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Solution to time dependent wave equation

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that Y(x,t) = cos(kx)exp(-iwt) is a solution to the time-dependent Schrodinger wave equation.

    where k is the wavenumber and w is the angular frequency

    2. Relevant equations

    Hamiltonian of Y(x,t) = ihbar d/dt Y(x,t)

    3. The attempt at a solution

    When I plug everything in and do the calculus, I get

    (hbar / 2m)*k^2 = w

    I've been paying around with it but for the life of me cannot figure out how the left hand side is equal to w even though the unit check suggests it could. So my question is, how can you simplify the left hand side to w

    (hbar / 2m)*k^2 = ?
     
  2. jcsd
  3. Sep 28, 2009 #2

    gabbagabbahey

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    Isn't there some relation between wavenumber and angular frequency given in your textbook?:wink:
     
  4. Sep 28, 2009 #3
    I can't find it. I know angular frequency is 2pi*nu and wavenumber is one over lambda. And I know nu*lambda = c, if that's the equation you mean, but still cannot get it to work....!

    (hbar / 2m)(1/lambda^2) = (h/(4pi*m*lambda^2)) =...?
     
  5. Sep 28, 2009 #4
    Last edited by a moderator: May 4, 2017
  6. Sep 28, 2009 #5
    Thanks! I got it now. A quick follow up: why is it assumed that the energy component for momentum (related to the wavenumber in this problem) is entirely kinetic?

    i.e.,

    p = mv => p^2 = m^2 v^2 = 2m (1/2) m v^2 = 2m E => p = sqrt(2mE)
     
  7. Sep 28, 2009 #6

    gabbagabbahey

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    A free particle (I assume your problem is for a free particle, although you didn't actually say so in your problem statement) is subject to zero external forces/potentials, so its energy (in non-relativistic QM) is entirely kinetic.
     
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