- #1

touqra

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In the infinite square well potential, the obtained wavefunction is,

[tex] \psi = \sqrt\frac{2}{a} sin \frac{n\pi x}{a} [/tex]

and we know that the Hamiltonian commutes with the momentum operator, which implies that the eigenfunctions for the Hamiltonian is exactly the same for the momentum operator. However, when I operate the momentum operator on the wavefunction above, I failed to get a momentum eigenvalue, together with the eigenfunction (wavefunction) again.

Is there something wrong with the wavefunctions, in that they are actually not the Hamiltonian's eigenfunction, or?

[tex] \psi = \sqrt\frac{2}{a} sin \frac{n\pi x}{a} [/tex]

and we know that the Hamiltonian commutes with the momentum operator, which implies that the eigenfunctions for the Hamiltonian is exactly the same for the momentum operator. However, when I operate the momentum operator on the wavefunction above, I failed to get a momentum eigenvalue, together with the eigenfunction (wavefunction) again.

Is there something wrong with the wavefunctions, in that they are actually not the Hamiltonian's eigenfunction, or?

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