Answer to Question #154868 in Quantum Mechanics for ali

Question #154868

Consider the wave function πœ“(π‘₯) = π΄π‘π‘œπ‘  π‘˜π‘₯ where A, k are constants. Is this the eigenstate of the operators 𝐻̂, 𝑝̂, 𝑝̂ 2 of a particle moving in a potential-free region. If so find the eigenvalues.


1
Expert's answer
2021-01-11T11:34:59-0500

To verify it, we will simply apply these operators (we know their "x" representation) :

  • "\\hat{p}\\psi=-i\\hbar \\frac{d}{dx}\\psi = i\\hbar Ak \\sin(kx)" . We clearly see that "\\psi, \\hat{p}\\psi" are not collinear (as sine and cosine are not proportional to each other), so "\\psi" is not an eigenstate of "\\hat{p}"
  • "\\hat{p}^2\\psi = \\hat{p}(\\hat{p}\\psi) = -i\\hbar\\frac{d}{dx}(i\\hbar Ak\\sin(kx)) = A \\hbar^2k^2\\cos(kx) =\\hbar^2k^2\\psi" . So "\\psi" is an eigenstate of "\\hat{p}^2" with an eigenvalue "\\hbar^2k^2" .
  • "\\hat{H} \\psi = (\\frac{\\hat{p}^2}{2m}+\\hat{V})\\psi =\\frac{1}{2m} \\hat{p}^2\\psi = \\frac{\\hbar^2k^2}{2m}\\psi" . Again "\\psi" is an eigenstate of "\\hat{H}" with an eigenvalue "\\frac{\\hbar^2k^2}{2m}" .

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