Quantum Mechanics Tunneling Physics 123 1192020 Lecture XVI

Wave Function • (Wave function Y of matter wave)2 d. V=probability to find particle

Properties of Wave function • Wave function respects the symmetry of the system •

Count knots 0 knots in the box symmetric 1 knot in the box antisymmetric

Particle in a finite potential well • Particle mass m in a finite potential

Particle in a finite potential well • Inside the box (region II) U 0

Particle in a finite potential well • Outside the box (regions I and III)

Wave functions 2 knots in the box Symmetric 1 knot in the box antisymmetric

Probability to find particle at x • • • Particle can be found outside

Probability to find particle at x Consider electron mc 2=0. 5 Me. V with

You can go through the wall!!! • It’s called tunneling effect • Probability of

Problem 39 -34 • A 1. 0 m. A current of 1. 0 Me.

Slides: 12

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Quantum Mechanics: Tunneling Physics 123 11/9/2020 Lecture XVI 1

Wave Function • (Wave function Y of matter wave)2 d. V=probability to find particle in volume d. V. In 1 -dimentional case probability P to find particle between x 1 and x 2 is • Unitarity condition (probability to find particle somewhere is one): • Schrödinger equation predicts wave function for a system • System is defined by potential energy, boundary conditions 11/9/2020 Lecture XVI 2

Properties of Wave function • Wave function respects the symmetry of the system • For example if the system is symmetric around zero • x -x • then wave function is either symmetric or antisymmetric around zero: 11/9/2020 Lecture XVI 3

Count knots 0 knots in the box symmetric 1 knot in the box antisymmetric 2 knots in the box Symmetric N-th state: (N-1) knots in the box N-odd – symmetric N-even - antisymmetric 11/9/2020 Lecture XVI 4

Particle in a finite potential well • Particle mass m in a finite potential well: • U(x)=0, if 0<x<L, • U(x)=U 0, if x<0 -or-x>L • Boundary conditions: U 0 I II 0 11/9/2020 Lecture XVI III x L 5

Particle in a finite potential well • Inside the box (region II) U 0 I • Possible solutions: sin(kx) and cos(kx) 11/9/2020 Lecture XVI II 0 III x L 6

Particle in a finite potential well • Outside the box (regions I and III) U 0 I • Possible solutions: exp(Gx) and exp(-Gx) 11/9/2020 Lecture XVI II 0 III x L 7

Wave functions 2 knots in the box Symmetric 1 knot in the box antisymmetric 0 knots in the box symmetric 11/9/2020 Lecture XVI 8

Probability to find particle at x • • • Particle can be found outside the box!!! E=U 0+KE KE must be positive KE=E-U 0, but U 0>E Energy not conserved? ! Fine print: Heisenberg uncertainty principle • Time spent outside the box is less than h/2 p divided by energy misbalance, then energy non-conservation is “virtual”=undetectable 11/9/2020 Lecture XVI 9

Probability to find particle at x Consider electron mc 2=0. 5 Me. V with U 0=2 e. V, E=1 e. V How much time does it spend outside the box? Characteristic depth of penetration x 0 : exp(-Gx)=exp(-x/x 0) 11/9/2020 Lecture XVI 10

You can go through the wall!!! • It’s called tunneling effect • Probability of tunneling P=|y|2=exp(-2 GL), Lwidth of the barrier • Transmission coefficient T~P=exp(-2 GL) 11/9/2020 Lecture XVI 11

Problem 39 -34 • A 1. 0 m. A current of 1. 0 Me. V protons strike 2. 0 Me. V high barrier of 2. 0 x 10 -13 m thick. Estimate the current beyond the barrier. p 11/9/2020 Lecture XVI 12