Quantum Mechanics Boundary conditions

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The discussion focuses on clarifying doubts regarding boundary conditions in quantum mechanics, specifically in Step 3 of a problem. Participants emphasize the importance of clearly stating the doubt to facilitate assistance. The solution involves using two separate wave functions for different regions, with each function representing waves in both directions. At the boundary, the wave functions must be equal, and this equality can be evaluated at the boundary point. The conversation highlights the necessity of matching both the wave functions and their derivatives at the boundary for a complete solution.
MaxJ
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Homework Statement
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Relevant Equations
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For this problem,
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The solution is,
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I have a doubt about Step number 3 about boundary conditions. Someone maybe be able to solve that doubt?

Kind wishes
 
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MaxJ said:
I have a doubt about Step number 3 about boundary conditions. Someone maybe be able to solve that doubt?
Not unless you tell us what your actual doubt is. The math is straightforward, essentially plug-and-chug.
 
Orodruin said:
Not unless you tell us what your actual doubt is. The math is straightforward, essentially plug-and-chug.
Sir, bless you.

Doubt is matching wave functions give A + B = C + D and how they get their expression for derivatives (also in step 3).
 
MaxJ said:
Doubt is matching wave functions give A + B = C + D and how they get their expression for derivatives (also in step 3).
It would be more conventional (and clearer) to consider 2 separate wave-functions, one for each region: ##\psi_1## for ##x<0## and ##\psi_2## for ##x \ge 0##.

Each wave-function contains 2 terms, representing waves moving in the +x and -x directions in that region.

So, adapting your solution’s notation:
##\psi_1(x) = A e^{ikx} + Be^{-ikx}##
##\psi_2(x) = C e^{iqx} + De^{-iqx}##

At ##x=0## we require that ##\psi_1 =\psi_2##. Simply evaluate ##\psi_1##and ##\psi_2## at ##x=0## and equate them.

Similarly for the dervatives.
 
Last edited:
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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