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General Solution to time independent schrodinger equation

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data

    This is really a maths problem I'm having.

    I need to get the general solution for the infinite square well in the form:

    u = A cos(kx) + B sin(kx)



    I found the general solution to be:

    u = A exp(ikx) + B exp(-ikx)

    Using Euler's formula:

    exp(ikx) = cos(kx) + isin (kx)
    exp(-ikx) = cos(kx) - isin(kx)

    Hence, general solution becomes :

    u = A cos(kx) + Bcos(kx)

    What am I missing out?
     
  2. jcsd
  3. May 6, 2010 #2

    Cyosis

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    Show how you got from the exponential form to the double cosine form. I have a feeling you think that A sin x- B sin x=0.
     
  4. May 6, 2010 #3
    Yes you're right! That was a mistake I've made:

    Instead I get:

    u = (A + B) cos(kx) + (A - B) i sin(kx)

    I still dont know how to get the correct form from this..
     
  5. May 6, 2010 #4

    Cyosis

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    The problem you're having is that you labeled the integration constants for both types of solutions the same. This causes confusion.

    Instead use [itex]u=C e^{ikx}+De^{-ikx}[/itex] and find a relation between C,D and A,B (just relabel A and B in post #3).
     
    Last edited: May 6, 2010
  6. May 6, 2010 #5
    I'm not sure what you mean. How would changing A & B to C & D make a difference?
     
  7. May 6, 2010 #6

    Cyosis

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    Because the A in your first equation is not necessarily the same as the A in your second equation (post #1). Using the same name for different constants gets you into trouble. Just use the u given in my post with constants C and D and find a relation between those and A and B.
     
  8. May 6, 2010 #7
    Forgive me for asking again. I seem to be confused over something so simple.

    I did as you said..I think and got:

    C exp(ikx) = A cos(kx) + iB sin(kx)

    D exp(-ikx) = A cos(kx) + iB sin(kx)
     
  9. May 6, 2010 #8

    vela

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    What people are trying to tell you is that [itex]Ae^{ikx}+Be^{-ikx} \ne A\cos(kx)+B\sin(kx)[/itex]. You can't get one from the other because they're not equal to each other. You can, however, show that [itex]Ce^{ikx}+De^{-ikx} = A\cos(kx)+B\sin(kx)[/itex] if you choose A and B correctly. Remember that C and D are arbitrary constants. Any combination of them will just be another arbitrary constant.
     
  10. May 6, 2010 #9
    My previous post was wrong, sorry, is it simply:

    C exp(ikx) = C cos(kx) + iC sin(kx)

    D exp(-ikx) = D cos(kx) - iD sin(kx)

    So,

    C+D = A

    (iC - iD) = B


    Therefore:

    C exp(ikx) + D exp(-ikx) = A cos(kx) + B sin(kx)
     
  11. May 6, 2010 #10

    vela

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    Yup, you got it.
     
  12. May 7, 2010 #11
    Thanks for the patience everyone
     
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