# Reflection Probability for Different Potentials - Quantum Mechanics

My question is above. Firstly, I don't actually know whether it is true or not (!). Secondly, if I were to try to prove it, then I have very little idea how to. The potential steps that I have always done are steps from a constant level to another constant level (Heavyside), whereas this is different.

I would imagine the answer is yes, but I'm not sure how to show it.

Can I approximate the curve by small steps (/sums of Heavyside functions), and then show that a larger Heavyside step gives a larger probability?

Thanks, Sam

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## migrated from math.stackexchange.comMay 25 at 20:58

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Seeing that this is not Physics, you should target the question at mathematical audience. What is the mathematical relation between $p_i$ and $V_i$? –  words that end in GRY May 25 at 4:50
I realise that this isn't physics, hence being in the maths part! =P The issue is that I can't determine the relation. :( –  Smiley Sam May 25 at 6:58
I mean this site is not Physics.SE: it's Mathematics.SE. And you don't have a clear mathematical question here. –  words that end in GRY May 25 at 17:16
Oh, I see what you mean. But no, this is a maths QM question, not a physics QM question. "Justify your answer carefully, either by giving a rigorous proof or by presenting a counterexample with explicit calculations of $p_1$ and $p_2$." That's maths, no physics. =P –  Smiley Sam May 25 at 20:27
Also, it's a past paper from the maths university course that I'm doing. –  Smiley Sam May 25 at 20:28

It is not necessarily true. For a zero potential $V_2$ you have $p_2=0$, whereas if $V_1$ is a rectangular pit, in general, $p_1>0$.
Ah yes, I thought that it was $V_i \ge 0$! It's clearly not true as said. How about the reverse inequality then, or when $V_1(x) \ge V_2(x) \ge 0$? –  Smiley Sam May 25 at 22:29