if the potential is bounded below $ V(x)=V(-x) \ge a $ for some real number a
and we can be sure that as $ x\rightarrow \infty $ we know that $ V(x) \ge 0 $ then does it mean that the energies for our one dimensional system will be positive ??
my idea WKB quntization $ N(E)= 2\int_{0}^{a} \sqrt {E_{n}-V(x)} $ here $ V(a)=E $ is a turnign point so if the potential is positive for big 'x' then the energies should be also positive otherwise the epxresion inside the integral would be complex am i right ?
