Question #155505

The wave function of a quantum particle of mass m is πœ“ (π‘₯) = 𝐴 cos(π‘˜π‘₯) + 𝐡 sin(π‘˜π‘₯) where 𝐴, 𝐡 π‘Žπ‘›π‘‘ π‘˜ are constants. (a) Assuming the particle is free (π‘ˆ = 0), show that πœ“ (π‘₯) is a solution of the π‘†π‘β„Žπ‘ŸΓΆπ‘‘π‘–π‘›π‘”π‘’π‘Ÿ equation. (b) Find the corresponding energy E of the particle.


1
Expert's answer
2021-01-17T14:45:09-0500

The wave function of a quantum particle satisfies the Schrodinger equation

βˆ’β„22md2ψ(x)dx2+U(x)ψ(x)=Eψ(x)-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+U(x)\psi(x)=E\psi(x)

In the case of free particle

βˆ’β„22md2ψ(x)dx2=Eψ(x)-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}=E\psi(x)d2ψ(x)dx2=d2dx2(Acos⁑(kx)+Bsin⁑(kx))=βˆ’k2(Acos⁑(kx)+Bsin⁑(kx))=βˆ’k2ψ(x)\frac{d^2\psi(x)}{dx^2}=\frac{d^2}{dx^2}\left(A\cos(kx)+B\sin(kx)\right)\\ =-k^2\left(A\cos(kx)+B\sin(kx)\right)=-k^2\psi(x)

Hence, the ψ(x)\psi(x) is a solution of the Schrodinger equation if


E=ℏ2k22mE=\frac{\hbar^2 k^2}{2m}

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