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Show that if the quantum state of a particle is descirbed by a single eigenfuncation i.e. the particle has a sharply defined eneryg, then the probability current density inside the well vanishes

ok Delta E = 0

[tex] \Psi (x,t) = \psi(x) \exp(\frac{-iEt}{m} [/tex]

[tex] j= \frac{\hbar}{2im} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) [/tex]

ok lets say that [tex] P(x,t) = | \Psi(x,t)|^2 = \psi^*(x) \psi(x)[/tex]

we also know that \frac{\partial}{\partial t} (\psi^* \psi)= -\frac{\hbar}{2im} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) [/tex]

and hence [tex] \frac{\partial}{\partial t} P(x,t) = \frac{\partial}{\partial t}( \psi^*(x) \psi(x) )= 0 = -\frac{\partial}{\partial t} j(x,t) = 0 [/tex]

so the position derivative of probability current wrt position is zero

but does that show that it vanishes??

thank you for your input