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Question concerning particle in a box

  1. Apr 4, 2006 #1
    I need to find the probability density function for a particle in a box of width L. The initial state is 1/sqrt(2) u1(x) + 1/sqrt(2) u2(x), where u1(x) is the ground state and u2(x) is the first excited state. Omega is E2-E1/h

    If I square this wave function, I get 1/2 u1(x)u1(x) +1/2 u2(x) u2(x) +u1(x)u2(x)

    The solution given is 1/2 u1(x) u1(x) + 1/2 u2(x) u2(x) + 2 cos (wt) u1(x) u2(x)

    My question is where does the 2 cos (wt) come from?
  2. jcsd
  3. Apr 4, 2006 #2

    Physics Monkey

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    Remember that energy eigenstates evolve in time like exp(- i En t / hbar), so you have to associate such a factor with each of your states. If you now you calculate the absolute square, the factor in question should come right out.
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