Reflection probability in Quantum Mechanics

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The discussion centers on the reflection probability formula in quantum mechanics, specifically questioning the derivation of the formula and its relationship to transmission probability, expressed as R + T = 1. Participants express uncertainty about the starting point for proving these concepts and the logical steps involved. There is a suggestion to refer back to the text for guidance on the derivation process. Overall, the conversation highlights a need for clarity on foundational principles in quantum mechanics. Understanding these probabilities is crucial for further exploration in the field.
MaxJ
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Homework Statement
below
Relevant Equations
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For this,
1723530842653.png

The solution is
1723530868613.png

1723530895562.png

I have doubt where they got the reflection probability formula from. Someone may know how to find it. I think that ##R + T = 1##. But I'm not sure where the transmission probability formula comes from either.

Kind wishes
 
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Did you try to do what they actually say in the text?

1723534387770.png
 
Orodruin said:
Did you try to do what they actually say in the text?

View attachment 349853
Sir, blessed.

I have a doubt where to start in the proof. Not sure what logic to use.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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