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: Normalize a wave function

  1. Apr 18, 2010 #1
    At time = 0 a particle is represented by the wave function

    [tex] \Psi(x,0) = \left\{ \begin{array}{ccc}
    A\frac{x}{a}, & if 0 \leq x \leq a, \\
    A\frac{b-x}{b-a}, & if a \leq x \leq b, \\
    0, & otherwise,
    \end{array} \right
    [/tex]


    where A, a, and b are constants.

    (a) Normalize [tex] \Psi [/tex] (that is, find A, in terms of a and b).

    (b) where is the particle most likely to be found, at t =0?

    (c) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b=a and b = 2a

    (d) what is the expectation value of x?
     
  2. jcsd
  3. Apr 18, 2010 #2
    My attempt:

    no bloody idea where to start,

    I know that to normalize a function,

    [tex] \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx =1 [/tex]

    But have no idea how I am supposed to incorporate that into my question...

    I need EXAMPLES
    I have no idea how to solve something i've never seen before!
     
  4. Apr 18, 2010 #3
    ok I'm just going to start guessing here,

    if someone would be kind enough to correct me on mistakes

    ----------------------------------------------------------

    since I want to find A,
    and the wave function [tex] \Psi(x,0) = A \frac{x}{a} [/tex]

    if I square it, it should = 1?
    [tex] \Psi(x,0)^2 = A^2 \frac{x^2}{a^2} ==1 [/tex]

    not too sure where to go from here
     
  5. Apr 18, 2010 #4
    the same with the middle equation,

    [tex]
    \Psi(x,0)^2 =(A\frac{(b-x}{b-a})^2) =1
    [/tex]
     
  6. Apr 18, 2010 #5
    found this

    http://en.wikipedia.org/wiki/Normalisable_wave_function

    so

    [tex] \frac{A^2}{a^2} \int_0^a x^2 dx =1 [/tex]

    the question says A, a and b are constants so i'm taking them outside the integral

    [tex] \frac{A}{b-a} \int_a^b (b-x)^2 dx =1 [/tex]
     
  7. Apr 18, 2010 #6
    You have the right idea but the two integrals added together should be equal to one I do believe. Also be sure to square the A/(b-a) on the second integral.
     
  8. Apr 18, 2010 #7
    When I am normalizing the wave function, do I add the equations from all the parts of the wave function?

    i.e

    [tex] \int_{-\infty}^{0} +\int_{0}^{a}+\int_{a}^{b}+\int_{b}^\infty [/tex]
     
  9. Apr 19, 2010 #8
    so adding that to concept to my question,

    i'd get

    [tex]|\Psi(x,0)|^2 = \int_{-\infty}^0 0^2dx + \frac{A^2}{a^2}\int_0^a x^2 dx + \frac{A}{(b-a)} \int_a^b (b-x)^2 dx + \int_b^\infty 0^2dx = [/tex]
     
  10. Apr 19, 2010 #9
    [tex]
    = |\Psi(x,0)|^2 = \frac{A^2}{a^2}\int_0^a x^2 dx + \frac{A^2}{(b-a)^2} \int_a^b (b-x)^2 dx
    [/tex]



    then uh simplifying and such i get,
    (since A^2 is in both equations)

    [tex] \left( A^2 \left[ \left( \frac{1}{a^2} \frac{x^3}{3} \right) + \left( \frac{1}{(b-a)^2} \int_a^b (b-x)^2 dx \right) \right] \right) =1 [/tex]
     
    Last edited: Apr 19, 2010
  11. Apr 19, 2010 #10
    That is almost correct except A/(b-a) should be squared as well and then all of that will be equal to one. Then you should do each integral then just solve for A.
     
  12. Apr 19, 2010 #11
    how do I integrate (b-x)^2 ?

    [tex] \int (b-x)(b-x)dx = \int (b^2 - 2bx + x^2 )dx= b^2 -\frac{2bx^2}{2} + \frac{x^3}{3} [/tex]

    ? then i evaluate that at [tex] & \left equation \right|_{a}^{b} [/tex]
     
  13. Apr 19, 2010 #12
    For b2, you can think of that as b2*x0. So when you take the integral of that, you would end up with b2*x1. Other than that, everything else looks correct.
     
  14. Apr 19, 2010 #13
    [tex]\int (b^2 - 2bx + x^2) dx = b^2x - \frac{2bx^2}{2} + \frac{x^3}{3} [/tex]

    [tex]

    b^2x - bx^2 +\frac{x^3}{3}

    [/tex]

    what now?
    I was shown that the answer to that integral was

    [tex] -\frac{(b-x)^3}{3} [/tex]

    but how do i get that???
    from

    [tex] \int_a^b (b-x)^2 dx [/tex]
    to
    [tex] \left -\frac{(b-x)^3}{3} \right|_a^b [/tex]
     
  15. Apr 19, 2010 #14
    They used u=(b-x) => du=-dx

    You are on a roll and almost there -- keep up the good work!
     
  16. Apr 21, 2010 #15
    embarassing

    but I can't integrate that, I don't know how

    can someone please explain to me how to integrate it?

    [tex]\int_a^b (b-x)^2 dx [/tex]

    :(
     
  17. Apr 21, 2010 #16
    You don't have to do it the way they did it. Just expand (b-x)2, and make three simple integrals. You should be able to do those.
     
  18. Apr 21, 2010 #17
    [tex]\int_b^a (b-x)^2dx \\[/tex]

    [tex]\int_b^a (b-x)(b-x)dx \\[/tex]

    [tex]\int_b^a (b^2 +x^2 -2bx)dx \\[/tex]

    [tex]b^2x + \frac{x^3}{3} + \frac{-2bx^2}{2} \\
    [/tex]

    i Don't see how that equation,

    [tex] b^2x + \frac{x^3}{3} + \frac{-2bx^2}{2} [/tex]
    equals

    [tex] \frac{(b-x)^3}{3} [/tex]

    can someone just explain please
     
  19. Apr 21, 2010 #18
    They do it by making the "u" substitution I wrote in the earlier post. The integral becomes -u2du

    Integrate that, then put the b-x back into u.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook