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Topic 5: Schrödinger Equation

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

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Topic 5: Schrödinger Equation

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  1. 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 HarmonicOscillator Potential (or parabolic) •  is more complex, E  n Example Infinite Well Solution

  2. Schrödinger Equation • 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 • Appendix: Complex Number Tutorial

  3. Wave Equation for Photons: Electric Field E 2nd space derivative 2nd time derivative Propose Solution: Calculate Derivatives: After Substitution: 

  4. Schrödinger Eqn. for Electrons+: Wave FunctionY 2nd space derivative 1st time derivative Propose Simple Solution for constant V: Calculate Derivatives: After Substitution: 

  5. Schrödinger Equation: Applications Now, find theeigenfunctions Y and eigenvalues E of the Schrödinger Equation for a particle interacting with different potential energy shapes. (assume no time dependence) • 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).

  6. Schrödinger Equation: Definitions Wave function  has NO PHYSICAL MEANING! BUT, the probability to find a particle in width dx is given by: • Normalization of  • Probability to find particle in all space must equal 1. • Solve for Y coefficients so that normalization occurs.

  7. Infinite Square Well Potential: Visual Solutions Wave and Probability Solutions Energy Solutions Yn(x) Yn2(x) n = 3 n = 2 n = 1

  8. Infinite Square Well: Solve general Y from S.Eqn. Inside Well: (V = 0) where Oscillatory ycannot penetrate barriers! Outside Well: (V = )

  9. Infinite Square Well: Satisfy B.C. and Normalization Satisfy boundary conditions Quantized Energy Solutions • Satisfy normalization using identity and Wave Solutions

  10. Finite Square Well Potential: Visual Solutions E Vo E3 E2 E1 Wave and Probability Solutions Energy Solutions “leaks” outside barrier High energy particles “escape” Yn(x) Yn2(x) n = 3 n = 2 n = 1 Energy vs. width: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07a.html Energy vs. height: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07b.html

  11. Finite Square Well: Solve general Y from S.Eqn. Inside Well: (V = 0) where Oscillatory Outside Well: (V = Vo) where ycan penetrate barriers! Decaying

  12. Finite Square Well: Example Problem (a) Sketch the wave function(x)for the n = 4 state for the finite square well potential. (b) Sketch the probability distribution2(x).

  13. Finite Square Well: Example Problem Sketch the wave function(x) corresponding to a particle with energy E in the potential well shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 2. • (x) oscillatesinside the potential well because E > V(x), and decays exponentially outside the well because E < V(x). • The frequency of (x) is higher in Region 1 vs. Region 2 because the kinetic energy is higher [Ek = E - V(x)]. • The amplitude of (x) is lower in Region 1 because its higher Ek gives a higher velocity, and the particle therefore spends less time in that region.

  14. Simple Harmonic Well Potential: Visual Solutions Wave and Probability Solutions Yn2(x) (different well widths) Yn(x) Energy Solutions n = 2 n = 1 n = 0

  15. Simple Harmonic Well: Solve Y from S.Eqn. NEW! Inside Well: where Y(x) is not a simple trigonometric function. Outside Well: where Y(x)is not a simple decaying exponential.

  16. Step Potential: Y(x) outside step Outside Step: V(x) = 0 where Y(x) is oscillatory Case 1 Case 2 Energy Y(x)

  17. Step Potential: Y(x) inside step Inside Step: V(x) = Vo where Y(x) is oscillatory for E > Vo Y(x) is decaying for E < Vo Case 1 Case 2 Energy E > Vo E < Vo Y(x) 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 - various: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05b.html

  18. Step Potential: Reflection and Transmission • At a step, a particle wave undergoes reflection and transmission (like electromagnetic radiation!) with probability rates R and T, respectively. R(reflection) + T(transmission) = 1 • Reflection occurs at a barrier(R  0), regardless if it is step-downor step-up. • R depends on the wave vector difference (k1 - k2) (or energy difference), but not on which is larger. • Classically, R = 0 for energy E larger than potential barrier (Vo).

  19. Step Potential: Example Problem A free particle of mass m, wave number k1 , and energyE = 2Vo is traveling to the right. At x = 0, the potentialjumps fromzero to –Voand remains at this value for positive x. Find the wavenumber k2 in the region x > 0 in terms of k1 and Vo. In addition, find the reflection and transmission coefficients R and T.

  20. Barrier Potential where Outside Barrier: V(x) = 0 Y(x) is oscillatory where Inside Barrier: V(x) = Vo Y(x) is decaying Energy Transmission is Non-Zero! Y(x) http://www.sgi.com/fun/java/john/wave-sim.html Single Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_12c.html

  21. 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) 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. 3 because kinetic energy is higher there. • 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). Non-resonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-5.html Resonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-6.html Double Barrier + : http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-0.html

  22. Scanning Tunneling Microscopy: Schematic Tip • STM is based upon quantum mechanical tunneling of electrons across the vacuum barrier between a conducting tip and sample. • To form image, tip is raster-scanned across surface and tunneling current is measured. VDC Bias voltage e- Distance s Constant current contour Sample e e e e Tunneling current  e -2ks

  23. STM: Ultra-High Vacuum Instrument Coarse Motion • Well-ordered, clean surfaces for STM studies are prepared in UHV. • Sample is moved towards tip using coarse mechanism, and the tip is moved using a 3-axis piezoelectric scanner. Sample Scanner Tip

  24. STM: Data of Si(111)7×7 Surface • STM topograph shows rearrangement of atoms on a Si(111) surface. • Adatoms appear as bright “dots” when electrons travel from sample to tip. empty STM 7×7 Unit 18 nm 7 nm = adatom

  25. Expectation Values and Operators By definition, the “expectation value” of a function is: • “Operate” on Y(x) to find expectation value (or average expected value) of an “observable.” ObservableSymbolOperator Position Momentum Kinetic energy Hamiltonian Total Energy x p K H E x

  26. Expectation Values: Example Problem Find <p>, <p2> for ground state Y1(x) of infinite well (n = 1) L 0 -L/2 +L/2 <p> = <p> = 0 by symmetry(odd function over symmetric limits) Note: The average momentum goes to zero because the “sum” of positive and negative momentum values cancel each other out.

  27. Expectation Values: Example Problem, cont. <p2> = = 1 by normalization <p2> =

  28. Complex Number Tutorial: Definitions • Imaginary number i given by:i2 = –1( i3 = –i, i4 = 1, i–1 = –i ) • Complex number z is composed of a real and imaginary parts. Cartesian Form: z = x + iy Polar Form: z = r(cosq+ i sinq) where r = (x2 + y2)1/2 and tanq = y/x Exponential Form: z = rei q Conjugate: z* = x– iy = rcosq– i rsinq = re– i q where (z*)(z) = (x – iy)(x + iy) = x2 + y2 (real!)

  29. Complex Number Tutorial: Taylor Series • Proof of equivalence for polar and exponential forms:

  30. Schrödinger Eqn.: Derivation of Space & Time Dependence Divide by y(x)0(t) Schrödinger Equation is 2nd Order Partial Differential Equation Assume y is separable [i.e. V(x) only] Substitution of y Partial derivatives are now ordinaryderivatives Space dependence ONLY Time dependence ONLY

  31. Schrödinger Eqn.: Derivation of Space & Time Dependence Left and right sides have only space (x) and time (t) dependence now Space: Set each side of equation equal to a constant C Time: Space Equation  need V(x) to solve! Check by substitution! Time Solution:

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