1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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,

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

    rearranged is,

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

    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


    User Avatar
    Science Advisor
    Gold Member


    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
  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

    how do i do this

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


    User Avatar
    Homework Helper

    You have the Schrodinger equation wrong. It should be:

    i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook