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let be the problem


however we have a problem, we do not know the potential but its inverse

$$ f^{-1}(x)=g(x) $$

we know $ g(x) $ but not $ f(x) $ what happense then since

a) the function $ f(x) $ may be multi-valued

b) the function $ f(x) $ may not exists even if we know $ g(x)$

howe can we analytically solve $ f^{-1}(x)=g(x)$ so we get $ f(x) $ ?? in the case $ f(x)=f(-x) $ is even.

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is $f^{-1}=1/f$ or is it the inverse function, i.e., $g(f(x))=x$? –  Arnold Neumaier Nov 22 '12 at 13:06
$ f^{-1) $ is the inverse function i mean $ f^{-1}(f(x))=x $ –  Jose Javier Garcia Nov 22 '12 at 13:54

1 Answer 1

up vote 3 down vote accepted

The potential in the Schroedinger equation must be single-valued and almost everywhere defined, so to have your problem weel-defined you must first decide which of several solutions is to be taken.

Apart from that, one doesn't need an explicit potential. If one calculates solutions numerically it is enough that one can compute the potential numerically at any given $x$.

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