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So, here's the question:

[tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex]

Where A, n, and X0 are constants.

Using Schrodinger's equation, find the potential U(x) and energy E such that the given wave function is an eigenfunction (we can assume that at x = infinity there is 0 potential).

So, in my futile attempt at beginning this I tried to use [tex] \hat p\psi(x) = p \psi(x) [/tex] but my partial of psi with respect to x left me with a difference, due to the product rule which left me with [tex] -i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p [/tex] which isn't an eigenvalue for momentum. Using an E operator makes no sense, because there's no t term so the derivative goes to 0 and it just doesn't make sense.

Forget the right track, is this the right station?

[tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex]

Where A, n, and X0 are constants.

Using Schrodinger's equation, find the potential U(x) and energy E such that the given wave function is an eigenfunction (we can assume that at x = infinity there is 0 potential).

So, in my futile attempt at beginning this I tried to use [tex] \hat p\psi(x) = p \psi(x) [/tex] but my partial of psi with respect to x left me with a difference, due to the product rule which left me with [tex] -i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p [/tex] which isn't an eigenvalue for momentum. Using an E operator makes no sense, because there's no t term so the derivative goes to 0 and it just doesn't make sense.

Forget the right track, is this the right station?

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