ekinetic = ∫ψ 0* ( x ) ⎜ − ψ x dx 2 ⎟ 0( ) ⎝ 2µ ∂ x ⎠ ∞
1 1 2 2 ⎛ α ⎞ − 2α x 2 ⎛ h ∂ ⎞ ⎛ α ⎞ − 2α x 2 = ∫⎜ ⎟ e dx ⎜− ⎟ e 2 ⎟⎜ π⎠ ⎝ 2µ ∂ x ⎠ ⎝ π ⎠ −∞ ⎝ 14
14
12∞
=−
h2 ⎛ α ⎞ µ ⎜⎝ π ⎟⎠
−α x 2 ∫ e (α x − α ) dx 2
0
h2 ⎛ α ⎞ =− ⎜ ⎟ µ ⎝π ⎠
12
h2 = 4µ
⎛ α π α π ⎞ h2 α − ⎜⎜ ⎟⎟ = ⎝4 α 2 α⎠ µ 4
kµ h k = h2 4 µ
7-5
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
P7.5) Evaluate the average kinetic and potential energies, E kinetic and E potential , for the second excited state (n = 2) of the harmonic oscillator by carrying out the appropriate integrations. ∞ 2 1 • 3 • 5 • • • (2n − 1) π We use the standard integrals ∫ x 2 n e − a x d x = and 0 2n +1 a n a ∞
∫
e
0
− ax 2
⎛π ⎞ dx = ⎜ ⎟ ⎝ 4a ⎠
1
2
E potential = ∫ψ 2* ( x ) V ( x )ψ 2 ( x ) dx ∞
⎛α ⎞ = ∫⎜ ⎟ 4π ⎠ −∞ ⎝
14
( 2α x
− 1) e
2
1 − α x2 2
12 ∞
=
1 ⎛α ⎞ k⎜ ⎟ 2 ⎝ 4π ⎠
∫ x ( 2α x 2
14
( 2α x
2
− 1) e
1 − α x2 2
dx
− 1) e −α x dx 2
2
−∞
12∞
⎛α ⎞ = k⎜ ⎟ ⎝ 4π ⎠
2
1 2⎛ α ⎞ kx ⎜ ⎟ 2 ⎝ 4π ⎠
∫ ( 4α
2
x 6 − 4α x 4 + x 2 ) e −α x dx 2
0
3 1 π ⎞ π ⎛ α ⎞ ⎛ 2 15 π = k⎜ − 4α 3 2 + 2 ⎟ ⎟ ⎜⎜ 4α 4 3 2α α 2 α α 2 α α ⎟⎠ ⎝ 4π ⎠ ⎝ 12
=
5 k 5h k = 4α 4 µ
⎛ h2 ∂ 2 ⎞ Ekinetic = ∫ψ 2* ( x ) ⎜ − ψ x dx 2 ⎟ 2( ) ⎝ 2µ ∂ x ⎠ ∞
⎛α ⎞ = ∫⎜ ⎟ 4π ⎠ −∞ ⎝
14
∞
=
⎛α ⎞ ∫−∞ ⎜⎝ 4π ⎟⎠
14
− 1) e
1 − α x2 2
⎛ h2 ∂ 2 ⎞ ⎛ α ⎞ ⎜− ⎟ 2 ⎟⎜ ⎝ 2 µ ∂ x ⎠ ⎝ 4π ⎠
( 2α x2 − 1) e
1 − α x2 2
1 − α x2 ⎛ h2 ⎞ ⎛ α ⎞ 2 4 2 − − + α α α x x e 2 11 5 ( ) 2 dx ⎜ ⎟⎜ ⎟ µ π 2 4 ⎝ ⎠ ⎝ ⎠
( 2α x
2
12∞
⎛ h2 ⎞ ⎛ α ⎞ = α ⎜ − ⎟⎜ ⎟ ⎝ µ ⎠ ⎝ 4π ⎠
∫ ( 2α x
⎛ h =α ⎜− ⎝ µ
2
( 2α x
− 1) e
2
1 − α x2 2
dx
14
− 1)( 2α 2 x 4 − 11α x 2 + 5 ) e −α x dx 2
0
12∞
⎛ h2 ⎞ ⎛ α ⎞ = α ⎜ − ⎟⎜ ⎟ ⎝ µ ⎠ ⎝ 4π ⎠
2
14
∫ ( 4α
3
x 6 − 24α 2 x 4 + 21α x 2 − 5 ) e −α x dx 2
0
⎞ ⎛ α ⎞ ⎛ 3 15 π 3 π 21 π 1 π ⎞ − 24α 2 3 2 +α 2 −5 ⎟ ⎟⎜ ⎟ ⎜⎜ 4α 4 3 2α α 2α α 2α α 2 α ⎠⎟ ⎠ ⎝ 4π ⎠ ⎝ 12
5 h 2α 5 k = = h 4 µ 4 µ
7-6
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
P7.6) Evaluate the average vibrational amplitude of the quantum harmonic oscillator about its equilibrium value, x , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3. x = ∫ ψ n* ( x ) xψ n d x = ∫ x ( ψ n ) d x . Because ( ψ n ) is an even function of x for ∞
∞
−∞
2
2
−∞
all n, the product x ( ψ n ) is an odd function of x. Therefore, x = 0 for all n. 2
P7.7) Evaluate the average linear momentum of the quantum harmonic oscillator, px , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3. ∞ 2 1 • 3 • 5 • • • (2n − 1) π We use the standard integrals ∫ x 2 n e − a x d x = and 0 2n +1 a n a ∞
∫ 0
e
− ax 2
⎛π ⎞ dx = ⎜ ⎟ ⎝ 4a ⎠
1
2
∞ ⎛ d ⎞ px = ∫ ψ n* ( x ) ⎜ −ih ⎟ψ n d x −∞ d x⎠ ⎝ 1 d ⎞ ⎛ α ⎞ − 12α x2 ⎛ α ⎞ − 2α x 2 ⎛ dx for n = 0, px = ∫ ⎜ ⎟ e ⎜ −i h ⎟⎜ ⎟ e −∞ π d x ⎠⎝ π ⎠ ⎝ ⎠ ⎝ 14
∞
14
∞ 2 ⎛α ⎞ px = ⎜ ⎟ ( −ih ) ∫ −α x e −α x d x −∞ ⎝π ⎠ Because the integrand is an odd function of x, px = 0 for n = 0. 12
14
⎛ 4α 3 ⎞ for n = 1, px = ∫ ⎜ ⎟ −∞ ⎝ π ⎠ ∞
xe
1 − α x2 2
14
⎛ d ⎞ ⎛ 4α 3 ⎞ h − i ⎟ ⎜ ⎟⎜ d x ⎠⎝ π ⎠ ⎝
xe
1 − α x2 2
dx
12
∞ ⎛ 4α 3 ⎞ 2 −α x 2 px = ⎜ dx ⎟ ( −ih ) ∫−∞ x (1 − α x ) e ⎝ π ⎠ Because the integrand is an odd function of x, px = 0 for n = 1.
⎛α ⎞ for n = 2, px = ∫ ⎜ −∞ 4π ⎟ ⎝ ⎠
14
∞
⎛α ⎞ px = ⎜ ⎟ ⎝ 4π ⎠
12
( 2α x
2
− 1) e
( −ih ) ∫−∞ ( 2α x 2 − 1) e−α x ∞
2
1 − α x2 2
( −2α
⎛ d ⎞⎛ α ⎞ ⎜ −i h ⎟⎜ ⎟ d x ⎠ ⎝ 4π ⎠ ⎝
14
( 2α x
x + 4α x + α x ) d x
2 3
∞ ⎛α ⎞ −α x 2 −4α 3 x 5 + 12α 2 x3 − 5α x ) d x px = ⎜ ( ⎟ ( −ih ) ∫−∞ e π 4 ⎝ ⎠ Because the integrand is an odd function of x, px = 0 for n = 3. 12
The result is general. px = 0 for all values of n.
7-7
2
− 1) e
1 − α x2 2
dx
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
P7.8) Evaluate the average of the square of the vibrational amplitude of the quantum harmonic oscillator about its equilibrium value, x 2 , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3. ∞ 2 1 • 3 • 5 • • • (2n − 1) π We use the standard integrals ∫ x 2 n e − a x d x = and 0 2n +1 a n a ∞
∫
e
0
− ax 2
⎛π ⎞ dx = ⎜ ⎟ ⎝ 4a ⎠
1
2
x 2 = ∫ ψ n* ( x ) ( x 2 )ψ n d x ∞
−∞
for n = 0, x
⎛ α ⎞ − α x2 ⎛ α ⎞ − α x2 = ∫ ⎜ ⎟ e 2 ( x2 ) ⎜ ⎟ e 2 d x −∞ π ⎝ ⎠ ⎝π ⎠
⎛α ⎞ =⎜ ⎟ ⎝π ⎠
x
x2
12
∞
x
2
−α x 2
2
14
⎛ 4α 3 ⎞ =∫ ⎜ ⎟ −∞ ⎝ π ⎠ ∞
2
12
x
⎛ 4α 3 ⎞ =⎜ ⎟ ⎝ π ⎠
∫
∞
−∞
12
⎛ 4α 3 ⎞ = 2⎜ ⎟ ⎝ π ⎠
xe
x e
−α x 2
12
x
x2
14
⎛ 4α 3 ⎞ x ( )⎜ π ⎟ ⎝ ⎠ 2
2α
∫ ( 4α ∞
2
−∞
FG IJ FG 4α H K H 12
2
2
∫
∞
0
xe
1 − α x2 2
dx
x 4 e −α x d x 2
π 3 3h = = α 2α 2 k µ
3 3
⎛ 4α 3 ⎞ d x = 2⎜ ⎟ ⎝ π ⎠
14
⎛α ⎞ =⎜ ⎟ ⎝ 4π ⎠ α =2 4π
1 − α x2 2
12
4
∞ ⎛ α ⎞ for n = 2, x 2 = ∫ ⎜ −∞ 4π ⎟ ⎝ ⎠ 2
1
FG IJ H K
for n = 1, x 2
14
1
⎛ α ⎞ ∞ 2 −α x2 ∫−∞ x e d x = 2 ⎜⎝ π ⎟⎠ ∫0 x e d x 12 α π h 1 1 =2 = = 2 π 2α 2 k µ 2 α α 12
2
14
∞
2
( 2α x 2 − 1) e
x − 4α x + x ) e 6
4
15 2 α 4
3
2
1 − α x2 2
−α x 2
π 3 − 4α 3 2 α 2 α
( x2 ) ⎛⎜⎝ 4απ ⎞⎟⎠
⎛α d x = 2⎜ ⎝ 4π π 1 π + 2 α 2 α
14
( 2α x2 − 1) e
12
⎞ ⎟ ⎠
∫ ( 4α ∞
0
2
1 − α x2 2
dx
x 6 − 4α x 4 + x 2 ) e −α x d x 2
I = 5 = 5h JK 2 α 2 k µ
P7.9) Evaluate the average of the square of the linear momentum of the quantum harmonic oscillator px2 , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3.
7-8
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
∫
We use the standard integrals ∞
∫ 0
2 ⎛π ⎞ e − ax dx = ⎜ ⎟ ⎝ 4a ⎠
1
∞
0
x 2n e−a x d x = 2
1 • 3 • 5 • • • (2n − 1) π and 2n +1 a n a
2
⎛ d2 ⎞ p x2 = ∫ ψ 0* ( x ) ⎜ − h 2 ⎟ψ 0 ( x ) dx d x2 ⎠ ⎝ For n = 0 ∞
⎛ α ⎞ − 2α x 2 ∫−∞ ⎜⎝ π ⎟⎠ e
=
14
1
12 ∞
⎛α ⎞ = −h ⎜ ⎟ ⎝π ⎠
⎛ 2 d2 ⎞⎛α ⎞ ⎜ −h ⎟⎜ ⎟ d x2 ⎠ ⎝ π ⎠ ⎝
14
∫e
2
−α x 2
2
1 − α x2 2
dx
− α ) dx
0
FG α IJ FG α HπK H 4
I JK
π α π α 1 − = h 2 = h kµ 2 2 α 2 α
12
= −h 2
(α x
e
For n = 1 ⎛ ∂2 ⎞ ψ 1 ( x ) dx p x2 = ∫ ψ 1* ( x ) ⎜ − h 2 ∂ x 2 ⎟⎠ ⎝ 14
∞
1 − α x2 ⎛ 4α 3 ⎞ 2 xe = ∫⎜ ⎟ π ⎠ −∞ ⎝
14
⎛ 2 ∂ 2 ⎞ ⎛ 4α 3 ⎞ ⎜ −h ∂ x 2 ⎟ ⎜ π ⎟ ⎝ ⎠⎝ ⎠
12 ∞
⎛ 4α 3 ⎞ = −h ⎜ ⎟ ⎝ π ⎠
∫ (α
2
= −h 2
2
4
2
2
−α x 2
−∞
3
12
2
2
2
=
dx
−∞
12 ∞
2
1 − α x2 2
x 4 − 3α x 2 ) e −α x dx
FG 4α IJ cα x − 3α x he dx HπK z F 4α IJ Fα LM π 3 OP − 3α LM = −h G H π K GH N α 4 α Q N 3
xe
3 2 3 h α = h kµ 2 2
7-9
π 1 α 2α
OPI QJK
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
For n = 2, ∞
p x2 =
⎛α ⎞ ∫−∞ ⎜⎝ 4π ⎟⎠
14
( 2α x 2 − 1) e
12 ∞
⎛α ⎞ = −h ⎜ ⎟ ⎝ 4π ⎠
∫ ( 2α x
2
2
1 − α x2 2
⎛ 2 ∂2 ⎞ ⎛ α ⎞ ⎜ − h x 2 ⎟ ⎜ 4π ⎟ ∂ ⎠⎝ ⎠ ⎝
14
( 2α x 2 − 1) e
1 − α x2 2
dx
− 1) e −α x α ( 2α 2 x 4 − 11α x 2 + 5) dx 2
−∞
12 ∞
⎛α ⎞ = −2 h 2 ⎜ ⎟ ⎝ 4π ⎠
∫ ( 2α x
2
− 1) e −α x α ( 2α 2 x 4 − 11α x 2 + 5) dx 2
0
12 ∞
⎛α ⎞ = −2 h 2 ⎜ ⎟ ⎝ 4π ⎠
−α x 4 6 3 4 2 2 ∫ e ( 4α x − 24α x + 21α x − 5α ) dx 2
0
3 π 21 π 1 π ⎞ ⎛ α ⎞ ⎛ 3 15 π = −2 h 2 ⎜ − 24α 2 3 2 +α 2 −5 ⎟ ⎟ ⎜⎜ 4α 4 3 2α α 2α α 2α α 2 α ⎟⎠ ⎝ 4π ⎠ ⎝ 5 5 = h 2α = h k µ 2 2 12
P7.10) Using your results for Problems P7.6 through P7.9, calculate the uncertainties in the position and momentum σ 2p = p 2 − p
2
and σ 2x = x 2 − x
2
for the ground state
(n = 0) and first two excited states (n = 1 and n = 2) of the quantum harmonic oscillator. Compare your results with the predictions of the Heisenberg uncertainty principle. 2 2 σ p2 = p 2 − p and σ x2 = x 2 − x
1 1 h kµ −0= h kµ 2 2 h h −0= σ x2 = 2 kµ 2 kµ For n = 0, σ p2 =
∆ p∆ x = σ p2σ x2 =
1 1 h≥ h 2 2
3 3 h kµ −0= h kµ 2 2 3h 3h σ x2 = −0= 2 kµ 2 kµ For n = 1, σ p2 =
∆ p∆ x = σ p2σ x2 =
3 1 h≥ h 2 2
7-10
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
5 5 h kµ −0= h kµ 2 2 h h 5 5 σ x2 = −0= 2 kµ 2 kµ For n = 2, σ p2 =
∆ p∆ x = σ p2σ x2 =
5 1 h≥ h 2 2
P7.11) The vibrational frequency of 1H35Cl is 8.963 x 1013 s–1. Calculate the force constant of the molecule. How large a mass would be required to stretch a classical spring with this force constant by 1.00 cm? Use the gravitational acceleration on Earth at sea level for this problem. 1 k ν= ; k = 4π 2 µν 2 2π µ 2 1.008 × 34.969 1.661×10−27 kg × ( 8.963 × 1013 s −1 ) k = 4 ×π × amu × 1.008 + 34.969 amu −2 k = 516 kg s F = kx = mg 2
m=
kx 516 kg s −2 × 10−2 m = = 0.525 kg g 9.81 m s −2
P7.12) Two 1.00-g masses are attached by a spring with a force constant k = 500 kg s–2. Calculate the zero point energy of the system and compare it with the thermal energy kT. If the zero point energy were converted to translational energy, what would be the speed of the masses? mm m µ = 1 2 = = 0.500 × 10−3 kg m1 + m2 2 E0 =
500 kg s −2 h k 1.055 ×10−34 J s = = 5.28 × 10−32 J −3 2 µ 2 0.500 ×10 kg
E0 5.28 × 10−32 J = = 1.27 × 10−11 −23 −1 kT 300 K ×1.381×10 J K 1 2 mv = 5.28 ×10−32 J 2 2 × 5.28 ×10−32 J v= = 7.27 ×10−15 m s −1 −3 2.00 × 10 kg
P7.13) Use
x 2 as calculated in Problem P7.8 as a measure of the vibrational
amplitude for a molecule. What fraction is
x 2 of the 127-pm bond length of the HCl
7-11
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules molecule for n = 0, 1, and 2? The force constant for the 1H35Cl molecule is 516 N m–1. Use your results from Problem P7.8 in this problem. For n = 0, 1
x2
⎛ h ⎞2 =⎜ ⎜ 2 k µ ⎟⎟ ⎝ ⎠ x2
bond length
=
⎛ ⎜ 1.055 ×10−34 J s =⎜ ⎜ 1.0078 × 34.9688 −1 ×1.66 ×10−27 kg amu −1 ⎜ 2 × 516 N m × 1.0078 + 34.9688 ⎝
1
⎞2 ⎟ ⎟ = 7.59 ×10−12 m ⎟ ⎟ ⎠
7.59 ×10−12 m = 5.97 ×10−2 −12 127 ×10 m
For n = 1
x2
⎛ ⎜ ⎛ 3h 3×1.055 × 10−34 J s ⎜ =⎜ = ⎟ ⎜2 kµ ⎟ ⎜ 1.0078 × 34.9688 −1 −27 −1 ⎝ ⎠ ⎜ 2 × 516 N m × 1.0078 + 34.9688 ×1.66 ×10 kg amu ⎝ 1 ⎞2
x2 bond length
=
1
⎞2 ⎟ ⎟ = 1.31×10−11m ⎟ ⎟ ⎠
1.31×10−11m = 0.103 127 ×10−12 m
For n = 2
x2
⎛ 5h =⎜ ⎜2 kµ ⎝
x2 bond length
=
1
⎞2 ⎟ ⎟ ⎠
⎛ ⎜ 5 ×1.055×10−34 J s =⎜ ⎜ 1.0078 × 34.9688 −1 −27 −1 ⎜ 2 × 516 N m × 1.0078 + 34.9688 ×1.66 ×10 kg amu ⎝
1
⎞2 ⎟ ⎟ = 1.70 ×10−11m ⎟ ⎟ ⎠
1.70 ×10−11m = 0.134 127 ×10−12 m
P7.14) 1H35Cl has a force constant k = N m–1 and a moment of inertia of 2.644 × 10–47 kg m2. Calculate the frequency of the light corresponding to the lowest energy pure vibrational and pure rotational transitions. In what regions of the electromagnetic spectrum do the transitions lie?
7-12
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
516 kg s −2 = 8.963 × 1013 s −1 −27 1.008 × 34.969 1.661× 10 kg µ amu × 1.008 + 34.969 amu This frequency lies in the infrared region of the eletromagnetic spectrum.
ν=
1 2π
k
=
1 2π
The lowest energy transition is J = 0 → J = 1. 1.055 × 10−34 J s ) ( h2 h2 = 2 ( J + 1) = 2 = = 4.20 × 10−22 J −47 2 2 2 µ r0 µ r0 2.644 × 10 kg m 2
∆Erot
∆Erot = 6.35 ×1011 s −1 h This frequency lies in the microwave region of the eletromagnetic spectrum.
ν=
P7.15) The vibrational frequency for N2 expressed in wave numbers is 2358 cm–1. What is the force constant associated with the N ≡ N triple bond? How much would a classical spring with this force constant be elongated if a mass of 1.00 kg were attached to it? Use the gravitational acceleration on Earth at sea level for this problem. 1 k ν = cν% = 2π µ so k = ( 2π cν% ) µ 2
k = ( 2π × 2.998 x1010 cm s −1 × 2358cm −1 ) × 2
14.01amu ×14.01amu 1.661×10−27 kg × 2 ×14.01amu amu
k = 2299 N m −1 x=
F mg 1.00 kg × 9.81m s −2 = = = 4.28 × 10−3 m k k 2299 N m −1
P7.16) The force constant for the 1H35Cl molecule is 516 N m–1. Calculate the vibrational zero point energy of this molecule. If this amount of energy were converted to translational energy, how fast would the molecule be moving? Compare this speed to the 3kT root mean square speed from the kinetic gas theory, v rms = for T = 300 K. m 516 N m −1 = 2.97 ×10−20 J 1.0078 × 34.9688 ×1.66 ×10−27 kg amu −1 1.0078 + 34.9688 The speed if converted to translational energy would be E=
v=
1 k 1 h = ×1.055 ×10−34 J s × µ 2 2
2E = m
2 × 2.97 ×10−20 J = 997 m s −1 −27 −1 (1.0078 + 34.9688 ) ×1.66 x10 kg amu
The average speed from the kinetic gas theory is
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Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
v rms = v v rms
=
3kT = m
3 ×1.381×10−23 J K −1 × 300 K = 456 m s −1 −27 −1 (1.0078 + 34.9688 ) ×1.66 ×10 kg amu
997 m s −1 = 2.19 456 m s −1
P7.17) A gas-phase 1H19F molecule, with a bond length of 91.7 pm, rotates in a threedimensional space. a) Calculate the zero point energy associated with this rotation. b) What is the smallest quantum of energy that can be absorbed by this molecule in a rotational excitation? a) There is no zero point energy because the rotation is not constrained. b) The smallest energy that can be absorbed is
2 × (1.055 x10−34 J s ) h2 h2 E= J ( J + 1) = 1(1 + 1) = 2 1.0078 ×18.9984 2I 2I 2× ×1.66 ×10−27 kg amu −1 × ( 91.7 x10−12 m ) 1.0078 + 18.9984 −22 E = 8.33 ×10 J 2
P7.18) A 1H19F molecule, with a bond length 91.7 pm, absorbed on a surface rotates in two dimensions. a) Calculate the zero point energy associated with this rotation. b) What is the smallest quantum of energy that can be absorbed by this molecule in a rotational excitation? a) There is no zero point energy because the rotation is not constrained. b) The smallest energy that can be absorbed is
1.055 ×10−34 J s ) ( h 2 ml2 h 2 E= = = 2I 2 I 2 × 1.0078 ×18.9984 ×1.66 ×10−27 kg amu −1 × 91.7 ×10−12 m 2 ( ) 1.0078 + 18.9984 E = 4.17 ×10−22 J 2
P7.19) The moment of inertia of 1H35Cl is 2.644 × 10–47 kg m2. Calculate J = 0, 5, 10, and 20 at 298 K. For which of these values of J is
7-14
Erot ≈ 1? kT
Erot for kT
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules 1.055 × 10−34 J s ) ( h2 h2 E= J ( J + 1) = 1(1 + 1) = J ( J + 1) = 2.089 × 10−22 J ( J + 1) 2I 2I 2 × 2.664 × 10−47 kg m 2 EJ =0 = 0 2
EJ =5 = 30 × 2.089 × 10−22 J = 6.267 × 10−21 J EJ =5 6.267 ×10−21 J = = 1.51 kT 1.381×10−23 J K −1 × 298 K EJ =10 = 110 × 2.089 × 10−22 J = 2.297 × 10−20 J EJ =10 2.528 × 10−20 J = = 5.55 kT 1.381×10−23 J K −1 × 298 K EJ = 20 = 20 × 21× 2.089 × 10−22 J = 8.774 × 10−20 J EJ = 20 8.774 × 10−20 J = = 21.2 kT 1.381×10−23 J K −1 × 298 K Erot ≈ 1 for J = 5. kT
P7.20) Using the Boltzmann distribution, calculate
nJ for 1H35Cl for the J values of n0
nJ go through a maximum as J increases? If so, what n0 can you say about the value of J corresponding to the maximum? Problem P7.19 at T = 298 K. Does
nJ − E − E kT = ( 2 J + 1) e ( J 0 ) = ( 2 J + 1) exp [ − EJ kT ] n0 n0 =1 n0 n5 = ( 2 × 5 + 1) exp ⎡⎣ − ( 6.267 × 10−21 J ) 1.381×10−23 J K −1 × 298 K ⎤⎦ = 2.42 n0 n10 = ( 2 ×10 + 1) exp ⎡⎣ − ( 2.297 × 10−20 J ) 1.381×10−23 J K −1 × 298 K ⎤⎦ = 0.082 n0 n20 = ( 2 × 20 + 1) exp ⎡⎣ − ( 8.774 × 10−20 J ) 1.381×10−23 J K −1 × 298 K ⎤⎦ = 2.60 ×10−8 n0 nJ goes through a maximum because it has a value greater than one for J = 5. You can n0 only conclude that Jmax < 10.
7-15
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
P7.21) By substituting in the Schrödinger equation for rotation in three dimensions, 1/ 2
⎛ 5 ⎞ 2 show that the rotational wave function ⎜ ⎟ ( 3cos θ − 1) is an eigenfunction of the π 16 ⎝ ⎠ total energy operator. Determine the energy eigenvalue. h2 − 2 µ r02
2 ⎡ 1 ∂ ⎛ ∂Y (θ , φ ) ⎞ 1 ∂ Y (θ , φ ) ⎤ ⎢ ⎥ = EY (θ , φ ) ⎜ sin θ ⎟+ 2 ∂θ ∂φ 2 ⎦ ⎠ sin θ ⎣ sin θ ∂θ ⎝
1/ 2 ⎡ ⎛ ⎧⎪⎛ 5 ⎞1/ 2 ⎫⎪ ⎞ ⎧ ⎫⎪ ⎤ 2 2 ⎪⎛ 5 ⎞ ∂ ⎨⎜ ∂ ⎨⎜ 3cos θ − 1) ⎬ ⎟ 3cos2 θ − 1) ⎬ ⎥ ⎢ ⎜ ( ( ⎟ ⎟ h2 ⎢ 1 ∂ ⎜ 1 ⎪⎝ 16π ⎠ ⎪⎭ ⎟ ⎪⎩⎝ 16π ⎠ ⎪⎭ ⎥ − + 2 sin θ ⎩ ⎥ 2 ⎢ 2 ⎜ ⎟ ∂θ ∂φ 2 µ r0 sin θ ∂θ sin θ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ 1/ 2
⎡ 1 ∂ ⎣⎢ sin θ ∂θ
1/ 2
2 ⎡ 6 {1 − cos2 } − 12 cos2 θ ⎤ = − h ⎛⎜ 5 ⎞⎟ ⎣ ⎦ 2 µ r02 ⎝ 16π ⎠
⎛ 5 ⎞ ⎜ ⎟ ⎝ 16π ⎠
=−
h2 2 µ r02
=−
h2 ⎛ 5 ⎞ ⎜ ⎟ 2 µ r02 ⎝ 16π ⎠
(
)
h2 ⎤ 2 − 6 cos sin = − θ θ ( ) ⎦⎥ 2µ r 2 0
(
1/ 2
⎛ 5 ⎞ ⎜ ⎟ ⎝ 16π ⎠ 1/ 2
)
The energy eigenvalue is
⎡ 1 ⎤ 3 2 ⎢⎣ sin θ ( 6 sin θ − 12 cos θ sin θ )⎦⎥ 1/ 2
6h 2 ⎛ 5 ⎞ ⎡( 6 − 18 cos2 θ ) ⎤ = ⎣ ⎦ 2 µ r 2 ⎜⎝ 16π ⎟⎠ 0
3h 2 h2 , corresponding to E = l (l + 1) with l = 2. l 2I µ r02
P7.22) Show by carrying out the necessary integration that the eigenfunctions of the Schrödinger equation for rotation in two dimensions, orthogonal.
FG H
IJ K c
2
z
zc
b
g
1 imlφ e and 2π
b
1 inlφ e , ml ≠ nl are 2π
gh
2π 1 1 2π e imlφ e − inlφ dφ = cos ml − nl φ + i sin ml − nl φ dφ 0 2π 0 2π 1 = cos 2π ml − nl − 1 + i sin 2π ml − nl = 0 2π because ml − nl is an integer
b
b
g
b
gh
g
P7.23) In this problem you will derive the commutator ⎡⎣lˆx , lˆy ⎤⎦ = ihlˆz . a) The angular momentum vector in three dimensions has the form l = i l x + j l y + k lz where the unit vectors in the x, y, and z directions are denoted by i, j , and k. Determine lx, ly, and lz by expanding the 3 × 3 cross product l = r × p. The vectors r and p are given by r = i x + jy + k z and p = i px + j p y + k pz . b) Substitute the operators for position and momentum in your expressions for lx and ly. Always write the position operator to the left of the momentum operator in a simple product of the two. c) Show that ⎡lˆx , lˆy ⎤ = ihlˆz . ⎣ ⎦
7-16
( 3cos
2
θ − 1)
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules a) i l= x
j y
px
py
k y z =i py pz
z pz
−j
x px
x z +k px pz
y py
= i ( y pz − z p y ) − j ( x pz − z p x ) + k ( x p y − y p x )
= i ( y pz − z p y ) + j ( z p x − x pz ) + k ( x p y − y px ) b)
⎛ ⎛ ∂ ∂ ⎞ ⎛ ∂ ∂ ⎞ ∂ ∂ ⎞ lˆ = i ⎜ −ih y + ih z + j ⎜ −i h z + i hx + k ⎜ −i h x + ih y ⎟ ⎟ ∂z ∂y⎠ ⎝ ∂x ∂z ⎠ ∂y ∂ x ⎟⎠ ⎝ ⎝ ⎛ ∂ ⎛ ∂ ⎛ ∂ ∂ ⎞ ˆ ∂ ⎞ ˆ ∂ ⎞ lˆx = −ih ⎜ y l y = −i h ⎜ z l z = −i h ⎜ x − z −x − y ⎟ ⎟ ∂ y⎠ ∂z ⎠ ∂ x ⎟⎠ ⎝ ∂z ⎝ ∂x ⎝ ∂y c)
⎡lˆx , lˆy ⎤ = lˆx lˆy − lˆy lˆx = ⎣ ⎦ ⎛ ∂ ∂ ⎞⎛ ∂f ∂f ⎞ 2 ⎛ ∂ ∂ ⎞ ⎛ ∂f ∂f ⎞ −h2 ⎜ y −z −z −z −z ⎟⎜ x ⎟+h ⎜x ⎟⎜ y ⎟ ∂ y ⎠⎝ ∂ z ∂x⎠ ∂ x ⎠⎝ ∂ z ∂y⎠ ⎝ ∂z ⎝ ∂z ∂ ⎛ ∂f ∂f ⎞ ∂ ⎛ ∂f ∂f ⎞ 2 ∂ ⎛ ∂f ∂f ⎞ 2 ∂ ⎛ ∂f ∂f ⎞ 2 = −h2 y −z −z −z −z ⎜x ⎟+ h z ⎜x ⎟+h x ⎜y ⎟−h z ⎜y ⎟ ∂z⎝ ∂z ∂x⎠ ∂y⎝ ∂z ∂x⎠ ∂z⎝ ∂z ∂y⎠ ∂x⎝ ∂z ∂y⎠ ∂2 f ∂f ∂2 f ∂2 f ∂2 f = −h2 y x 2 + h2 y + h 2 yz + h 2 zx − h2 z 2 ∂z ∂x ∂ z∂x ∂ y∂z ∂ y∂x 2 ∂2 f ∂2 f ∂2 f 2 2 ∂f 2 2 2 ∂ f xz x yz z − − − + = h h h h ∂ z2 ∂ y∂z ∂y ∂ z∂x ∂ y∂x ∂f ∂f − h2 x h2 y ∂x ∂y
+ h 2 xy
⎛ ∂ ∂ ∂ ⎞ ∂ ⎞ 2⎛ ˆ ⎡ˆ ˆ ⎤ ⎣ l x , l y ⎦ = − h ⎜ x ∂ y − y ∂ x ⎟ = i h × i h ⎜ y ∂ x − x ∂ y ⎟ = i hl z ⎝ ⎠ ⎝ ⎠
P7.24) For molecular rotation, the symbol J rather than l is used as the quantum number for angular momentum. A 1H35Cl molecule whose bond length and force constant are 127 pm and 516 N m–1, respectively, has the rotational quantum number l = 10 and vibrational quantum number n = 0. a) Calculate the rotational and vibrational energy of the molecule. Compare each of these energies with kT at 300 K.
7-17
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules b) Calculate the period for vibration and rotation. How many times does the molecule rotate during one vibrational period? a) Erot
J ( J + 1) h 2 = = 2µ r 2
10 ×11× (1.055 ×10−34 J s ) 2×
2
2 1.0078amu × 34.9688amu 1.661×10−27 kg × × (1.275 ×10−10 m ) 1.0078amu + 34.9688amu amu
Erot = 2.55 × 10−20 J Erot 2.55 × 10−20 J = = 6.15 kT 1.381× 10−23 J K −1 × 300 K 1⎞ k 1 516 N m −1 ⎛ = ×1.055 ×10−34 J s × Evib = ⎜ n + ⎟ h 1.0078amu × 34.9688amu 1.661×10−27 kg 2⎠ µ 2 ⎝ × 1.0078amu + 34.9688amu amu Evib = 2.97 ×10−20 J 2.97 ×10−20 J Evib = = 7.17 kT 1.381×10−23 J K −1 × 300 K b) Calculate the period for vibration and rotation. How many times does the molecule vibrate during one rotational period? 1 2π Erot = Iω 2 ; ω = 2πν = Trot 2 Trot =
=
2π
ω
=
2π 2 Erot I 2π −20
2 × 2.55 × 10 J 2 1.0078amu × 34.9688amu 1.661×10−27 kg × × (1.275 ×10−10 m ) 1.0078amu + 34.9688amu amu
= 1.43 ×10−13s
1.0078amu × 34.9688amu 1.661×10−27 kg × µ 1 amu 1.0078amu + 34.9688amu = 1.12 ×10−14s Tvib = = 2π = 2π −1 ν 516 N m k −13 T 1.43 ×10 s = 12.8 times in one rotational period. It vibrates rot = Tvib 1.12 ×10−14s
P7.25) At what values of θ does Y20 (θ ,φ ) = ⎛⎜
1/ 2
5 ⎞ ⎟ ⎝ 16π ⎠ nodes points, lines, planes, or other surfaces?
7-18
( 3cos θ − 1) have nodes? Are the 2
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
b g
c
h
Y20 θ , φ has nodes when 3 cos 2 θ − 1 has nodes. This occurs for
θ = 0.955 and π − 0.955 radians or 54.7 and 125.3 degrees. These surfaces are cones.
P7.26) The wave functions px and dxz are linear combinations of the spherical harmonic functions, which are eigenfunctions of the operators Hˆ , lˆ 2, and lˆz for rotation in three dimensions. The combinations have been chosen to yield real functions. Are these functions still eigenfunctions of lˆz ? Answer this question by applying the operator to the functions. ∂ 3 3 L$ z p x = −ih sin θ cos φ = ih sin θ sin φ . This shows that px is not an ∂θ 4π 4π eigenfunction of Lˆz . ∂ L$ z d xz = −ih ∂θ
15 15 sin θ cos θ cos φ = ih sin θ cos θ sin φ . This shows that dxz is 4π 4π not an eigenfunction of Lˆ . z
1/ 2
5 ⎞ P7.27) Show that the function Y (θ ,φ ) = ⎛⎜ ⎟ ⎝ 16π ⎠ interval 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π . 0 2
∫∫ (Y (θ ,φ ) ) 0 2
2
2π
( 3cos θ − 1) is normalized over the 2
π
2 ⎛ 5 ⎞ 2 dτ = ⎜ ⎟ ∫ dφ ∫ sin θ ( 3cos θ − 1) dθ ⎝ 16π ⎠ 0 0
π
⎛ 5 ⎞ 4 2 = 2π ⎜ ⎟ ∫ sin θ ( 9cos θ − 6cos θ + 1) dθ π 16 ⎝ ⎠0
∫∫ (Y (θ ,φ ) ) 0 2
2
2π
π
2 ⎛ 5 ⎞ 2 dτ = ⎜ ⎟ ∫ dφ ∫ sin θ ( 3cos θ − 1) dθ ⎝ 16π ⎠ 0 0
π
⎛ 5 ⎞ 4 2 = 2π ⎜ ⎟ ∫ sin θ ( 9cos θ − 6cos θ + 1) dθ 16 π ⎝ ⎠0 π
5⎛ 9 5 ⎛ 18 ⎞ ⎞ = ⎜ − cos5 θ + 2cos3 θ − cosθ ⎟ = ⎜ − 4 + 2 ⎟ = 1 8⎝ 5 ⎠0 8 ⎝ 5 ⎠
P7.28) Is it possible to know simultaneously the angular orientation of a molecule rotating in a two dimensional space and its angular momentum? Answer this question by ⎡ ∂ ⎤ evaluating the commutator ⎢φ , − ih ⎥ . ∂φ ⎦ ⎣
7-19
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules
d ⎡⎣φ f (φ ) ⎤⎦ df (φ ) ⎡ ∂ ⎤ φ = i hf ( φ ) ⎢φ , − ih ∂φ ⎥ f (φ ) = −ihφ d φ + ih dφ ⎣ ⎦
⎡ ∂ ⎤ ⎢φ , − ih ∂φ ⎥ = ih ⎣ ⎦ Because the commutator is not equal to zero, it is not possible to simultaneously know the angular orientation of a molecule rotating in a two-dimensional space and its angular momentum.
P7.29) At 300 K, most molecules are in their ground vibrational state. Is this also true nJ =1 n and J =5 for the H35Cl molecule n J =0 n J =0 –10 whose bond length is 1.27 × 10 m. Make sure that you take the degeneracy of the levels into account. for their rotational degree of freedom? Calculate
nJ =1 2J +1 = e nJ = 0 1
h 2 J ( J +1) 2 I kT
⎡ ⎤ 2 ⎢ ⎥ 1.055 ×10−34 J s ) 1(1+1) ( = 3 exp ⎢ − ⎥ ⎢ 2 × 1.0078 × 34.9688 amu ×1.66 ×10−27 kg amu −1 × (1.27 ×10−10 m )2 ×1.381×10−23 J K −1 × 300 K ⎥ ⎥⎦ 1.0078 + 34.9688 ⎣⎢ nJ =1 = 2.708 nJ = 0 nJ = 5 2 J + 1 = e nJ = 0 1
h 2 J ( J +1) 2 I kT
⎡ ⎤ 2 ⎢ ⎥ 1.055 ×10−34 J s ) 5 ( 5+1) ( = 11 exp ⎢ − ⎥ ⎢ 2 × 1.0078 × 34.9688 amu ×1.66 x10−27 kg amu −1 × (1.27 ×10−10 m )2 ×1.381×10−23 JK −1 × 300 K ⎥ ⎢⎣ ⎥⎦ 1.0078 + 34.9688 nJ = 5 = 2.369 nJ = 0
P7.30) Draw a picture (to scale) showing all angular momentum cones consistent with l = 5. Calculate the half angles for each of the cones.
7-20
Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules ml = +5 ml = +4 ml = +3 ml = +2 ml = +1 ml = 0 ml = -1 ml = -2 ml = -3 ml = -4 ml = -5
The half angles of the cones measured from the positive and negative z axis are ⎛ ⎞ 5 ⎟ = 0.420 radians θ ml =5 = cos−1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ ⎛ ⎞ 4 ⎟ = 0.752 radians θ ml =4 = cos−1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ ⎛ ⎞ 3 ⎟ = 0.991radians θ ml =3 = cos−1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ ⎛ ⎞ 2 ⎟ = 1.20 radians θ ml =2 = cos−1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ ⎛ ⎞ 1 ⎟ = 1.39 radians θ ml =1 = cos −1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ ⎛ ⎞ 0 ⎟ = 1.57 radians θ ml =0 = cos −1 ⎜ ⎜ 5 ( 5 + 1) ⎟ ⎝ ⎠ as well as π minus these values for the negative values of ml.
7-21