1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability current density

  1. Nov 14, 2006 #1
    Consider the infinite square well potentail over the range 0<x<a with energy eigenfunctions [itex]Psi _{n} (x,t)[/itex]

    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
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Probability current density
  1. Probability density (Replies: 6)