Rectangular Potential well if hbar -> 0

[divB]

Rectangular Potential well if hbar --> 0

Hi,

I have a very simple question: My professor said that if \hbar \rightarrow 0 in the transmission coefficient does not yield 0 (case E < V0) as classical results would expect.

I think the correct transmission coefficient is solved in Wikipedia: http://en.wikipedia.org/wiki/Rectangular_potential_barrier#E_.3C_V0

But if I calculate \lim_{\hbar \rightarrow 0} T I get 0.

Can anybody tell me what is wrong about it?

Thank you,
divB

 

[Matterwave]

Your transmission coefficient is more correct. Wikipedia has assumed that k1 and k2 on the left and right side of the barrier is the same. Your transmission coefficient would be necessary if the barrier did not fall all the way back to the original position (or fell more).

I think hbar going to 0 should yield classical results...I don't know why it wouldn't yield classical results as your professor suggests.

 

[divB]

Hi Matterwave,

Thank you. you replied while I was editing my post. Sorry. Yes, I saw that and therefore re-edit my post.

The reason is that the coefficient from my professor is not for a rectangular potential but for abrupt change. In case of the rectangular potential I have in the first and in the third region k1 and in the middle k2 in the lecture notes, that means that the result is the same in this case.

Concerning the main problem: This is the statement in the lecture notes:

T contains ħ in its expression. In the limit of ħ → 0, we may expect that the expression will coincide with the classical limits. Namely, T=0 for E < U0 (no tunneling) and T =1 for E > U0 (no reflection). But this is not the case.

But indeed, also Wikipedia says this but no further comments on it. The article describing the semiclassical approach:

http://en.wikipedia.org/wiki/Quantum_tunneling

If we take the classical limit of all other physical parameters much larger than Planck's constant, abbreviated as \hbar \rightarrow 0, we see that the transmission coefficient correctly goes to zero. This classical limit would have failed in the unphysical, but much simpler to solve, situation of a square potential.

Regards,
divB

 

[Matterwave]

Hmm, interesting. I have not heard about this before. I don't know why this would be the case, other than the fact that square wells are unphysical since they are not differentiable everywhere (classically, that would lead to a discontinuous, infinite, force at the turning points).

Perhaps a more knowledgeable member can answer your question. :)

 

[divB]

Maybe I just calculate the limit the wrong way?

In limes I can disregard any constants, can't I?

So

T= \frac{1}{1+\frac{V_0^2\sinh^2(k_1 a)}{4E(V_0-E)}}

becomes

T= \frac{1}{1+\sinh^2(k_1 a)}

and as k_1=\sqrt{2m (V_0-E)/\hbar^{2}}

T= \frac{1}{1+\sinh^2(\sqrt{2m (V_0-E)/\hbar^{2}} a)}

and in the limit for \hbar:

T= \frac{1}{1+\sinh^2(\frac{1}{\hbar})}

Is this true for here?

But it is obvious that this must be zero. Or am I completely wrong??

Regards,
divB

 

[divB]

Sorry, again me.

Maybe I should not take the simple limit of T itself?

On the one hand, this text suggests that I need to take the limit of T directly:
http://img690.imageshack.us/img690/7789/rectpot.png

On the other hand I found this source: http://physics.usask.ca/~dick/tunnel.pdf, page 4 takes first the limit E --> U0.
But I do not understand the result nor why he does this :-(

But if [mm]\hbar[/mm] goes to zero the result is zero anyway ...

regards,
divB