Answer to Question #236590 in Quantum Mechanics for dumela

Question #236590
Consider the Schrödinger equation for a particle in the presence of the potential
of the form
1
Expert's answer
2021-09-13T12:17:17-0400

h2d2ψ(x)2mdx2+V(x)ψ(x)=Eψ(x)\frac{h^2d^2\psi(x)}{2m{dx^2}}+V(x)\psi(x)= E\psi(x)


and

ψ(x)=Axex2c2\psi(x) = Ax{e^\frac{-x^2}{c^2}} having E=0E = 0

Therefore; h2d2ψ(x)2mdx2+V(x)ψ(x)=0\frac{h^2d^2\psi(x)}{2m{dx^2}}+V(x)\psi(x)= 0


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