Schrödinger Equation (why are U and -x^2 =0?)

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The discussion revolves around the Schrödinger Equation and the specific condition where U and -x^2 equal zero. The original poster expresses confusion about why U cannot simply be set to zero, as it is defined as a function of x within the problem constraints. There is an emphasis on the independence of x, which can take values between -L and L. The poster seeks clarification on their reasoning and the validity of their procedure. Overall, the conversation highlights the complexities of interpreting the equation and the conditions involved.
Que7i
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New poster has been reminded to use LaTeX for math equations (not pictures of their work)
Homework Statement
A subatomic particle with mass m is in a one-dimensional potential well. The potential energy is infinite for x < −L and for x > +L, while for −L < x < L, the potential energy is given by:U(x) =(−ℏ^2x^2)/(mL^2(L^2 − x^2)
The particle is in a stationary state described by the wave function 𝜓(x) = A(1 − x^2/L^2)
for −L < x < +L and by 𝜓(x) = 0 elsewhere. (Assume A is a positive, real constant.)
What is the total energy E of the system in terms of ℏ, m, and L?
Relevant Equations
-ℏ/(2m)d^2𝜓/dx^2+U𝜓=E𝜓
I've already found out how to do it, but, either I got lucky and my procedure is wrong or I just don't get it. Why are U and -x^2 equal 0?
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You cannot pick ##U = 0##, it is given in the problem to be a particular function of x. Furthermore ##x## is an independent variable that can take any value between ##-L## and ##L##.
 
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Que7i said:
I've already found out how to do it, but, either I got lucky and my procedure is wrong or I just don't get it. Why are U and -x^2 equal 0?
I can't tell what your reasoning is from your work. Could you explain to us what you're doing?
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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