Chapter 5: Problem 53
In Section \(5.5\). it was shown that the infinite well coergis follow simply from \(\Lambda=h / p\), the fornulafor kinelic encrgy. \(p^{2} / 2 m\); and a famous standing-wave condition \(\lambda=2 U_{n} .\) The arguments are perfectly valid when the potential energy is 0 (inside the well) and \(L\) is strictly conskant, but they can also be useful in other cases. The length \(L\) allowed the wave should be roughly the distance between the classical tuming points, where there is no kinetic eneigy left. Apply these arguments to the oscillator potential energy, \(U(x)=\frac{1}{2} \kappa x^{2}\). Find the location \(x\) of the classical tuming point in terms of \(E\); use nwice this distance for L: then insert this into the infinite well eneigy fonnula so that \(E\) appears on both sides. Thus far, the procedure Ieally only deals with kinctic energy. Assume, as is tue for a classical oscillator, that there is as much potential eneigy, on average, as kinetic energy. What do you obtain for the quantized energies?
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