How to Normalize a Wave Function in a Potential Well?

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The discussion focuses on normalizing a wave function defined as Ae^(ikx)*cos(πx/L) within a one-dimensional potential well. The user is unsure whether to normalize using the bounds -L/2 to L/2 or -L/4 to L/4. It is clarified that normalization should occur over the range where the wave function is non-zero, which is -L/2 to L/2, leading to A=√2/L. After normalization, the user can find the probability of the proton being between -L/4 and L/4 by integrating over those bounds, ensuring the resulting probability is less than 1. The discussion emphasizes the importance of proper normalization in quantum mechanics.
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Homework Statement


I have the wave function Ae^(ikx)*cos(pix/L) defined at -L/2 <= x <= L/2. and 0 for all other x.

The question is:
A proton is in a time-independent one-dimensional potential well.What is the probability that the proton is located between x = − L/4 and x = L/4 ?

Homework Equations


∫ψψ* = 1

The Attempt at a Solution


I know i have to normalize this first. Should I be normalizing with the bounds being -L/2 and L/2 or should the bounds be -L/4 and L/4.
A^2∫cos^2(πx/L)dx =1
(1/2)A^2[x+ L/2π (sin(2πx/L)] evaluated at some bounds.

I actually evaluated it both ways one answer gives me A=√2/L at L/2,-L/2 and the other one gives me A = √2
π/L which makes more sense (at -L/4,L/4)

But i just want to make sure i am approaching this right way.
Thank you.
 
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Actually, the normalization in one dimension is formally over all space, i.e. from -∞ to +∞, but since the wavefunction is zero outside the box, you would be adding a whole bunch for zeroes if you went outside it. Therefore it suffices to integrate from -L/2 to +L/2, i.e. over all space where there is non-zero probability.
 
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I actually re did it and got A=√2/L. Thank you for the clarification. Now to find the probability I can just integrate over the other bounds right?
 
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Right. Make sure the number you get is less than 1.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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