Quantum Mechanics Part 2A PDF

Part 2A 1 Outline 2.1 The behavior of wave at the boundary , y 2.2 Confining a particle 2.3 Schrodinger Equation : Free particle (1D) 2.4 Schrodinger Equation: Probability Density, Probability of detection and Normalization (1D)...

Part 2A 1 Outline 2.1 The behavior of wave at the boundary , y 2.2 Confining a particle 2.3 Schrodinger Equation : Free particle (1D) 2.4 Schrodinger Equation: Probability Density, Probability of detection and Normalization (1D) p(x) p(x , x) 2 2.1 Behavior of wave at the boundary In classical physics (nonrelativistic, nonquantum): In nonrelativistic quantum mechanics: O The behavior of a particle can be explained according The basic equation to be solved is a second-order differential to Newton’s laws. equation known as the Schrodinger equation. 1) Particle interaction: describe in terms of force. 1) Particle interaction: describe in terms of potential energy (rather than the force). 2) According to Newton’s law, some relationships of 2) Schrodinger Eq.’s solution gives the wave function of the particle, e.g. location (x) and velocity (v), can be particle, which carries information about the behaviors of wave obtained. particle. 3) F = ma, predicts how an object’s behavior changes Example of information: - The probability density of finding the particle due to applied forces. - The solutions of the time-independent Schrödinger Eq. for electrons in an atom correspond to quantized energy levels. - Quantum tunneling of a particle that tunnels through an energy barrier that would be classically impossible. 3 Note 1: Wave function of the particle carries information about the behaviors of wave particle. E For example: Free Electrons: Not bound to any atom and is not under the influence of external forces (fields). The wavefunction of a free e- in 1D can be described by solution of the time-independent Schrödinger equation for a free particle. (no potential acting on the electron) (angular wavenumber) -> 2π comes from expressing the wave's oscillation in terms of right-moving left-moving radians. -> Unit, e.g., radians per meter. -> wavenumber without the angular (radian) factor. -> Unit - e.g., cycles per meter. 4 2.1 Behavior of wave at the boundary X !" = !# $" < $# de Broglie waves of electrons Light wave in air Surface wave a) The wavelength (!) in air in a) Electrons moving from ‘region 1’ of a) In shallower depth, its region 3 is the same as the original constant zero potential to ‘region 2’ of wavelength is smaller, but its wavelength of the incident wave in constant negative potential "! also have amplitude is larger. region 1. transmitted and reflected components. b) At region 3 (the same depth as in b) The amplitude (A) in region 3 is 1), the λ returns to its original value, b) At ‘region 3’, =? less than the amplitude in region 1, but the amplitude of the wave in 3 is because some of the intensity is smaller (some intensity was reflected). ‘ we focus on here.’ reflected at A and at B. (Note: Amplitude of Tsunami Propagation) 5 2.1 Behavior of wave at the boundary 1) In region 1, electrons move inside a narrow metal rod that is at ground potential (V = 0). (Assume one electron first) 2) Region 2 is connected to the negative terminal of a battery, keeping it at uniform potential of −&$. 3) Region 3 is connected to region 1 at ground potential. X 4) The gaps between the tubes be made so small that we can regard the changes in potential at A and B as occurring suddenly. % In region 1: Electron has ’kinetic energy (()’, ‘momentum (* = +,()′ and ‘de Broglie wavelength (. = &)’. In region 2: Potential energy for the electron is / = 0". (Assume that K in region 1 is higher than U) Electrons move into In region 2 with a smaller kinetic energy, smaller momentum, and thus greater wavelength. (1 ∝ 3) In region 3: Electron gains back the lost kinetic energy and moves with its original kinetic energy K and thus with their original wavelength. Similar to the light wave or the water wave, the amplitude of the de Broglie wave in !" = !# region 3 is smaller than in region 1 because some of the electrons are reflected. $" < $# 6 2.1 Behavior of wave at the boundary In region 2: Potential energy for the electron is / = 0" D' X Like light waves, at suitable (( > /), electron can go through the forbidden region with decreasing amplitudes. !" = !# $" < $# Suppose we increase the battery voltage so that the potential In some case, the penetration of the electrons (wave- energy in region 2 is greater than the initial kinetic energy like behavior) into the forbidden region is also possible in region 1 ( < /. and is related to the uncertainty principle, i.e., the probability of tunneling events occurring What would happen ?. 7 2.1 Behavior of wave at the boundary Concept - --calculation Continuity at the Boundaries When a wave, e.g. light, water surface wave, matter wave, crosses a boundary, The mathematical function that describes the wave must have two properties at each boundary: 1. The wave function must be continuous. 2. The slope of the wave function must be continuous, except when the boundary height is infinite. A discontinuous wave function (not allowed) Sine curve (c) and exponential curve (d): Continuous wave function with a both wave function and discontinuous slope. slope are continuous. (not allowed, except the boundary is of infinite height. ) 8  1) Wave function is continuous at boundary A. 2.1 Behavior of wave at the boundary ⑳ Thus, wave fnc. In region 2 should be From the Fig., the wave particle in ‘region 1’ is E( F = G( sin ()* − K( (1) +! E# = G# sin(2MF/!# − K# ) At the boundary A (x = 0) G# = 11.5, !# = 4.97 cm, K# = −65.3° E# F = E( F In ‘region 2’, !( is 10.5 cm. the boundary ‘A’ is at x = 0 cm At the boundary ‘B’ is at x = L = 20 cm G# sin(−K# ) = G( sin(−K( ) Find K( and G( =? −G# sin(K# ) = −G( sin(K( ). (2) 2) The slope in region 1 match the slope in region 2 at boundary A. ]^ E = Csin(2MF/! − K) The slopes à dy/dx = (2π/λ)C cos(2πx/λ − φ) At the boundary A (x = 0) X =0 X =L !" !" Sol ! cos(&$ ) = ! cos &!. (3) #! $ #" ! Use two boundary conditions. 9 2.1 Behavior of wave at the boundary −G# sin(K# ) = −G( sin(K( ). (2) To find ;& () () From Eq.(2) G cos(K# ) +" # = G cos +! ( K(. (3) sin(K# ) G( = G# sin(K( ) Eq.(2)/Eq.(3) sin(−65.3) #! #" G( = (11.5) )*+ &$ = )*+ &! sin(−45.8) !" !" ,$ )*+ &$ = ,! )*+ &! a, = bc. d ,$ From Eq.(1) )*+ &! = )*+ &$ ,! %$ ,$ Thus,