So our teacher claimed that if we have a Potential of the form $V(x)= x^\nu$ then the Energy is of the form $E={2\nu \over \nu+2}$ Can anyone break up the math for this problem?
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Gert
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Siddhartha Dam
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1Do you mean the ground state energy? If $\nu$ is an odd integer the energy is not bounded below. – Keith McClary Oct 08 '15 at 03:32
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Yes the bound state energy – Siddhartha Dam Oct 08 '15 at 04:31
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$x^\nu$ in imaginary if $x<0$ and $\nu$ is $\frac{1}{2}$. – Keith McClary Oct 08 '15 at 04:42
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V actually varies with x in,this manner.. So you can ignore the non trivial solutions – Siddhartha Dam Oct 08 '15 at 09:55
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Assuming you consider only even powers of $x$ (for odd powers there are no bound states) the behaviour you describe comes from the Born-Sommerfeld condition. This is described in the Wikipedia article Old quantum theory, and there is a related question on this site.
Suppose $E$ is the energy of your system, and $\pm X$ are the turning points where $E = V(X)$, then for your system the Born-Sommerfeld condition states:
$$ \int_{-X}^{+X} p \, dx = nh $$
where the momentum $p$ is given by:
$$ p(x) = \sqrt{2m(E - V(x))} $$
John Rennie
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