1. Quantum Mechanics: Tunneling
Physics 123
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Lecture XVI

2. 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

3. 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

4. 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
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Lecture XVI

5. 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
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Lecture XVI

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

7. 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
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Lecture XVI

8. Wave functions 2 knots in the box Symmetric 1 knot in the box
antisymmetric
0 knots in the box
symmetric
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Lecture XVI

9. 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
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Lecture XVI

10. 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)
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Lecture XVI

11. 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)
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Lecture XVI

12. 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