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genius
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I was looking for questions to practice normalizing a wave function, so I visited the following online pdf, http://people.physics.tamu.edu/syeager/teaching/222/hw1solution.pdf. The first question was to find the normalization constant, A of ψ(x) = A cos (2πx/L) for (−L/4) ≤ x ≤ (L/4). After attempting to solve it on my own I checked my answer verses that of the pdf and was surprised to see they arrived at A^2=4/L. To see their entire solution, visit the pdf.
Here was my way of solving it:
1 = ∫A^2 cos^2 (2πx/L)dx
A^2 ∫cos^2(2πx/L)dx =(A^2)((2πx/2L)+(1/4)(sin(4πx/L))
= (A^2) ((πx/L)+(sin(4πx/L)/4)
= (A^2) ((π((L/4)-(-L/4))/L)+sin((4π((L/4)-(-L/4)))/L)/4)
= (A^2)((π(L/2)/L)+sin((4π(L/2))/L)/4)
= (A^2)((π/2)+(sin(2π))/4)
so
A^2(π/2)=1
That is what I got, and from there you would get A, so I was wondering how the solution showed A^2=4/L instead of A^2=2/π?
Here was my way of solving it:
1 = ∫A^2 cos^2 (2πx/L)dx
A^2 ∫cos^2(2πx/L)dx =(A^2)((2πx/2L)+(1/4)(sin(4πx/L))
= (A^2) ((πx/L)+(sin(4πx/L)/4)
= (A^2) ((π((L/4)-(-L/4))/L)+sin((4π((L/4)-(-L/4)))/L)/4)
= (A^2)((π(L/2)/L)+sin((4π(L/2))/L)/4)
= (A^2)((π/2)+(sin(2π))/4)
so
A^2(π/2)=1
That is what I got, and from there you would get A, so I was wondering how the solution showed A^2=4/L instead of A^2=2/π?
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