# Apparent contradiction between calculations and intuition?

I am rather confused because it would seem that mathematical conclusions I have drawn here goes against my physical intuition, though both aren't too reliable to begin with.

We have a potential step described by $$V(x)=\begin{cases}0& x\le0\\V_0 & x>0\end{cases}$$

and a wavefunction $\psi(x)$ that satisfies the equation $${\hbar^2\over 2m}{\partial^2 \over \partial x^2}\psi(x)+V(x)\psi(x)=E\psi(x)$$

I wish to find the probability of reflection. By continuity constraints at $x=0$ I have arrived at the reflection amplitude being $$R={k-q\over k+q}$$ where $k=\sqrt{2mE\over \hbar^2}$ and $q=\sqrt{2m(E-V_0)\over \hbar^2}$ then we let $V_0\to -\infty$ giving $q\to \infty$, so $R\to -1$

$\implies |R|^2\to 1$

But I would have guessed that $|R|^2$ should vanish at the limit so that the incident wave is totally transmitted!

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## migrated from math.stackexchange.comJul 18 '12 at 11:32

This question came from our site for people studying math at any level and professionals in related fields.

Recall that if we have an incident wave

$$\frac{1}{\sqrt{k_1}}e^{ik_1 x}$$

from left (region 1, $x<0$, with constant potential $V_1$) that is partially transmitted

$$\frac{T}{\sqrt{k_2}}e^{ik_2 x}$$

to the right (region 2, $x>0$, with constant potential $V_2$), and partly reflected back to region 1,

$$\frac{R}{\sqrt{k_1}}e^{-ik_1 x}$$

then the the reflection coefficient is well-known to be

$$R~=~ \frac{k_1-k_2}{k_1+k_2},$$

where

$$k_i ~:=~\frac{\sqrt{2m(E-V_i)}}{\hbar}.$$

OP question is related to the fact that the reflection probability $|R|^2$ is invariant under permutation $V_1\leftrightarrow V_2$. Roughly speaking, quantum mechanically, the reflection probability $|R|^2$ is the same whether the incident wave meets a potential barrier/wall, or a potential abuss/well!

Intuitively, one would probably have guessed that the wave tends to go the region with smallest potential $V_i$. Classically, this is because one is forgetting momentum conservation of the wave, and implicitly allowing the wave to layoff/absorb momentum at $x=0$ to/from the environment. Quantum mechanically, the momentum conservation is implemented by the requirement that the left and right derivative of the wave-function at $x=0$ should be the same. Momentum conservation implies that when the wave meets a potential step (either up or down), then a fraction of the wave must always be reflected to preserve momentum.

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