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One-dimensional Schrodinger Equation

  1. Jun 12, 2004 #1
    one-dimensional Schrodinger's Equation

    Hi !
    I wonder how to solve one-dimensional Schrodinger's Equation :

    [tex] \frac{d^2 \psi (x)}{dx^2}\ = \ -(\frac{2 \pi}{\lambda})^2 \ * \ \psi (x) [/tex]

    I've to obtain [tex]\psi (x) [/tex] , when
    [tex]-(\frac{2 \pi}{\lambda})^2 [/tex] is known
    Can you solve it as an example, when [tex]-(\frac{2 \pi}{\lambda})^2 = 2 [/tex] ? Please...
     
    Last edited: Jun 12, 2004
  2. jcsd
  3. Jun 12, 2004 #2
    Based on my rather basic knowledge, making
    [tex]
    - \left( \frac{2 \pi}{ \lambda} \right)^2 = 2
    [/tex]
    would make the psi function an exponential function rather than having the wave characteristics you want.The solution would be:
    [tex]
    \psi(x) = e^{ \sqrt{2} x } .
    [/tex]
    But you want a periodic wave function. By taking
    [tex]
    \frac{2 \pi}{\lambda} = \sqrt{2}
    [/tex]
    (i.e. without the minus sign), the solution would be a wave equation and complex:
    [tex]
    \psi(x) = e^{ \sqrt{2} i x } .
    [/tex]
     
    Last edited: Jun 12, 2004
  4. Jun 12, 2004 #3

    Tom Mattson

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    You shouldn't commit to a specific value of -(π/λ)2. Just define it to be a literal constant, say, -k2. Then your differential equation is:

    ψ"+k2ψ=0

    Surely you have the solution of that one somewhere!
     
  5. Jun 12, 2004 #4

    Tom Mattson

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    No, an exponential function arises when the first derivative is negatively proportional to the wavefunction itself. The problem here involves the second derivative.

    The solution to his equation is indeed sinusoidal, but let's let him work it out. :wink:
     
  6. Jun 12, 2004 #5
    In fact [tex]$\frac{{2\pi }}
    {\lambda } = k$
    [/tex] when k is the wavenumber, so the k^2 substution can be made even without the knowledge of how it helps solve the DE.
     
  7. Jun 13, 2004 #6

    Tom Mattson

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    Yes, my choice in symbols was not accidental. :wink:
     
  8. Jun 13, 2004 #7
    Is it the correct answer ?:
    [tex]\psi (x) = e^{ikx + \varphi_R + i \varphi_I} = e^{\varphi_R} \cdot e^{i(kx + \varphi_I)} = A(\cos (kx + \varphi_I) + i \sin (kx + \varphi_I))[/tex]
     
  9. Jun 13, 2004 #8

    Tom Mattson

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    Yes, though you might find it more convenient to replace the constants A and φI with A and B as follows:

    ψ(x)=Acos(kx)+Bsin(kx),

    and applying the boundary conditions from there.
     
  10. Jun 13, 2004 #9
    Ok - thanks :biggrin:
     
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