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Normalization of wave function in x, y and z

  1. Oct 27, 2007 #1
    Hello,

    How do I find the normalization constant for psi(x,y,z) = N exp -(x/2+y/2+z/2) ??

    I did the following:

    [tex]\int(psi^* psi)dx dy dz = 1 [/tex]

    the integral bounds are from -infinity to infinity and the * means the complex conjugate.The integral is so weird that I couldn't find N. I used maple to evalulate the integral and it gave me something like infinity multiplied by some other things.

    Please help.
    thanks
     
  2. jcsd
  3. Oct 27, 2007 #2

    nrqed

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    Welcome to the forums!

    The equation for the normalization you wrote is entirely correct.
    However, I am sure that you have the wrong expression for the wavefunction because the wavefunction you wrote down is non normalizable. Please double check the problem because I am sure this is not the correct wavefunction. Are you sure it was not [tex] N exp -(\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2})[/tex]
    instead?
     
  4. Oct 27, 2007 #3
    That's what I thought too, but I double checked and I have it written in two sources. The only difference between what I typed here and the original is that the x, y and z are in absolute, that is: |x|, |y|, and |z| and that each is divided by a constant, a, b and c respectiviely.
     
  5. Oct 28, 2007 #4

    dextercioby

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    I see. What does [itex] \int_{-\infty}^{\infty} e^{-|x|} \ {} dx [/itex] equal to ?
     
  6. Oct 28, 2007 #5
    [tex]-e^(-|x|)[/tex] evaluated from -infinity to infinity which is zero?
     
  7. Oct 28, 2007 #6
    Oops, I guess the derivative of |x| is : |x|/x
    But that gives an infinity/infinity which is not defined!
     
    Last edited: Oct 28, 2007
  8. Oct 28, 2007 #7

    dextercioby

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    Do you know how to explicitate the || ?
     
  9. Oct 28, 2007 #8
    I am not sure what you mean by explicitate.
     
  10. Oct 28, 2007 #9

    dextercioby

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    Simply means to write what |x| is equal to on [itex]\mathbb{R} [/itex].
     
  11. Oct 28, 2007 #10
    yes. the value inside the absolute can be +ve or -ve, but the absolute value makes it always +ve regardless of the sign of x, so :

    |x| = x when x >=0 and -x when x <0

    is that what you meant>?
     
  12. Oct 28, 2007 #11

    dextercioby

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    Yes, now can you compute the integral i wrote above ?
     
  13. Oct 28, 2007 #12
    I get -infinity or :

    - (|x|/x * e^ -|x|) eval. bw infinity and -infinity so =
    - (1*e^-infinity - {- e^infinity}) = infinity
     
  14. Oct 28, 2007 #13
    Lorna, try splitting the integral that dextercioby gave you into two pieces.
     
  15. Oct 28, 2007 #14
    I am not sure I know how to do it, that's why I'm asking. I get:

    answer (if x >=0) = -e^-x (eval from -inf to +inf)
    answer (if x < 0) = e^x (eval from -inf to +inf)

    I think I'm missing something here
     
  16. Oct 29, 2007 #15
    Why are you integrating both halves of the function from -infinity to +infinity? The negative side should be integrated from -infinity to 0, and the positive side should be integrated from 0 to +infinity.
     
  17. Oct 29, 2007 #16
    Wow Manchot, now I see how it works! Thanks a lot for all who helped.
     
  18. Oct 29, 2007 #17

    nrqed

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    Ahhh! You did not mention absolute values in your initial post!!!!!

    You should not have an overall minus sign in front.

    It's simply this:


    for x>0, we have [tex] e^{- |x|} = e^{-x} [/tex]

    and for x<0, we have [tex] e^{- |x|} = e^{x} [/tex]

    Now you simply have to break up the x integral into
    [tex] \int_{\infty}^{\infty} = \int_{-\infty}^0 + \int_0^{\infty} [/tex] and in each integral replace the exponential of the absolute value by its expression given above.
     
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