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Re-arranging the Schrodinger equation

  1. Apr 11, 2010 #1
    I just have a small question,

    In my book it says that the schrodinger equation,

    [tex]
    i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
    [/tex]

    rearranged is,

    [tex]
    \frac{\partial\Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial\Psi ^2 psi}{\partial x^2} - \frac{i}{\hbar}V\Psi
    [/tex]

    how does the complex number, move over, and in the numerator? instead of the denominatior?

    I can see how [tex] A\hbar = B\hbar ^2 becomes A = B \hbar [/tex]

    but I don't understand how

    [tex] A i = B + V\Psi becomes A = iB - i V\hbar [/tex]

    could someone please explain to me the mathematical rules behind rearranging complex numbers in equations,

    or give me some links that explain it, (in simple terms) please :P
     
  2. jcsd
  3. Apr 11, 2010 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    [tex]\frac{1}{i}=-i[/tex]

    This is one of the properties of imaginary numbers. I don't exactly recall a proof for this...hopefully someone else can answer your question in more detail.
     
  4. Apr 11, 2010 #3
    Do you know what I would search for If i wanted to understand how to manipulate i?

    I tried the wikipedia log of complex numbers, but it is jungle of crap that is too hard to understand
    tl;dr
     
  5. Apr 11, 2010 #4
    Like this?
    [tex]\frac{1}{i} = \frac{1}{i}\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i[/tex]
     
  6. Apr 11, 2010 #5
    JaWiB

    how do i do this

    [tex]
    A i = B + V\Psi to A = iB - i V\Psi
    [/tex]
     
  7. Apr 11, 2010 #6
    I don't think what you have is correct.
    [tex]
    i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
    [/tex]
    If you multiply both sides by [tex]i/\hbar[/tex], you get
    [tex]
    -\frac{\partial\Psi}{\partial t} = i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} + \frac{i}{\hbar}V\Psi
    [/tex]
    or
    [tex]
    \frac{\partial\Psi}{\partial t} = -i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi
    [/tex]
     
  8. Apr 11, 2010 #7
    mmmm well that's closer then to what I had,

    thank you i'll ask my tutors tomorrow
     
  9. Apr 11, 2010 #8

    Cyosis

    User Avatar
    Homework Helper

    You have the Schrodinger equation wrong. It should be:

    [tex]
    i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
    [/tex]
     
  10. Apr 11, 2010 #9
    so Jawib's way does work!??
     
  11. Apr 11, 2010 #10
    I need to take some math papers...
    I'm not as good as the rest of the physics majors at mathematics,

    I've only done 1 math paper and 5 physics ones

    I didn't even think to multiply both sides by i/h, like Jawib said :(

    study study study
     
  12. Apr 11, 2010 #11
    If you had kept at it, you would have got it eventually. You gave up. If you give up a lot that is a problem. Half-hearted studying won't get you anywhere.
     
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