Potential step and tunneling effect

[Maximilian2]

We know that thanks to the tunnel effect, in the case of a finite potential step and considering a stationary state, when a plane wave encounter the step the probabability that the wave-particle coming from -∞ (where potential is V=0) will be ≠ 0, in particular the wave function will be exponential decay. We can also calculate the probability flux (J) through the potential step and the result is J=0. In my book i read that taking into account all these results, the interpretation that we can give is that considering many particles, a certain percentage will cross the step and after a definite amount of time it will turns back before setting out in the direction where it came from, this vision allow us to justify why J=0. Here is my question: once (and if) the wave-particle cross the potential step, shouldn't continue its path without turning back? There is a cause that force it to reverse the direction and that can be explain from an "intuitive" point of view?

[swansont]

You can think of it as reflection or transmission. The particle will bounce off or tunnel. Tunneling particles don’t return, they continue on their path. The probability flux isn’t zero.

[Maximilian2]

But in this case the exponential decay probability inside the step would be destroyed, because the same ammount of particles that cross the barrier go to infinity. (I forgot to mention explicity that I'm considering the case where E < V0, with E the energy of the wave-particle).

[Sriman Dutta]

4 hours ago, Maximilian2 said:

But in this case the exponential decay probability inside the step would be destroyed, because the same ammount of particles that cross the barrier go to infinity. (I forgot to mention explicity that I'm considering the case where E < V0, with E the energy of the wave-particle).

Assumin          V=0 when -a<x<a   

                          V=V0 elsewhere        (V0>0)

 

There are two cases:

Case 1: E < V0 (bound state)

In this case, the wavefunction has certain discrete energy levels, the number of which depends on the strength (aka shallowness) of the potential. The discrete states can be obtained after some numerical calculus, as there is no direct analytical method.

Also the wavefunction is not zero outside (-a,a). This is the striking feature of quantum tunneling. If you integrate the square modulus of the wavefunction in (a,infinity) or (-infinity, -a), that effectively gives you the probability to find the particle in that region. And surprisingly it's non-zero, which means there is some probability for the particle to exist outside the potential(finite) barrier in spite of insufficient energy.

 

Case 2: E> V0 (scattering state)

In this condition, the particle exists like a free guy, but with just a potential acting outside (-a,a). Here if you assume that you are directing the particle from one side, then it can be shown that there exists some probability that the particle is reflected back. Obviously, the transmission probability is more, and it increases with increasing E.

Edited November 23, 2020 by Sriman Dutta

[swansont]

30 minutes ago, Sriman Dutta said:

Assumin          V=0 when -a<x<a   

                          V=V0 elsewhere        (V0>0)

 

That's not a scenario for tunneling. That's a particle in a box.

Tunneling has a barrier of length L, with V = V₀, and V=0 elsewhere.

A commonality here is that the wave function extends into the barrier in both of these cases, but V=V₀ everywhere but the box leaves no opportunity to tunnel

4 hours ago, Maximilian2 said:

But in this case the exponential decay probability inside the step would be destroyed, because the same ammount of particles that cross the barrier go to infinity. (I forgot to mention explicity that I'm considering the case where E < V0, with E the energy of the wave-particle).

I don't understand what you mean here. A particle that crosses the barrier does go to infinity. Why is that a problem? The probability of crossing the barrier is small.

[Sriman Dutta]

1 minute ago, swansont said:

That's not a scenario for tunneling. That's a particle in a box.

Tunneling has a barrier of length L, with V = V₀, and V=0 elsewhere.

A commonality here is that the wave function extends into the barrier in both of these cases, but V=V₀ everywhere but the box leaves no opportunity to tunnel

yes, to be strictly speaking, this is not formally tunnelling. A better example of it can be the delta potential.

However, even in this finite potential well, the wavefunction behaves like that. I am actually trying to draw the analogy here between the probability of finding the particle on the other side of potential barrier and the probability of finding in the regions outside (-a,a) where E<V0, yet still it manages to get there.

[Maximilian2]

Sorry l'll try to be more clear. I took the picture that rapresent the behaviour of the probability below, the thing I can't understand is that if a particle cross the point x=0 (where there is the step potential) why the probability decrease exponentially after that point? Once the particle overcome the obstacle there shouldn't be any reason that cause a less probable localization of the particle with x -> +∞ if not that the particle stop or reverse, and consequently shouldn't we find the probability constant after x=0? I wanted to know the cause or if i was minunderstanding some concept. 

[swansont]

It's harder to tunnel through a thicker barrier. The probability drops off exponentially.

What you haven't shown is the wave function after the barrier ends.