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Contents: Copenhagen Interpretation of Young’s Double slit The Quantum atom Heisenberg uncertainty principle The Einste

Quantum mechanics and the Copenhagen Interpretation. Contents: Copenhagen Interpretation of Young’s Double slit The Quantum atom Heisenberg uncertainty principle The Einstein Bohr debate Quantum Mechanics. Copenhagen Interpretation – Demo laser.

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Contents: Copenhagen Interpretation of Young’s Double slit The Quantum atom Heisenberg uncertainty principle The Einste

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  1. Quantum mechanics and the Copenhagen Interpretation • Contents: • Copenhagen Interpretation of Young’s Double slit • The Quantum atom • Heisenberg uncertainty principle • The Einstein Bohr debate • Quantum Mechanics

  2. Copenhagen Interpretation – Demo laser • Electrons interfere even when sent one at a time (why?) • Copenhagen:  = Schrödinger wave function of electron • “Probability waves” interfere (2 = probability) TOC

  3. Heisenberg Uncertainty Principle Let’s find an electron Photon changes the momentum of electron x   ph/ (smaller , bigger p) xp>h/4 x – uncertainty of position p - uncertainty of momentum Et>h/4 E - uncertainty of energy (Range) t - uncertainty of time (Range) if x = 3.5 ± 0.2 m, x = 0.4 m – if you are using h/4π TOC

  4. Heisenberg Uncertainty Principle xp>h/4 Et>h/4 • Strange quantum effects: • Observation affects reality • Energy is not conserved (for t) • Non determinism • Quantum randomness • Quantum electrodynamics Example: What is the uncertainty in the position of a 0.145 kg baseball with a velocity of 37.0 ± 0.3 m/s? TOC

  5. Example: What is the uncertainty in the position of a 0.145 kg baseball with a velocity of 37.0 ± 0.3 m/s. The uncertainty of momentum is (0.145 kg)(0.6 m/s) = 0.0870 kg m/s And now we use xp> h/4: x(0.0870 kg m/s) > (6.626E-34 Js)/(4), x> 6.1E-34 m so x is > 6.1E-34 m So not really very much.

  6. HL Paper 2 B4 3 d, p. 36 2012 ???????? They use ½ Range…

  7. xp>h/4 Et>h/4 Example: The electron stays in the first excited state of hydrogen for a time of approximately Δt = 1.0 x 10-10 s Determine the uncertainty in the energy of the electron in the first excited state.

  8. HL Paper 2 B3 1 e, p. 34 2011

  9. Whiteboards: Heisenberg Uncertainty 1 | 2 TOC

  10. For what period of time is the uncertainty of the energy of an electron ΔE = 2.5 x 10-19 J? Et>h/4 (2.5 x 10-19 J)t>h/4 t = 2.1 x 10-16 s W t = 2.1 x 10-16 s

  11. You know an electron’s position is ±.78 nm, what is the minimum uncertainty of its velocity? (4) xp>h/4 p = mv m = 9.11 x 10-31 kg x = 2x0.78 x 10-9 m = 1.56E-9 m (1.56 x 10-9 m)p>(6.626 x 10-34 Js)/4 p =3.38 x 10-26 kg m/s p = mv (3.38 x 10-26 kg m/s) = (9.11 x 10-31 kg)v v = 37102 ≈ 3.7 x 104 m/s W v = 3.7 x 104 m/s

  12. The Einstein-Bohr debate • Einstein objected to quantum randomness • “God does not play dice” • Attacked either Heisenberg uncertainty, or complementarity TOC

  13. The Einstein-Bohr debate Einstein would challenge Bohr at conferences Front: Bohr, Heisenberg, Pauli, Stern, Meitner, Ladenburg TOC For example…

  14. Gedanken experiment (to disprove complementarity) s d • Detect which slit the electron went through with light beam (particle behaviour) • If interference pattern appears, then we have both wave and particle behaviour • Complementarity says it must be either Electron beam TOC

  15. Bohr would take a walk E. Fermi, N. Bohr

  16. Bohr’s reply s d • No interference pattern would happen • The light that detected the electron would change its momentum • To have interference, electrons must be monochromatic •  = h/p • Complementarity is intact Electron beam TOC

  17. Quantum mechanics Bohr always prevailed (God apparently does play dice) • Three “types” of physics • Newtonian/classical - big stuff • Relativity - speeds close to c • Quantum mechanics - physics of the atom • Correspondence principle TOC

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