# I Probability current in positive finite square potential

Tags:
1. Apr 7, 2016

### Joker93

Hello!
I want to prove that the probability current is a continuous entity at the boundaries of the square for the situation of 0< E< Vo in the problem where V is zero except a finite region in space where it is +Vo and we consider an incoming particle from the left(for example).
I thought that in the region of the non-zero potential we have an evanescent wave(described by real exponentials), so the probability current there is zero because there is no time change for the probability of finding the particle inside the region of non zero potential energy and thus the continuity equation gives us that the current there must be zero. But this is clearly wrong because:
If the current there is zero then the current in every region would be zero since each current is independent of time and space, which means total reflection and no tunneling in the more general case. Otherwise the current would not be continuous.
Clearly, i am getting something wrong!

2. Apr 7, 2016

### drvrm

i wish to know about those waves .......are there such waves with virtual exponentials?

3. Apr 7, 2016

### Joker93

They occur in every kind of wave. In quantum mechanics they are the reason that quantum tunneling occurs. See the following:

It is the wave between 0 and L.

4. Apr 7, 2016

Anyone...?!

5. Apr 7, 2016

### drvrm

inside the barrier you have wave function and calculate the current then it will come out to be zero as its solution of a time independent schrodinger equation and tunneling is a transition process involving time dependent solutions -i was looking at the issue on the net and found one piece in phys org. which you may see

ref.http://phys.org/news/2015-05-physicists-quantum-tunneling-mystery.html

6. Apr 7, 2016

### Joker93

Yeah, but in a scattering from the left problem, why would there be a non-zero current on the right of the square if the current in the middle is zero? Also, if the current on the right side is non zero and exists forever, then wouldn't all the probability of finding the particle transfer to the right of the square? But we know this is not the case!
In case you just know popular science kind of quantum mechanics, then ignore this! :D

7. Apr 7, 2016

### Jilang

I can see how this would be problematic as the particle is not spending "real" time in the barrier and therefore has no defined velocity there. Perhaps you might consider the currents instead at each of the boundaries in the region V = 0.

8. Apr 7, 2016

### Joker93

I tried to do that and have found the four boundary conditions and found the currents. When i applied the boundary conditions(via including the relations between the coefficients), the currents were all the same and proportional to the square of the coefficient of the wavefunction on the right of the potential. But, if the current in the middle region is zero(as i predicted but have not found a way to actually show it with mathematics, so i could be wrong), then every current is zero, and thus the coefficient is zero, which means that the wavefunction on the right is zero, so the transmission coefficient is zero, which from solutions in the internet i know is wrong. The transmission coefficient is not zero in general. So, something is wrong in my reasoning. I think it might be that the current in the middle region is not zero. But, how would i prove it? I can't get the intuition behind it either. If it is an evanescent wave, then why wouldn't the current there be zero?

9. Apr 7, 2016

### Jilang

The current isn't zero, but it isn't real. You won't detect one there.

10. Apr 7, 2016

### Joker93

But why isn't it zero? Is it not an evanescent wave?
And what do you mean by "you won't detect one there"? You mean a particle? And if yes, then why? |Ψ|^2 will be real.

11. Apr 7, 2016

### Joker93

I have calculated it! It is non-zero and the same as the currents in the other two regions and it is real, as one would expect it. Thanks for the replies!

12. Apr 7, 2016

### Jilang

13. Apr 8, 2016

### Joker93

For the current being real? Yes, it's by definition real. If you get its conjugate from the relation J=ihbar/2m(Ψ*dΨstar/dx-Ψstar*dΨ/dx.) then you get the same answer as not getting the conjugate, which means it is real.

14. Apr 8, 2016

### Joker93

For the current being real? Yes, it's by definition real. If you get its conjugate from the relation J=ihbar/2m(Ψ*dΨstar/dx-Ψstar*dΨ/dx.) then you get the same answer as not getting the conjugate, which means it is real.

15. Apr 8, 2016

### Jilang

i am wondering now about the wavefunction inside the barrier... Is it real or complex?

16. Apr 8, 2016

### Joker93

It is complex. If it was real, then the probability current in that region would be zero, which is not.
My advice is that you should work out the problem yourself! It offers a lot of insights if you haven't worked at a scattering problem using currents!

17. Apr 9, 2016

### Jilang

Hi Adam, yes I stayed up really late last night doing just that and have pages filled with i's and k's. I thought to try to find the wavefunction in the barrier, applied the continuity conditions, made some substitutions to eliminate the amplitudes reflected wave in the first region and the transmitted wave in the third region and ended up with a fine old mess! However it was enough to see that the amplitudes in that region would be complex.

18. Apr 9, 2016

### Joker93

Great! Thanks for helping me and it was a pleasure helping you too!