Discussion
Using the normalized trial wave function \(\psi(x) = \left(\frac{2b}{\pi}\right)^{1/4}e^{-bx^{2}}\), an upper bound for the ground state energy of a one-dimensional harmonic oscillator, oscillating with frequency ω, is :
Answer: B
The upper bound is \(\frac{1}{2}\hbar\omega\).
The variational energy for this Gaussian trial function is:
\[ \begin{aligned} \langle H \rangle &= \frac{\hbar^{2}b}{2m} \\ &\quad + \frac{m\omega^{2}}{8b} \end{aligned} \]
Minimising with respect to \(b\) gives \(b = \frac{m\omega}{2\hbar}\) and \(\langle H \rangle = \frac{1}{2}\hbar\omega\). The bound is exact here because a Gaussian of this width is the true ground state of the harmonic oscillator.
\(\frac{3}{2}\hbar\omega\) is the first excited state energy, and \(\frac{3}{4}\hbar\omega\) and \(\hbar\omega\) are values the variational energy takes for non-optimal widths, not the best bound.
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