Question #203210

Show that the wavefunction Ψ(x) = N sin kx + iN cos kx is an eigen function of the momentum operator and determine its eigen value.


1
Expert's answer
2021-06-07T09:32:11-0400

The momentum operator is p^=ix\hat{p} =- i\hbar\dfrac{\partial}{\partial x}. By definition, the eigen function has the following property:


p^Ψ(x)=λΨ(x)\hat{p}\Psi(x) = \lambda\Psi(x)

where λ\lambda is the corresponding eigenvalue. Substituting the function and peforming the differentiation, obtain:

p^Ψ(x)=ix(Nsin(kx)+iNcos(kx))=i(Nkcos(kx)iNksin(kx))==iNkcos(kx)Nksin(kx)\hat{p}\Psi(x) = - i\hbar\dfrac{\partial}{\partial x} (N\sin(kx) + iN\cos(kx)) = - i\hbar(Nkcos(kx) - iNk\sin(kx)) =\\ =-i\hbar Nkcos(kx) - \hbar Nk\sin(kx)

Factoring out k-\hbar k , have:


p^Ψ(x)=k(Nsin(kx)+iNcos(kx))=kΨ(x)\hat{p}\Psi(x) = -\hbar k(N\sin(kx) + iN\cos(kx)) = -\hbar k\Psi(x)


Thus, this finction is indeed the eigen function of the momentum operator. Comparing the last equation with the first one, find the eigenvalue:


λ=k\lambda = -\hbar k

Answer. Ψ(x)\Psi(x) is the eigenfunction of the momentum operator with the eigenvalue λ=k\lambda = -\hbar k.


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