Quantum Mechanics: Tunneling Physics 123 1/11/2019 Lecture XVI

Wave Function (Wave function Y of matter wave)2 dV=probability to find particle in volume dV . In 1-dimentional case probability P to find particle between x1 and x2 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 1/11/2019 Lecture XVI

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: 1/11/2019 Lecture XVI

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 1/11/2019 Lecture XVI

Particle in a finite potential well Particle mass m in a finite potential well: U(x)=0, if 0<x<L, U(x)=U0, if x<0-or-x>L Boundary conditions: x L U0 I II III 1/11/2019 Lecture XVI

Particle in a finite potential well Inside the box (region II) Possible solutions: sin(kx) and cos(kx) x L U0 I II III 1/11/2019 Lecture XVI

Particle in a finite potential well Outside the box (regions I and III) Possible solutions: exp(Gx) and exp(-Gx) x L U0 I II III 1/11/2019 Lecture XVI

Wave functions 2 knots in the box Symmetric 1 knot in the box antisymmetric 0 knots in the box symmetric 1/11/2019 Lecture XVI

Probability to find particle at x Particle can be found outside the box!!! E=U0+KE KE must be positive KE=E-U0, but U0>E Energy not conserved?! Fine print: Heisenberg uncertainty principle Time spent outside the box is less than h/2p divided by energy misbalance, then energy non-conservation is “virtual”=undetectable 1/11/2019 Lecture XVI

Probability to find particle at x Consider electron mc2=0.5 MeV with U0=2eV, E=1eV How much time does it spend outside the box? Characteristic depth of penetration x0 : exp(-Gx)=exp(-x/x0) 1/11/2019 Lecture XVI

You can go through the wall!!! It’s called tunneling effect Probability of tunneling P=|y|2=exp(-2GL), L-width of the barrier Transmission coefficient T~P=exp(-2GL) 1/11/2019 Lecture XVI

Problem 39-34 A 1.0 mA current of 1.0 MeV protons strike 2.0 MeV high barrier of 2.0x10-13m thick. Estimate the current beyond the barrier. p 1/11/2019 Lecture XVI