Answer to Question #202006 in Quantum Mechanics for Vinod Kumar

Question

A particle of mass m and zero energy has a wave function Ψ (x)= Nx exp -x^2/16 , where

N is a constant. Determine the potential energy V (x) for the particle.

Answer 1

$Ψ (x)= Nx e^{ -x^2/16}$

In order to find the energy potential in which the particle is moving, we must insert

$\Psi(x,t)$ in Schrodinger Equation,

$i\hbar\dfrac{\partial \Psi}{\partial t}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2 \Psi}{\partial x^2}+V\Psi$

$\dfrac{\partial \Psi}{\partial t}=0$

$\dfrac{\partial \Psi}{\partial x}=Ne^{-x^2/16}+Nxe^{-x^2/16}\times\dfrac{-2x}{16}$

$\dfrac{\partial \Psi}{\partial x}=Ne^{-x^2/16}\bigg(1-\dfrac{x^2}{8}\bigg)$

$\dfrac{\partial^2 \Psi}{\partial x^2}=Ne^{-x^2/16}\dfrac{x}{4}+N\dfrac{x}{8}e^{-x^2/16}\bigg(1-\dfrac{x^2}{8}\bigg)$

$\dfrac{\partial^2 \Psi}{\partial x^2}=\dfrac{\Psi(x)}{4}+\dfrac{\Psi(x)}{8}\bigg(1-\dfrac{x^2}{8}\bigg)$

$\dfrac{\partial^2 \Psi}{\partial x^2}=\dfrac{\Psi(x)}{32}({12-x^2})$

Substituting these values in Schrodinger equation

$V\Psi(x)=\dfrac{\hbar^2}{2m}\dfrac{\Psi(x)}{32}({12-x^2})$

$V(x)=\dfrac{\hbar^2}{64m}({12-x^2})$