1. P460 - dev. wave eqn.1 Developing Wave Equations Need wave equation and wave function for particles. Schrodinger, Klein-Gordon, Dirac not derived. Instead forms were guessed at, then solved, and found where applicable E+R Ch 3+5 and Griffiths Ch 1 try to show why choice is reasonable…….. Start from 1924 DeBroglie hypothesis: “particles” (those with mass as photon also a particle…)have wavelength = h/p What is wavelength of K = 5 MeV proton ? Non-rel p=sqrt(2mK) = sqrt(2*938*5)=97 MeV/c =hc/pc = 1240 ev*nm/97 MeV = 13 Fermi p=50 GeV/c e or p gives.025 fm (size of p: 1 F)

2. P460 - dev. wave eqn.2 Wave Functions Particle wave functions are similar to amplitudes for EM waves…gives interference (which was used to discover wave properties of electrons) probability to observe =|wave amplitude| 2 =|  x,t)| 2 particles are now described by wave packets if  = A+B then  2 = |A| 2 + |B| 2 + AB* + A*B giving interference. Also leads to indistinguishibility of identical particles t1 t2 vel= - (t2-t1) merge Can’t tell apart

3. P460 - dev. wave eqn.3 Wave Functions Describe particles with wave functions  x) =  a n sin(k n x) Fourier series (for example) Fourier transforms go from x-space to k-space where k=wave number= 2 . Or p=hbar*k and Fourier transforms go from x-space to p-space position space and momentum space are conjugate the spatial function implies “something” about the function in momentum space

4. P460 - dev. wave eqn.4 Heisenberg Uncertainty Relationships Momentum and position are conjugate. The uncertainty on one (a “measurement”) is related to the uncertainty on the other. Can’t determine both at once with 0 errors p = hbar k electrons confined to nucleus. What is maximum kinetic energy?  x = 10 fm  p x = hbarc/(2c  x) = 197 MeV*fm/(2c*10 fm) = 10 MeV/c while = 0 Ee=sqrt(p*p+m*m) =sqrt(10*10+.5*.5) = 10 MeV electron can’t be confined (levels~1 MeV) proton Kp =.05 MeV….can be confined

5. P460 - dev. wave eqn.5 Heisenberg Uncertainty Relationships Time and frequency are also conjugate. As E=hf leads to another “uncertainty” relation atom in an excited state with lifetime  = 10 -8 s |  t)| 2 = e -t/  as probability decreases  t) = e -t/2  e iM (see later that M = Mass/energy)  t ~  E = h    6 s -1  is called the “width” or and can be used to determine ths mass of quickly decaying particles ·if stable system no interactions/transitions/decays

6. P460 - dev. wave eqn.6 Bohr Model From discrete atomic spectrum, realized something was quantized. And the bound electron was not continuously radiateing (as classical physics) Bohr model is wrong but gives about right energy levels and approximate atomic radii. easier than trying to solve Scrod. Eg…. Quantized angular momentum (sort or right, sort of wrong) L= mvr = n*hbar n=1,2,3... (no n=0) kinetic and potential Energy related by K = |V|/2 (virial theorem) gives ·a 0 is the Bohr radius =.053 nm = ~atomic size

7. P460 - dev. wave eqn.7 Bohr Model E n = K + V = E 0 /n 2 where E 0 = -13.6 eV for H. = -m/2*(e*e/4  hbar) 2 Bohr model quantizes energy and radius and 1D angular momentum. Reality has energy, and 2D angular momentum (one component and absolute magnitude) for transitions easily extend Bohr model. He + atom, Z=2 and E n = 4*(-13.6 eV)/n 2 (have (zZ) 2 for 2 charges) reduced mass  =mamb/(ma+mb) if other masses E n =  /(me )*E 0 (zZ/n) 2 Atom mass E(n=1 ) e p.9995me -13.6 eV  p 94 MeV 2.6 keV  MeV 1.6 keV bb quarks q=1/3 2.5 GeV.9 KeV