up to its next-higher
53. In Section 5.5, it was shown that the infinite well energies
follow simply from $\lambda = h/p$; the formula for kinetic energy,
$p^2/2m$; and a famous standing-wave condition, $\lambda = 2L/n$.
The arguments are perfectly valid when the potential
energy is 0 (inside the well) and L is strictly constant, but
they can also be useful in other cases. The length L
allowed the wave should be roughly the distance between
the classical turning points, where there is no kinetic
energy left. Apply these arguments to the oscillator poten-
tial energy, $U(x) = \frac{1}{2}kx^2$. Find the location $x$ of the clas-
sical turning point in terms of $E$; use twice this distance
for $L$; then insert this into the infinite well energy formula,
so that $E$ appears on both sides. Thus far, the procedure
really only deals with kinetic energy. Assume, as is true
for a classical oscillator, that there is as much potential
energy, on average, as kinetic energy. What do you
obtain for the quantized energies?
