Transmission Coefficient of a Rectangular Square Potential Barrier

[SOMEBODYCOOL]

If you have a rectangular square potential barrier of some height, say \lambda/L, and thickness L, what is the transmission coefficient and what is its value in the limit that L goes to 0?

Thus you have the height of the barrier going to infinity, while the width goes to zero... Assuming some fixed incident energy, the probability amplitude decays exponentially as it enters the wall, but since the barrier is thinner, there is less room for that decay. It seems obvious to me that the exponential wins out, but I'm not really sure... Would the transmission be zero for such a barrier? The alternative, I suppose, would be 1, meaning that the barrier is so infinitely thin that the particle doesn't even see it, which doesn't make a lot of sense either.

 

[mfb]

It will go to a constant.
$$T=|t|^2= \frac{1}{1+\frac{V_0^2\sinh^2(k_1 L)}{4E(V_0-E)}}$$
where $$k_1=\sqrt{2m (V_0-E)/\hbar^{2}}$$
Plugging in $V_0 = \lambda/L$ we get
$$k_1 L=L \sqrt{2m (\lambda/L-E)/\hbar^{2}} \approx \sqrt{2m \lambda L/\hbar^2}$$
This goes to zero for decreasing L. As $\sinh(x) \approx x$ for small x we get
$$T \approx \frac{1}{1+\frac{V_0^2 \,2m \lambda L}{4E(V_0-E) \hbar^2}} = \frac{1}{1+\frac{m}{2\hbar^2}\frac{\lambda^3 }{E(\lambda-EL)}} \approx \frac{1}{1+\frac{m}{2\hbar^2}\frac{\lambda^2 }{E}}$$