UNIVERSITY PHYSICS II CHAPTER 27 An introduction to Quantum Mechanics 827. 1 The Heisenberg uncertainty principles Nature is bilateral: particles are waves and waves are particles. The particle aspect carries with it the traditional concepts of position and momentum; The wave aspect carries with it the concepts of wavelength and frequency. Nature places natural limits on the precision of our measurements: some knowledge and information forever is shrouded from our prying eyes
1 §27.1 The Heisenberg uncertainty principles Nature is bilateral: particles are waves and waves are particles.The particle aspect carries with it the traditional concepts of position and momentum; The wave aspect carries with it the concepts of wavelength and frequency. Nature places natural limits on the precision of our measurements; some knowledge and information forever is shrouded from our prying eyes
8 27.1 The Heisenberg uncertainty principles 1. The position-momentum uncertainty principle Single slit diffraction of electrons P asine=d a0 hh P2 a0p =h 8 27.1 The Heisenberg uncertainty principles The position uncertainty: 4y The momentum uncertainty in y direction: P 4n=P,=p44=h If we account for the secondary maxima 4x42h·,2h△z△p2≥h
2 1. The position-momentum uncertainty principle Single slit diffraction of electrons： . . . . . .. . a sinθ 1 = λ pz h aθ 1 = aθ 1 pz = h §27.1 The Heisenberg uncertainty principles θ 1 p r y z The position uncertainty： ∆y = a ∆py = py = pz θ1 ∆y ⋅ ∆py = h p pz py θ 1 The momentum uncertainty in y direction: θ 1 p r y §27.1 The Heisenberg uncertainty principles z If we account for the secondary maxima ∆x ⋅ ∆px ≥ h ∆y ⋅ ∆py ≥ h ∆z⋅ ∆pz ≥ h
s 27.1 The Heisenberg uncertainty principles The Heisenberg uncertainty principle states that there exists a fundamental limit to the extent to which we can simultaneously determine the position and corresponding momentum component of a wave-particle any given direction 2. The energy-time uncertainty principle Recall mathematical form of a monochromatic wave Y(x, t)=Acos(kx-at) 2丌 Acos(x-2riv) 8 27.1 The Heisenberg uncertainty principles Substitution for a and v h E=hv P 2丌 Y(x, +)=AcoS(Px 2兀E The role of e and t in the equation are mathem- atically identical to the role ofx and p ∠EMt2h 3.The implications of the uncertainty principles o the uncertainty principle indicate that it is not exact describing the microscopic particle by classical theory
3 The Heisenberg uncertainty principle states that there exists a fundamental limit to the extent to which we can simultaneously determine the position and corresponding momentum component of a wave-particle in any given direction. 2. The energy-time uncertainty principle 2 ) 2 cos( ( , ) cos( ) A x t x t A kx t πν λ π Ψ ω = − = − Recall mathematical form of a monochromatic wave §27.1 The Heisenberg uncertainty principles λ E hν p h x = , = Substitution for λ and ν ) 2 2 ( , ) cos( Et h p x h x t A x π π Ψ = − The role of E and t in the equation are mathematically identical to the role of x and px. ∆E∆t ≥ h 3.The implications of the uncertainty principles 1 the uncertainty principle indicate that it is not exact describing the microscopic particle by classical theory. §27.1 The Heisenberg uncertainty principles
8 27.1 The Heisenberg uncertainty principles @the uncertainty principle gives an limitation using classical model. Implications of the position-momentum uncertainty principle In classical physics, the position and momentum component can be measured with arbitrary precision in principle, but in quantum mechanics, this is not strictly true. because of the nature of particles. Example 1: Why atoms do not collapse? The minimum momentum of an electron is h h p≈4≈≈ 8 27.1 The Heisenberg uncertainty principles The total energy of the electron is E=KE+PE= P2 h 2n4 2mr24 The value of r minimizes the total energy satisfied de h (一-2)=0 dr 2m r=tEo It is same order of magnitude as the fictitious of the bohr model
4 2the uncertainty principle gives an limitation using classical model. ¾Implications of the position-momentum uncertainty principle In classical physics, the position and momentum component can be measured with arbitrary precision in principle, but in quantum mechanics, this is not strictly true, because of the nature of particles. Example 1:Why atoms do not collapse? The minimum momentum of an electron is r h x h p ≈ p ≈ ≈ ∆ ∆ §27.1 The Heisenberg uncertainty principles The total energy of the electron is r e mr h r e m p E KE PE 0 2 2 2 0 2 2 2 4πε 2 4πε = + = − = − The value of r minimizes the total energy satisfied 2 2 0 2 0 2 3 2 4 ) 0 1 ( 4 ) 2 ( d 2 d me h r r e m r h r E πε πε = = − − − = It is same order of magnitude as the fictitious of the Bohr model. §27.1 The Heisenberg uncertainty principles
s 27. 1 The Heisenberg uncertainty principles Example 2: You measure the diameter of a shotgun pellet of mass 1.0 g to be 1.00+0.01 mm with a micrometer. What is the minimum uncertainty of its momentum if the magnitude of its momentum is measured simultaneously? Solution: h6.626×10 4 =7×10-29kgm/s 1.0×10 pp=m,v=10×10-×100=0, 1kg msY It is exceedingly small. As a result, for such a macroscopic particle, the uncertainty principle has no effect 8 27.1 The Heisenberg uncertainty principles Example 3: The accelerating potential difference of TV kinescope is 9kv, the diameter of exit of electron gun is 0. 1 mm, is it reasonable to describe the electrons as classical particles? Solution: The uncertainty of the speed A4=4x·m4v≥h h 6.626×10-34 ν =727m/s mx9.1×10-×0.1×10 The speed of the electrons =,2AE=2e4=2×1.6×10°×103 ≈56×107m/s 9.11×10
5 Example 2: You measure the diameter of a shotgun pellet of mass 1.0 g to be 1.00±0.01 mm with a micrometer. What is the minimum uncertainty of its momentum if the magnitude of its momentum is measured simultaneously? Solution: 7 10 kg m/s 1.0 10 6.626 10 29 5 34 = × ⋅ × × = = − − − x h px ∆ ∆ It is exceedingly small. As a result, for such a macroscopic particle, the uncertainty principle has no effect. §27.1 The Heisenberg uncertainty principles 3 1 p p 1.0 10 100 0.1 kg m s − − p = m v = × × = ⋅ ⋅ Example 3: The accelerating potential difference of TV kinescope is 9kV, the diameter of exit of electron gun is 0.1 mm, is it reasonable to describe the electrons as classical particles? Solution: 7.27m/s 9.11 10 0.1 10 6.626 10 31 3 34 = × × × × = = = ⋅ ≥ − − − m x h v x p x m v h x x x ∆ ∆ ∆ ∆ ∆ ∆ The uncertainty of the speed The speed of the electrons 5.6 10 m/s 9.11 10 2 2 2 1.6 10 10 7 31 19 3 ≈ × × × × × = = = − − m e V m KE v ∆ §27.1 The Heisenberg uncertainty principles
8 27.1 The Heisenberg uncertainty principles Example 4: Minimum energy of a particle in a box -zero point energy Solution: For a particle in a box of length L (4p)2=(p-p) (p2-2p-p2) P-p=p h (4p)2=p2≥( The microscopic In particle cannot E ≠0 be rest! 2m 2mL s 27.1 The Heisenberg uncertainty principles Example 5: a proton is known to be in the nucleus of an atom. The size of the nucleus is about 1.0x10- 4m. Find a Pmin=?; b vmin=? For nonrelativistic situation c is it reasonable to express its momentum in classical expression? Solution: a. According to the uncertainty principle h6.626×10 4p 10×104=6.6×10-20kgm/s b The minimum speed of the proton 6.6×10 40×107m/s 1.67×10 6
6 Example 4: Minimum energy of a particle in a box —zero point energy 2 2 2 2 2 2 2 ( 2 ) ( ) ( ) p p p p pp p p p p av av = − = = − − ∆ = − 0 2 2 ( ) ( ) 2 2 2 2 2 2 = ≥ ≠ = ≥ mL h m p E x h p p ∆ ∆ For a particle in a box of length L §27.1 The Heisenberg uncertainty principles Solution: The microscopic particle cannot be rest! Example 5: A proton is known to be in the nucleus of an atom. The size of the nucleus is about 1.0×10-14 m. Find a. pmin=?; b. vmin=? For nonrelativeistic situation. c. is it reasonable to express its momentum in classical expression? Solution: a. According to the uncertainty principle 6.6 10 kg m/s 1.0 10 6.626 10 20 14 34 = × ⋅ × × = = − − − x h p ∆ ∆ b. The minimum speed of the proton 4.0 10 m/s 1.67 10 6.6 10 7 27 20 min = × × × = = − − m p v §27.1 The Heisenberg uncertainty principles
8 27.1 The Heisenberg uncertainty principles c. The fraction of the speed of light 4.0×10 =0.13 c3.0×10 y ≈1.0 It is reasonable to express its momentum in classical expression. s 27.1 The Heisenberg uncertainty principles >Implications of the energy-time uncertainty principle a. The mass of fundamental particles According to special relativity Em=mc2→AEt=c2m According to the energy-time uncertainty principle c2AmAt≥h The mean lifetime of a free neutron is 888 s then 6.626×10 =829×10-k c2r(3×105)2(88》
7 c. The fraction of the speed of light 1.0 1 / 1 0.13 3.0 10 4.0 10 2 2 8 7 ≈ − = = × × = v c c v γ It is reasonable to express its momentum in classical expression. §27.1 The Heisenberg uncertainty principles ¾Implications of the energy-time uncertainty principle a. The mass of fundamental particles According to special relativity E mc ∆E c ∆m 2 rest 2 rest = ⇒ = According to the energy-time uncertainty principle c ∆m∆t ≥ h 2 8.29 10 kg (3 10 ) (888) 6.626 10 54 8 2 34 2 − − = × × × = = c t h m ∆ ∆ The mean lifetime of a free neutron is 888 s, then §27.1 The Heisenberg uncertainty principles
8 27.1 The Heisenberg uncertainty principles The mass of the neutron can be determined quite precisely because it is a relatively long lived particle The values of the masses of very short-lived particles will have intrinsic spread because of the Heisenberg uncertainty principle. b. The nature of a vacuum vacuum is not empty The idea of a vacuum in quantum physics is quite different from classical idea of a vacuum as simply a volume with nothing in it. s 27.1 The Heisenberg uncertainty principles Classical mechanics KE=E-PE=P>0 2n Relativity quantum mechanics E=Pc+mc, E=tpc+mc Dirac sea: positive energy states are empty, negative energy states are filled fully. E E=mc2 E=0 E=0 E=-mc2 E=-mc2 AE=E-E pc +m c P 8
8 The mass of the neutron can be determined quite precisely because it is a relatively longlived particle. The values of the masses of very short-lived particles will have intrinsic spread because of the Heisenberg uncertainty principle. b. The nature of a vacuum —vacuum is not empty The idea of a vacuum in quantum physics is quite different from classical idea of a vacuum as simply a volume with nothing in it. §27.1 The Heisenberg uncertainty principles 0 2 2 = − = ≥ m p Classical mechanics KE E PE 2 2 2 2 4 2 2 2 4 E = p c + m c , E = ± p c + m c Relativity quantum mechanics E+=mc2 E- =-mc2 E=0 E+=mc2 E- =-mc2 E=0 Dirac sea: positive energy states are empty, negative energy states are filled fully. 2 2 2 4 2 2 2 4 E = E − E = p c + m c + p c + m c ∆ + − §27.1 The Heisenberg uncertainty principles
8 27.1 The Heisenberg uncertainty principles The quantum mechanical vacuum is a seething sea of particle-antiparticle pairs, called virtual particles, since their existence is ephemeral The pairs of virtual particles well up out of nothing, live for very short time, and then disappear(annihilate each other) According to energy-time uncertainty principle ∠EAt≥h For a virtual electron-positron pair h h t 4E 2mc 6.626×10 =404×10-2ls 2(9.11×10)(3×103) 8 27. 1 The Heisenberg uncertainty principles C. The verification of experiment: H. B. G. Casimir effect: Lamb shift: Example: Light of wavelength 632.8 nm is incident on an extremely fast shutter that chops the beam into pulses. The shutter stays open for only 1.5x10-9s. what is the approximately minimum range of wavelengths m in the light pulses that pass the shutter
9 The quantum mechanical vacuum is a seething sea of particle-antiparticle pairs, called virtual particles, since their existence is ephemeral. The pairs of virtual particles well up out of nothing, live for very short time, and then disappear(annihilate each other). According to energy-time uncertainty principle ∆E∆t ≥ h 4.04 10 s 2(9.11 10 )(3 10 ) 6.626 10 2 21 31 8 2 34 2 − − − = × × × × = = = mc h E h t ∆ ∆ For a virtual electron-positron pair §27.1 The Heisenberg uncertainty principles c. The verification of experiment: H. B. G. Casimir effect: Lamb shift: Example: Light of wavelength 632.8 nm is incident on an extremely fast shutter that chops the beam into pulses. The shutter stays open for only 1.5×10-9 s. What is the approximately minimum range of wavelengths ∆λ in the light pulses that pass the shutter? §27.1 The Heisenberg uncertainty principles
827. 1 The Heisenberg uncertainty principles ∠EMt≥h→AEM=h h6.626×10 ∠E= =44×10-25J t15×10 since E=hy=hC→ AE- hc A therefore hc (44×1025)(6328×10- ≈8.9×10-13m (6.626×10-3)(3.00×103) 827.2 Particle-waves and the wavefunction 1. The strange behavior of microscopic particles 波粒子 少膀目 10
10 4.4 10 J 1.5 10 6.626 10 25 9 34 − − − = × × × = = ≥ ⇒ = t h E E t h E t h ∆ ∆ ∆ ∆ ∆ ∆ since ∆λ λ ∆ λ ν 2 hc E c E = h = h ⇒ = therefore 8.9 10 m (6.626 10 )(3.00 10 ) (4.4 10 )(632.8 10 ) 13 34 8 25 9 2 − − − − ≈ × × × × × = = hc ∆Eλ ∆λ §27.1 The Heisenberg uncertainty principles §27.2 Particle-waves and the wavefunction 1. The strange behavior of microscopic particles
