- #1
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I must demonstrate a certain result, and I have found how to do it, provided
[tex]\lim_{x\rightarrow \pm \infty} xV(x)[\psi(x)]^2 =0[/tex]
The text (/%?$! Gasiorowicz) always do things like saying expressions such as the one above is 0 but he never gave a clear explanation as to why that is. Sure psi goes to zero at infinity (as required by the probability interpretation [normalization]) but what assures us that xV(x) doesn't go to infinity "faster"?
[tex]\lim_{x\rightarrow \pm \infty} xV(x)[\psi(x)]^2 =0[/tex]
The text (/%?$! Gasiorowicz) always do things like saying expressions such as the one above is 0 but he never gave a clear explanation as to why that is. Sure psi goes to zero at infinity (as required by the probability interpretation [normalization]) but what assures us that xV(x) doesn't go to infinity "faster"?
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