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: 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

    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,

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



    [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?


    [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
    = |\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

    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]


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


    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???

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

    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} \\

    i Don't see how that equation,

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

    [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