Let's see how the variational method works in another application. Let's assume we didn't know the ground-state energy of the quantum harmonic oscillator and use the variational method to determine it. First, the Hamiltonian is

\[\begin{equation*}\hat{H}=\frac{\hat{p}^{2}}{2 m}+\frac{m \omega^{2}}{2} \hat{x}^{2} \tag{10.115}\end{equation*}\]

Note that this potential is symmetric about \(x=0\) and the position extends over the entire real line: \(x \in(-\infty, \infty)\). This motivates the guess for the ground-state wavefunction

\[\begin{equation*}\psi(x ; a)=N\left(a^{2}-x^{2}\right), \tag{10.116}\end{equation*}\]

for a normalization constant \(N\) and \(a\) is the parameter that we will minimize over. Now, this wavefunction is only defined on \(x \in[-a, a]\).

(a) First calculate the normalization factor \(N\), as a function of \(a\).

(b) Now, calculate the expectation value of the Hamiltonian on this state, \(\langle\psi|\hat{H}| \psiangle\), as a function of the parameter \(a\).

(c) Finally, minimize the expectation value over \(a\) to provide an upper bound on the ground-state energy, \(E_{0}\). How does this estimate compare to the known, exact value?