- #1
ehrenfest
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Homework Statement
The time-independent Schrodinger equation solutions for an infinite well from 0 to a are of the form:
[tex] \psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a) [/tex]
If you move the well over so it is now from -a/2 to a/2, then you can replace x with x-a/2 and get the new equations right?
If I try to actually apply these new boundary conditions to the sin function, I get something different, however:[tex] \psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a) [/tex]
How can these two techniques give different results?