Solution to 2008 Problem 65

The energy levels of the harmonic oscillator are
\begin{align*}E_n = \left(n + \frac{1}{2} \right)\hbar \omega\end{align*}
where n = 0,1,2,3,\ldots. So, the difference in energy between adjacent levels is always \hbar \omega. So I is correct.
The potential energy is m \omega^2 x^2/2 which is quadratic, not linear, in x, so II is incorrect.

The ground-state corresponds to n = 0, and it has energy \hbar \omega/2. This is equal to the kinetic energy plus the potential energy. By the virial theorem, \langle V\rangle = \langle T \rangle = \hbar \omega/4 for this state, so the kinetic energy is not zero, and III is incorrect.

The wavefunction of the ground state of the harmonic oscillator is
\begin{align*} %blah dog cat\psi(x) = a e^{-b x^2}\end{align*}
where a and b are positive constants. So, there is a nonzero probability for the particle to be found at any x \in \left(-\infty,\infty \right). So, IV is correct.
So, answer (C) is correct.

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