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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 ?

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Comment to the question(v1): The finite potential well [with appropriate (possible shifted) definition of zero-energy level] seems to be a counterexample. –  Qmechanic Oct 10 '12 at 9:12
    
in this case it would have a finite amount of negative energies , however I still think that as $ n $ increases the big energies will be always positive. –  Jose Javier Garcia Oct 10 '12 at 9:53
    
Yes, that is true, as e.g. discussed in this Phys.SE post. –  Qmechanic Oct 10 '12 at 12:35
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