VIEWS: 27 PAGES: 24 POSTED ON: 11/10/2009 Public Domain
Topic 5: Schrödinger Equation • Wave equation for Photon vs. Schrödinger equation for Electron+ • Solution to Schrödinger Equation gives wave function – 2 gives probability of finding particle in a certain region • Square Well Potentials: Infinite and Finite walls – oscillates inside well and is zero or decaying outside well, E n2 • Simple Harmonic Oscillator Potential (or parabolic) – is more complex, E n • Step Potential of Height V0 – is always affected by a step, even if E > V0 – For E > V0, oscillates with different k values outside/inside step. – For E < V0, oscillates outside step and decays inside step. • Barrier Potential of Height V0 – oscillates outside and decays inside barrier. • Expectation Values and Operators Phys320 - Baski Page 1 Schrödinger Equation: Applications • Now, find the eigenfunctions and eigenvalues E of the Schrödinger Equation for a particle interacting with different potential energy shapes. 2 ( x) 2 2m x 2 V ( x) ( x) E ( x) • Possible potential energies V(x) include: • Infinite and Finite square wells (bound particle). • Simple Harmonic or parabolic well (bound particle). • Step edge (free particle). • Barrier (free particle). Phys320 - Baski Page 2 Infinite Square Well Potential: Visual Solutions Wave and Probability Solutions n=3 Energy Solutions n(x) n2(x) n=2 n=1 2 2 k2 En n2 2m 2mL2 2 2 nx n x sin L L Phys320 - Baski Page 3 Infinite Square Well : Probability Example An electron is in the n = 3 excited state of a 1-D infinite square well (width L). Draw the wave function and probability distribution of the electron. Solve the probability of finding the electron at x = 0.5L in a width Dx = 0.02L. (x) P(x) L/2 L x L/2 L x Because Dx is so small, use P(x)Dx = 2 (x)Dx with for infinite well potential. Pn 3 ( x)Dx 2 3 ( x)Dx n Pn 3 (x L )Dx 2 Phys320 - Baski 2 2 3 x sin Dx L L 0.04 Page 4 2 2 3 L 3 sin (0.02 L) 0.04 sin 2 L 2L 2 Simple Harmonic Well Potential: Visual Solutions Wave and Probability Solutions n=2 n(x) n2(x) (different well widths) Energy Solutions n=1 n=0 1 En n 2 Page 5 Phys320 - Baski Step Potential: (x) inside step Inside Step: V(x) = Vo " x k2 x 2 where (x) is oscillatory for E > Vo (x) is decaying for E < Vo 2m k22 2 E Vo Case 1 Energy Case 2 E > Vo (x) E < Vo Scattering at Step Up: http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_06b.html Scattering at Well - wide: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05d.html Scattering at Well Phys320 - Baski - various: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05b.html 6 Page Barrier Potential Outside Barrier: x k1 x V(x) = 0 (x) is oscillatory " 2 where 2m k12 E 2 Inside Barrier: x k2 x where V(x) = Vo (x) is decaying " 2 k2 2 2m 2 Vo E 0 Energy Transmission is Non-Zero! (x) Te 2 k2 a http://www.sgi.com/fun/java/john/wave-sim.html Single Barrier: Phys320 - Baski http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_12c.html Page 7 Barrier Potential: Example Problem Sketch the wave function (x) corresponding to a particle with energy E in the potential shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 3. (x) Region 2 E Vo Region 1 Region 3 x • (x) oscillates in regions 1 and 3 because E > V(x), and decays exponentially in region 2 because E < V(x). • Frequency of (x) is higher in Region 1 vs. Region 3 because kinetic energy is higher there [Ek = E - V(x)]. • Amplitude of (x) in Regions 1 and 3 depends on the initial location of the wave packet. If we assume a bound particle in Region 1, then the amplitude is higher there and decays into Region 3 (case shown above). Phys320 - Baski Page 8 Topic 6: Atomic Physics • Hydrogen Atom: 3D Spherical Coordinates – = (spherical harmonics)(radial) and probability density P – E, L2, Lz operators and resulting eigenvalues • Angular momenta: Orbital L and Spin S – Addition of angular momenta – Magnetic moments and Zeeman effect – Spin-orbit coupling and Stern-Gerlach (Proof of electron spin s) • Periodic table – Relationship to quantum numbers n, l, m – Trends in radii and ionization energies Phys320 - Baski Page 9 Group IV Group VI Group III n 1 2 3 4 5 6 7 l = 0 (s) l = 1 (p) l = 2 (d) l = 3 (f) Phys320 - Baski Group V Page 10 Noble Gas Alkali Halogen Periodic Table Hydrogen Atom: 3D Spherical Schrödinger Equation “Rewritten” Schrodinger Eqn.: ˆ p2 r , , Q Veff r , , Q En r , , Q 2 where ˆ p 2 2 1 2 2 r r r r Eigenfunctions: nlm r , , Rn r Ylm , Laguerre Polynomials Spherical Harmonics Eigenvalues: z E0 En n2 2 where E0 1 ke 2 2 2 13.6eV Page 11 Phys320 - Baski Hydrogen Atom: 3D Spherical Schrödinger Equation 3 Quantum Numbers (3-dimensions) n = energy level value (average radius of orbit) n = 1, 2, 3 … l = angular momentum value (shape of orbit) l = 0, 1, 2, … (n – 1) m = z component of l (orientation of orbit) m = -l,(-l + 1) ..0,1,2,..+l How many quantum states (n, l, m) exist for n = 3? Is there a general formula? Phys320 - Baski Page 12 Probability Density: Cross Sections http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter40/multiple3/deluxe-content.html Rank the states (1s to 3d) from smallest to largest for the electron’s most PROBABLE radial position. For which state(s) do(es) the most probable value(s) of the electron's position agree with Phys320 - Baski the Bohr model? Page 13 Orbital Momentum L: Vector Diagram For l = 2, find the magnitude of the angular momentum L and the possible m values. Draw a vector diagram showing the orientations of L with the z axis. l2 L L l (l +1) 6 2(2+1) 2h h h 2h z L = 6h = 2.45h m=2 55º 24º or 2.45 m=1 m=0 m l to l 0, 1 2 Lz m 0, 1 , 2 m = 1 m = 2 Can you draw the vector diagram for l = 3? For j = 3/2? Phys320 - Baski Page 14 Angular Momentum Addition: General Rules • General Case: J1 J 2 Vectors J tot J1 J 2 J tot j j 1 Quantum Numbers j j1 j2 , j1 j2 1,... j1 j2 m j j , j 1,...j 1, j Example: j1 = 3/2, j2 = 3/2 jmax 3 3 3 and jmin 2 2 j 3, 2, 1, 0 3 2 3 0 2 m j 3, 2, 1, 0,1, 2, 3 for j 3 Phys320 - Baski Page 15 “Anomalous” Zeeman Effect: Spin-Orbit + Zeeman Spin-Orbit j = 3/2 l = 1 s = 1/2 Zeeman mj = 3/2 mj = 1/2 mj = –1/2 mj = –3/2 mj = 1/2 mj = –1/2 j = 1/2 mj = 1/2 l = 0 s = 1/2 j = 1/2 Phys320 - Baski mj = –1/2 Page 16 • Quantum numbers mj (j-j coupling) for HIGHER Z elements. Topic #7: Solid State Physics • Types of Solids – Ionic, Covalent, and Metallic. • Classical Theory of Conduction – Current density j, drift velocity vd, resistivity . • Band Theory and Band Diagrams – Energy levels of separated atoms form energy “band” when brought close together in a crystal. – Fermi Function for how to “fill” bands. – Metal, Insulator, and Semiconductor band diagrams. – Donor and Acceptor dopants (Hall Effect). • Devices – pn junction, diode, LED, solar cell, laser. Phys320 - Baski Page 17 Classical Theory of Conduction: Resistivity vs. Temp. • Temperature dependence of resistivity. E e e m 1 J ne vd ne (a ) n ne2 • Metal: Resistance increases with Temperature. FE ma • Why? Temp , n same (same # conduction electrons) • Semiconductor: Resistance decreases with Temperature. • Why? Temp , n (“free-up” carriers to conduct) Phys320 - Baski Page 18 Band Theory: “Bound” Electron Approach • For the total number N of atoms in a solid (1023 cm–3), N energy levels split apart within a width DE. – Leads to a band of energies for each initial atomic energy level (e.g. 1s energy band for 1s energy level). Two atoms Six atoms Solid of N atoms Electrons must occupy different energies due to Phys320 - Pauli Exclusion principle. Baski Page 19 Band Diagram: Metal “Fill” the energy band with electrons. EC,V Fermi “filling” function Energy band to be “filled” EC,V EF T=0K Moderate T EF • At T = 0, energy levels below EF are filled with electrons, while all levels above EF are empty. • Electrons are free to move into “empty” states of conduction band with only a small electric field E, leading to high electrical conductivity! • At T > 0, electrons have a probability to be thermally “excited” from below the Fermi energy to above it. Phys320 - Baski Page 20 Band Diagram: Semiconductor with moderate Egap T>0 Conduction band (Partially Filled) EF Valence band (Partially Empty) EC EV • At T = 0, valence band is filled with electrons and conduction band is empty, leading to zero conductivity. • At T > 0, electrons thermally “excited” from valence to conduction band, leading to partially empty valence and partially filled conduction bands. What happens to the conductivity for T > 0? How - Baski Phys320 would the band diagram look for lower & higher temperatures? Page 21 Band Diagram: Donor Dopant in Semiconductor • Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) • For group IV Si, add a group V element to “donate” an electron and make n-type Si (more negative electrons!) • “Extra” electrons donated from donor energy level ED just below EC. – Resultant electrons in conduction band increase conductivity by increasing free carrier density n. n-type Si EC EF EV ED • Fermi level EF moves up because there are more carriers. Fermi Baski Phys320 -Function & Doping: http://jas.eng.buffalo.edu/education/semicon/fermi/bandAndLevel/fermi.html Page 22 pn Junction: Band Diagram • At equilibrium, Fermi levels (or charge carrier densities) must equalize. • Hence, electrons move from n to p-side (diffusion process). • Depletion zone occurs at junction where immobile charged ion cores remain. • Results in a built-in electric field (103 to 105 V/cm), which opposes further diffusion. pn regions “touch” & free carriers move EC EF EV n-type electrons EF p-type pn regions in equilibrium EC EF EV –– – ++–– – + + – + ++–– – + ++–– ++ Depletion Zone Page 23 PIN junction: Phys320 - Baski http://jas.eng.buffalo.edu/education/pin/pin/# Semiconductor: Dopant Density via Hall Effect • Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a semiconductor. • How? Place semiconductor into external B field, push current along one axis, and measure induced Hall voltage VH along perpendicular axis. Carrier density n = (current I) (magnetic field B) (carrier charge q) (thickness t)(Hall voltage VH) • Derived from Lorentz equation FE (qE) = FB (qvB). Hole + charge FB qv B Electron – charge Phys320 - Baski Page 24