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A silly question regarding general solutions to second order DE's

  1. Mar 27, 2008 #1
    It's been a while since I've had DE, and it seems I have forgotten something basic, yet crucial. In solving the Time-Independent Schrodinger Wave Equation in one dimension, I have the following:

    [tex]\left( -\frac{\hbar}{2m} \frac{\partial^2}{\partial x^2}+V\right) \Psi(x)=E\Psi(X)[/tex]

    [tex]\frac{\partial^2}{\partial x^2}\Psi(x)+K^2\Psi(x)=0[/tex]

    where V=0 and

    [tex] K^2=\frac{2mE}{\hbar^2}[/tex]

    And my question is: are the following solutions equal?

    [tex]\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}[/tex]

    [tex]\Psi(x)=C_1Cos(Kx)+C_2Sin(Kx)[/tex]

    I know that Euler's Formula is [tex]e^{i\theta x}=Cos(\theta x)+iSin(\theta x)[/tex], however inserting this into the first solution above does not result in the second. Thanks!

    IHateMayonnaise
     
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2

    Dick

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    There is no difference between those two forms of the solution. But they aren't 'equal'. When you go from one to the other you need to change the values of C1 and C2. For example C1 in the real form is equal to C1+C2 in the complex form, C2 in the real form is i*(C1-C2) in the complex form.
     
    Last edited: Mar 27, 2008
  4. Mar 27, 2008 #3
    Well, when solving The Schrodinger equation, how do I know which form to take? Can I just put it in one form or the other and solve for the coefficients?
     
  5. Mar 27, 2008 #4

    Dick

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    You can use them interchangeably. E.g. 2*exp(iKx)+3*exp(-iKx)=(2+3)*cos(Kx)+i*(2-3)*Sin(Kx) (check that using Euler's formula). They are really just plain equal. Depends on whether you have a preference for where to put the i's.
     
    Last edited: Mar 27, 2008
  6. Mar 27, 2008 #5
    Well what you have there makes sense to me, what confuses me is why there isn't an i in front of the sine term in the second solution

    [tex]\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}[/tex]

    [tex]\Psi(x)=C_1Cos(Kx)+C_2Sin(Kx)[/tex]

    Since

    [tex]\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}=C_1(Cos(Kx)+iSin(Kx))+C_2(Cos(Kx)-iSin(Kx)) = (C_1+C_2)Cos(Kx)+(C_1-C_2)iSin(Kx) [/tex]

    and the "i" is still there!! Why?? (I get the part about the coefficients)
     
  7. Mar 27, 2008 #6

    Dick

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    Because to express a completely general COMPLEX solution to the Schrodinger equation, C1 and C2 need to be COMPLEX constants. In BOTH forms. You can only restrict them to be real (for example) if you have have special boundary conditions and use the freedom to choose an arbitrary phase and normalization for psi.
     
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