Question

Consider the following one-dimension simple harmonic oscillator Hamiltonian operator
equation$
Use a trial wave function equation with a variational parameter equation to estimate the upper bound to the ground state energy. 

19 Jan 2026
Answer :
Word Count : 521
We are asked to estimate the ground state energy of a one-dimensional harmonic oscillator using the variational method with the trial wave function [ \psi(x) = N \exp\Big(-\frac{x^2}{2\alpha^2}\Big), ] where (\alpha) is the variational parameter. Let’s solve this step by step, manually. --- 1. Normalize the trial wave function The normalization condition is: [ \int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1 ] [ \int_{-\infty}^{\infty} N^2 \exp\Big(-\frac{x^2}{\alpha^2}\Big) dx = 1 ] We use the standard Gaussian integral: [ \int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}} ] Here, (a = 1/\alpha^2), so [ \int_{-\infty}^{\infty} e^{-x^2/\alpha^2} dx = \alpha \sqrt{\pi} ] Thus, [ N^2 (\alpha \sqrt{\pi}) = 1 \quad \Rightarrow \quad N = \frac{1}{\sqrt{\alpha \sqrt{\pi}}} = \frac{1}{\pi^{1/4} \sqrt{\alpha}} ] --- 2. Write the Hamiltonian expectation value The Hamiltonian is: [ \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 ] The expectation value is: [ E(\alpha) = \langle \psi | \hat{H} | \psi \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{H} \psi(x) dx ] We
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