= Solution
The <boundary conditions> on the positive half-line, absent from the transcription, select the <half-harmonic oscillator>. The trial state $xe^{-\alpha x}$ with $\alpha>0$ satisfies both endpoints and has finite <norm> and energy. <Integration by parts>, with vanishing endpoint term, gives
$$
\Lambda(\alpha)=\frac{\int_0^\infty[|\widetilde\psi'|^2+x^2|\widetilde\psi|^2]dx}{\int_0^\infty|\widetilde\psi|^2dx}.
$$
Using $\int_0^\infty x^ne^{-2\alpha x}dx=n!/(2\alpha)^{n+1}$, the three required <integrals> are
$$
\int_0^\infty\widetilde\psi^2dx=\frac1{4\alpha^3},\qquad\int_0^\infty(\widetilde\psi')^2dx=\frac1{4\alpha},\qquad\int_0^\infty x^2\widetilde\psi^2dx=\frac3{4\alpha^5}.
$$
Thus the <exponential trial state for a half-harmonic oscillator> gives $\Lambda(\alpha)=\alpha^2+3/\alpha^2$. Its <derivative> vanishes at $\alpha^4=3$, and the <arithmetic-geometric mean inequality> confirms the global minimum:
$$
\boxed{\alpha_*=3^{1/4},\qquad\lambda_0\leq\lambda_{\rm estimate}=2\sqrt3\simeq3.46410.}
$$
For comparison, the exact lowest half-line state is proportional to $xe^{-x^2/2}$ and has <eigenvalue> three. The exponential trial family gives a valid upper bound rather than an exact <eigenvalue>.
Back to article page
