joex444
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So, here's the question:
\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}})
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 \hat p\psi(x) = p \psi(x) but my partial of psi with respect to x left me with a difference, due to the product rule which left me with -i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p 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?
\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}})
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 \hat p\psi(x) = p \psi(x) but my partial of psi with respect to x left me with a difference, due to the product rule which left me with -i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p 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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