Quantum Mechanical Tunneling

Tunnelling through step barier

V(x)=V0 for 0<x<a and V(x)=0 for all other x.

For regions I, III, the solutions will be oscillatory exponentials. At the boundary the function and the derivative are continuous.

κ=2mE2, ΨI,III=AI,IIIexp(iκx)+BI,IIIexp(iκx).

A is the incoming wave, from to +. B is going the opposite direction.

Initial Condition

Incoming from and no reflective wall at +. Then BIII=0.

Energy below Barrier

In region II it will be a decaying exponential. k=2m(V0E)/.

Energy above Barrier

In region II it will be a oscillatory exponential. k=2m(EV0)/.

Probability Flux

Region I. j=m[ΨddxΨ]=mκ(|A|2|B|2). jinc=κm|A|2 is the incoming wave velocity. jref=κm|B|2 is the reflected wave velocity.

Region III. j=κm|C|2 is the transmitted wave velocity.

Reflection Coefficient. R=jrefjinc=|B|2|A|2.

Transmission Coefficient. T=jtransjinc=|C|2|A|2.

R+T=1 for a non-absorbing wall.

Classical Expectations

If E<V0 then we expect R=1 and T=0. If E>V0 then we expect T=1 and R=0.

For E<V0, T=[1+V02sinh2ka4E(V0E)]1.

For E>V0, T=[1+V02sin2ka4E(EV0)]1.

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:19

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