130_notes.dvi

Inside the barrier,

κ=

√

2 m(V 0 −E)

̄h^2

.

x

incident wave

reflected

E

V 0

V(x)

transmitted

x

0

0

e + Re

ikx -ikx Teikx

Ae +Be

Energy

κ −κx

This is actually the same as the (unbound) potential well problem withthe substitution

k′→iκ

in the center region.

Theamplitude to be transmittedis

T=e−^2 ika

2 kκ

2 kκcosh(2κa)−i(k^2 −κ^2 ) sinh(2κa)

.

We can compute the probability to be transmitted.

|T|^2 =

(2kκ)^2

(k^2 +κ^2 )^2 sinh^2 (2κa) + (2kκ)^2

→

(

4 kκ

k^2 +κ^2

) 2

e−^4 κa

Anapproximate probabilityis sometimes useful.

|T|^2 ≈e−^2 κ(2a)=e

− 2

∫a

−a

√ 2 m

̄h^2 [V(x)−E]

dx

Classically the transmission probability would be zero. In Quantum Mechanics, the particle
isallowed to violate energy conservation for a short timeand so has a chance to tunnel
through the barrier.

Tunneling can be applied to cold emission of electrons from a metal, alpha decay of nuclei, semicon-
ductors, and many other problems.