# Step potential - question about reflection coefficient

## Main Question or Discussion Point

Hi. I've been trying to understand the phenomenon of step potential, when energy of particle E is higher than the potential V.
Then we have solution on both sides of boundary in the form of wave functions....
Is the reflection coefficient in this case simply (E-V)/E ??
Can anyone show me a more formal explanation, however not as formal as in books (so that i understand :)
What does the quantum mechanical coefficient really tell us (in comparison with classical thinking)?
If we think about it classically I think we can be sure that particle gets transmitted at boundary, but loses some energy.
From the QM point of view, on the other hand I guess we have a probability that particle gets reflected?

Please tell me if my reasoning is correct:) cheers!

Related Quantum Physics News on Phys.org
jtbell
Mentor
Is the reflection coefficient in this case simply (E-V)/E ??
No, it's

$$R = {\left( \frac {\sqrt{E} - \sqrt{E - V}} {\sqrt{E} + \sqrt{E - V}} \right)}^2$$

Can anyone show me a more formal explanation, however not as formal as in books (so that i understand :)
I don't know any other way to justify this particular equation except by solving the Schrödinger equation on both sides of the step boundary, and applying the boundary conditions at the boundary, as described http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/models/barrier/barsola.htm.

What does the quantum mechanical coefficient really tell us (in comparison with classical thinking)?
If we think about it classically I think we can be sure that particle gets transmitted at boundary, but loses some energy.
From the QM point of view, on the other hand I guess we have a probability that particle gets reflected?
Exactly. If you have a particle coming in from the "low" side of the step, it has probability R of ending up on that side of the step, moving in the opposite direction; and probability 1 - R of ending up on the other side of the step, continuing in the same direction.

Last edited by a moderator:
Thanks:]