In finding solutions to the time independent Schrodinger equation we have to normalize [itex] \psi [/itex] to find the constant A. So we get [tex] \int_{0}^{a} |A|^{2} sin^{2}(kx) dx = |A|^2 \frac{a}{2}=1 [/tex](adsbygoogle = window.adsbygoogle || []).push({});

For A we then get [itex] |A|^2 = \frac{2}{a} [/itex]. Griffiths says that this only determines the magnitude of A but it's simplest to pick the positive real root. I know how to work with complex numbers generally but I'm a little confused as to what the imaginary root would be anyway. It looks to me like [itex] A= \pm \sqrt{ \frac{2}{a} } [/itex]. Since we aren't square rooting a negative I don't see where the [itex] i [/itex] comes in. Also, I know the magnitude of a complex number is real, so are we saying that A itselfcouldbe complex? I don't want to make any assumptions about this. Thanks!

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# Infinite square well solution

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