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I Time independent Schrodinger equation results (1D)

  1. May 8, 2017 #1
    okay so i need some help interpreting some of the results,
    so (-ħ2/2m)Ψ''=E-V0Ψ;
    So i set k2= 2m*(E-V0)/ħ2
    and so : Ψ''=-k2Ψ
    so if V0=0 or is smaller than E, k2 is positive;
    *need for help starts here*
    Ψ=Aeikx+Be-ikx;
    another result for this would also be only eikx so is the second term only there to represent the possibility of the wave to travel in both directions?
    now for the trignometric result is the same, are both the sines and cosines there to represent both directions?

    ok so when V0 is bigger than E i can just set k2=-2m*(E-V0)/ħ2 and so
    Ψ=Aekx+Be-kx
    there is no trig result correct?
     
  2. jcsd
  3. May 8, 2017 #2

    BvU

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    No. The equation is time independent. The Schroedinger equation is linear, so if 1 is a solution and 2 is a solution, then A*1 + B*2 is also a solution. And you need two integrations constants to describe a particular solution of a second order differential equation.
    No. It's just a rewrite of the other solution.
    You can rewrite this one in hyperbolic sine/cosine. Not really trigonometric, but close.
    And: these solutions can only apply piecewise, because they go to infinity far away.
     
  4. May 8, 2017 #3
    So if i found another function f(x) that was a solution to the differential equation it should be included?
    About the direction part, i remember an problem where the wave/particle came from left to right and encountered a step with a certain lenght, there was a reflected wave on the left of the beggining of the step and one inside the step but on the right of the step there wasn't, and that translated into:
    Left of the step:
    e^ikx +re^-ikx
    Inside:
    Ae^iqx +Be^-ikx
    Right of the step:
    te^ikx
    And so what i understood from it was that re^-ikx and Be^-iqx where about the reflected waves and since to the right of the step there was nothing "reflecting" the wave so there was no -ikx term on it; is this a wrong way of looking at it?
    I dont know if this was clear enough i can provide drawings and more info about the problem if necessary (1:30 am here, so tommorow)
    Thanks for your answer!!
     
  5. May 9, 2017 #3

    BvU

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    Just fine - but I think I remember that for ##k## the only requirement is on ##k^2##
    Check the full time-dependent solution for ##\pm \hbar k \pm \omega t##
     
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