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[tex]H\phi(x)=E_{n}\phi(x) [/tex] with [tex]H=-D^{2}\phi+V(x)\phi [/tex]

my question is how i would choose the potential so we have that the energies are the root of the equation [tex]f(x)=0 [/tex]

i try using the WKB approach to calculate the function:

[tex]\phi(x)=e^{iS(x)/\hbar} [/tex] with [tex]s^{2}=E_{n}-V(x) [/tex]

with that i get a functional equation for the potential V(x), my problem is how i introduce the condition that the energies are the roots of f(x)...