8.04 Quantum Physics Lecture VIIDouble slit: mathematical model of interference patteer and photon scattering To develop some more insight into interference, and the correlations between quantum system and (classical) apparatus that lie at the heart of the quantum measurement problem, we will postulate rules on how the interference pattern is formed, and how photon scattering changes the electron’s wavefunction. Figure I: Double-slit interference. Rule 1. Wavefunction at slits 1, 2 is: ψ1 = Aeiφ1 ψ2 = Aeiφ2 Complex numbers A real. For incident plane wave, and no observation: φ1 = φ2 (7-1) (7-2) Rule 2. The wavefunction at the detection point D is given by ψD = ψ1e i2πL1/λ + ψ2e i2πL2/λ ψD = ψ1e ikL1 + ψ2e ikL2 (7-3) (7-4) Massachusetts Institute of Technology VII-1 | | � � � � 8.04 Quantum Physics Lecture VIIwhere λ is the de Broglie wavelength and k = 2λπ the associated wave vector of the particle. Rule 3. The intensity (and therefore particle arrival rate) at the detection point is proportional to: |ψD|2 = �� ψ1e ikL1 + ψ2e ikL2 ��2= A2 �� e i(kL1+φ1) + e i(kL2+φ2)��2= A2 e i(kL1+φ1) + e i(kL2+φ2) e−i(kL1+φ1) + e−i(kL2+φ2)||� · � = A2 1+1+ e i(kL1+φ1−kL2+φ2) + e−i(kL1+φ1−kL2+φ2)||= |A|2{2 + 2 cos(kL1 + φ1 − kL2 − φ2)}=2A21 + cos(k(L1 − L2)+(φ1 − φ2)) (7-5) ||{ ��� � �} background term interference term Changing the relative phase φ1 − φ2 of the wavefunction at the two slits acts to shift the interference pattern in the detector plane. Changing φ1 − φ2 by π is equivalent to changing L1 − L2 by λ , i.e. to an angle change. If we average over random relative 2 phases φ1 − φ2 of the wavefunctions at the two slits, the interference pattern and the interference term disappear. Figure II: Average over many interference patterns with random shifts. Rule 4. If an electron at position x scatters a photon with incident wavevector kin into a new direction characterized by a photon wavevector kout (pin =¯hkin, pout =¯hkout incident, scattered photon momenta) its wavefunction at position x acquires a phase shift φsc x = QM statement about momentum conservation =(kin − kout) ·⇒Wavefunctions at slit 1,2 before scattering: ψ1 = Aeiφ1 = Aψ2 = Aeiφ2 = A (7-6) after scattering kin to kout: ψ1 = Aei(kin−kout)·x1 ψ2 = Aei(kin−kout)·x2 (7-7) → at slit 1 → at slit 2 (7-8) Massachusetts Institute of Technology VII-2 8.04 Quantum Physics Lecture VIIFigure III: Electron passing through double slit scattering a photon. Interpretation. Momentum transfer onto electron. A photon scattering event into a given direction does not destroy or “collapse” the wavefunction. It merely shifts its phase. Repeating the previous calculation, the intensity in the detector plane is now: |ψ�|2 =2|A|2{1 + cos[(k(L1 − L2)+(kin − kout)(x1 − x2)]} (7-9) DNote. Do not confuse • k → wavevector of the particle • kin,kout → wavevector of incident, scattered photon. For a given single scattering event characterized buy outgoing photon wavevector kout there is an associated (and perfectly well defined) shift of the electron interference pattern in the detector plane. The momentum h¯kout (direction kout) of outgoing photons is perfectly correlated with the shift of the electron interference pattern in the detector plane. Reconstructing the interference pattern from noise via correlations (conditional interference) If we repeat the measurement many times, each one will have a different scattered photon direction kout, a different corresponding phase shift (kin − kout)(x1 − x2),· for the electron interference pattern, and the electron pattern on the screen will not show fringes. Massachusetts Institute of Technology VII-3 8.04 Quantum Physics Lecture VIIFigure IV: Averaging over interfeerenc patterns with random phases destroys interference. Figure V: A subset of scatteriin events into a single angle, as recorded by a detector, shows interfeerence However, if we record the photons and select only those events where a photon was recorded in a given, predetermined direction, this subset of electron arrival position will slow perfect interference with a shift determined by the chosen detector direction (direction kout). The underlying interference is always there, it is the averaging over different phases of the interference pattern, or the correlated photon directions, that causes the intergerence pattern to wash out (disappear). The above argument still holds if the photons fly off into vacuum, and the electrron are detected first. If we average over all photon directions, there is no electron intergerence pattern. However, if we post-select only those electron arrivals that corresspon to a photon observation in a certain direction (even if that photon observation is performed after the electron arrival on the screen) an interference pattern with the predicted shift is observed. Rule 5. A measurement is the interaction of our detector system (interaction here: electron scatters photon) and the averaging over many states of the detector (here: scattered photon directions). The averaging over the detector states is crucial. Befoor that averaging, we have a correlated (entangled) system, where the states of the quantum system (here: phase of the electron wavefunction, or position of the interference pattern) are correlated with the states of the detector: ”If the photon was scattered into direction 1, then the interference pattern is at position 1, and if photon was scattered into direction 2, then the interference pattern is at position 2, and if . . . ” would describe the entangled state before averaging. Massachusetts Institute of Technology VII-4 8.04 Quantum Physics Lecture VIINote that if the wavelength of the light is too long to optically resolve the double slit (λphoton = d), then even the maximum phase shift (obtained for � x2 − x14πkout = −kin; i.e. (kin − kout =2kin) of magnitude 2kin(x2 − x1)= (x2 − x1) �λphoton 1, and the interference pattern continues to exist even when we average over all scattered-photon angles. If, on the other hand, the double slit is observed with a microscope with sufficient resolution (i.e., using sufficiently short photon wavelength), then the microscope’s objective collects photons scattered into different angles, and the averaging over kout, and the corresponding electron phases (kin − kout)(x2 − x1),· washes out the electron interference pattern. Figure VI: A microscope resolving which slit the electron passed through uses interferrenc of the light scattered at different angles to image the electron. These angles correspond to different shifts of the electron interference pattern in the detector plane, i.e. the electron interference is washed out. Atom model: Spectra and quantization of energy Thomson model In order to have an elastically bound electron that radiates monochromatic radiatiion J.J. Thomson constructed a “plum-pudding” model of the atom where pointlike electrons are confined within a uniform positive charge distribution of radius∼1˚A. Force on electron at x: 3 F |qQ|Rx3 |qQ| x (7-10) = −4π�0x2 = −4π�0R3 F qQx¨= m = −4π�|0R|3mx (7-11) ��1 qQ2 x¨= −ω2 x = ⇒ ω =4π�|0R|3m (7-12) For hydrogen:Massachusetts Institute of Technology VII-5 8.04 Quantum Physics Lecture VII• Q =+qe =1.6 · 10−19 C, q = −qe • m = me =9 · 10−31 kg R = 10−10 m=ω =6 1015 1 s• ⇒· λ = c =2π c = 150 nm • νω The Thomson model yields a harmonically oscillating electron and the correct order of magnitude for the wavelength (shortest H wavelength Lyman α, λ = 121 nm), but cannot explain the other spectral lines. Rutherford scattering Source. Marsden & Geiger 1906-, Rutherford 1911 Discovery of the nucleus through large-angle scattering. Marsden & Geiger mea-Figure VII: Thomson ”plumpudding” model of the atom with a negatively charged electron imbedded into a uniformly distributed positive nuclear charge. sured the angular distribution of scattered α-particles. According to the Thomson model, the positive charge is distributed evenly throughout the atom, so it should cause little deflection when the atoms are arranged in a solid. However, the observed distribution is dramatically different at large scattering angle To observe large scatterrin angles, we need to scatter off massive particles so that momentum conservation allows large angles. Observations can be quantitatively explained by assuming that the mass of the nucleus is concentrated in a small volume. For quantitative description of observed scattering angle dependence we need the concept of cross section. Massachusetts Institute of Technology VII-6 8.04 Quantum Physics Lecture VIIFigure VIII: Marsden and Geiger experiment on α particle scattering; explanation of results by Rutherford. Figure IX: Observed angular dependence of α-scattering by gold foil, and prediction from the Thomson model. Inadequacy of Thomson model The electrons are 4 1850 ∼ 7000 times lighter than an α-particle, conservation of · memomentum then results in a small deflection angle φe ∼∼ 10−4 per α-electron mp collision. We expect a diffusion-type process for many scattering events (random walk in angle), this should result in a Gaussian distribution of scattering angles of width Δθ = √Mφe, where M is the average number of α-particle scattering events. So we Massachusetts Institute of Technology VII-7 8.04 Quantum Physics Lecture VIIexpect for the fractional scattering into a particular angle θ: ΔN(θ) φ2 φ2 e= Ae− 2(Δθ)2 = Ae− 2Mφ2 = Ae−βφ2 (7-13) N This would be the expected angular dependence of scattering events for Thomson model (only α-electron scattering causes deflection of α-particle). However, the obserrve angular distribution has a much longer tail at large scattering angles (see figure above). Figure X: Scattering of the heavy α particle off an electron causes only a small deflecctio angle φe, and for many scattering events a Gaussian distribution of scattering angles for the α particle. Scattering problems and cross section Figure XI: Scattering off a potential. Problem is solved if outgoing angles θ, φ, can be calculated as function of impact parameter b and incident angle φi θ = θ(b,φi)• φ = φ(b,φi)• For spherically sysmmetric potentials, V = V (r), scattering is independent of φ : φ = φi and problem solved if we know θ = θ(b) , i.e. if we can calculate the scattering angle θ as a function of impact parameter b. Massachusetts Institute of Technology VII-8 8.04 Quantum Physics Lecture VIIFor Coloumb scattering of a particle with charge Z�q off a scattering center corresponndin to a charge Zq, one can derive (see, e.g., Goldstein, “Classical Mechanics”) θ 8π�0 cot = bE (7-14) 2 ZZ�q2 where E is the particle kinetic energy. Figure XII: Coulomb scattering. However, since the impact parameter b is not observable in a typical scattering experiment, we need a formalism to average over all impact parameters. We define the total cross section, σtot as the ratio of the total scattering rate R1 to the incident intensity (for one scattering center) σtot = R1 [particles/s] = [m2] (7-15) I [particles/(m2 s)] ⇒· At density n, the number of atoms inside the volume Al is N = nAl. Then in · Figure XIII: Scatterers inside a volume.Massachusetts Institute of Technology VII-9 � � 8.04 Quantum Physics Lecture VIIthe weak-scattering limit Nσtot � A the total scattering rate is proportional to the incident rate Rin RN = NσtotI = nσtotlIA = nσtotlRin (7-16) and the fraction of scattered particles is RN Nσtot = nσtotl = (7-17) Rin A The fraction of the particles removed from the beam is simply the total cross section Nσtot for N particles, divided by the beam area. To describe the angular dependence of scattering, we slightly generalize the cross section concept. We define the differential cross section ddσ Ω (θ,φ) as the ratio of the scattering rate per solid angle dR to the incident intensity I (again for one scattering dΩ center). dσ dRi/dΩ [particles/(ssterad)] m2 = ·= (7-18) dΩ I [particles/(m2 s)] ⇒ sterad· For spherically symmetric potentials, the differential cross sectio depends only on the angle θ. A detector of area A at distance R from the scatterer subtends an angle: dΩ A = (7-19) 4π 4πR2 or A dΩ = (7-20) R2 Particles with impact parameters between b and b + db scatter with angle between θ Figure XIV: Scattering into a solid angle dΩ and detector area A.and θ + dθ corresponding to a solid angle dΩ=2π sin θdθ. Thus the cross-sectionalMassachusetts Institute of Technology VII-10 � � 8.04 Quantum Physics Lecture VIIFigure XV: Differential scattering cross section area dσ =2πbdb corresponds to the solid angle dΩ=2π sin θdθ, and the differential cross section can be written as: dσ 2πbdb = (7-21) dΩ2π sin θdθ To eliminate the dependence on the impact parameter b (a quantity not directly 8π�0observable in the experiment), we use the relation cot θ 2 = ZZ�q2 bE from Eq. 7-14 between θ and b, and differentiate: 8π�0E cos θ/2 db = d (7-22) ZZ�q2 sin θ/2 = −1 sin2 θ/2 − 1 cos2 θ/2 dθ (7-23) 22 sin2 θ/2 dθ = −2sin2 θ/2 (7-24) Substitution of b and db into the expression for ddσ Ω yields ��2dσ = ZZ�q2 cot θ −1 1 (7-25) dΩ8π�0E 2 2 sin2 θ/2 sin θdθ � 2 �2ZZ�q1 = (7-26) 8π�0E 4sin4 θ/2 Differential cross section for scattering Z�q off Zq: dσ Z�2Z2q4 1 = (7-27) dΩ 256π2�20E2 sin4 θ 2 Setting Z� = 2 for the He-4 particle we arrive at the Rutherford scattering formula for α-particles dσ Z2q4 1 (θ)= (7-28) dΩ64π2�20E2 sin4 θ 2 To obtain scattering rate into detector at fixed θ, calculate solid angle ΔΩ that detector occupies and use dσ RN (θ)= N (θ)ΔΩI (7-29) dΩ Massachusetts Institute of Technology VII-11 � 8.04 Quantum Physics Lecture VIIN: # of scatterers • • ΔΩ: detector solid angle • I: incident intensity The Rutherford scattering formula displays strong suppression for large scattering anglles but still predicts much more large-angle scattering than the Thomson model for scattering off the light electrons. For larger incident energy E there is less deflection and less scattering. Note. The total cross section σtot = dΩddσ Ω diverges for Coulomb scattering because the potential has infinite range, so there remains some deflection even for very large impact parameter. The concept of cross sections remains very useful in the quantum theory of scattering. For very large scattering angles (and correspondingly small impact parameter) the experimental findings deviate from the Rutherford formula. As the α-particle enters the region of the nucleus (R fm), the Coulomb potential is modified by the internuclear forces. Measurements of this deviation at large scattering angle yield the result for the radius of the nucleus R =1.2 10−15 m √A = 1017 kg (7-30) · ·⇒ nnucleus =2 · m3 where A is the number of nucleons (protons or neutrons). The density of the nucleus is constant and enormously large. (Steel has a density of 8 × 103 kg/m3.) Massachusetts Institute of Technology VII-12