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

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

Heisenberg Uncertainty Principle TOC Let’s find an electron Photon changes the momentum of electron  x   p  h / (smaller, bigger p) xp > h/4xp > h/4  x – uncertainty of position  p - uncertainty of momentum Et > h/4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

Heisenberg Uncertainty Principle TOC xp > h/4Et > h/4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 kg baseball with a velocity of 37.0 ± 0.3 m/s?

Example: What is the uncertainty in the position of a kg baseball with a velocity of 37.0 ± 0.3 m/s. The uncertainty of momentum is (0.145 kg)(0.6 m/s) = kg m/s And now we use  x  p > h/4  :  x( 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.

HL Paper 2 B4 3 d, p ???????? They use ½ Range…

xp > h/4Et > h/4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 s Determine the uncertainty in the energy of the electron in the first excited state.

HL Paper 2 B3 1 e, p

Whiteboards: Heisenberg Uncertainty 11 | 22 TOC

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

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

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

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

Gedanken experiment (to disprove complementarity) TOC 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 s d Electron beam

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

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

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