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The wave fuction for particle in a box

  1. Oct 4, 2013 #1
    Whwn i am doing exercise, i dont know how to solve the follow question by myself Although i hv the ans, i want to complete it by myself. Plz give me some tips only so that i can finish this question, please also tell me how do you know the question should be solved in this way.

    The question requires me to solve the wave equation for a pacticle in a infinity potential well, from x=-l/4 to x= 3l/4
    I let the wave equ be the form
    Asinkx + B cos kx

    and i got.
    Asin(-kl/4) + B cos (-kl/4) = 0
    Asin(3kl/4) + Bcos(-3kl/4) = 0

    and than i cant do anymore
    it is different from to x=0 to x= a case
    What should i do? Any tips?
    Is there any general solution for the wave fuction of the pacticle in a box from a point x to y?
  2. jcsd
  3. Oct 4, 2013 #2


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    Change the variable to u=x+1/4.
  4. Oct 4, 2013 #3
    thinks i am now trying to calculate
    Could you plz tell me why u know that the varible should be changed? Due to the experience? I am not good at physics but i love it should i do more exerience to develope this sence?
  5. Oct 4, 2013 #4
    i dont get it
    where should i change varible
    do u mean that i should differinate the fuction?
    I cant normalize it caz i dont know A abd B
  6. Oct 4, 2013 #5


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    clem meant "let u= x+ l/4" (it's a little difficult to distinguish between "l" ('ell') and "1" ('one') depending upon which font you use.) The only purpose of that is to change the boundaries from x= -l/4 and x= 3l/4 to u= 0 and u= 1 which makes the arithmetic slightly easier.
  7. Oct 4, 2013 #6


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    Script ell is easier to read, one reason I use UTF.
  8. Oct 4, 2013 #7
    Thank you guys. I got the answer. This skill is really useful for dealing with waves in the box.
    The answer is : fuction = A sin (kx + kl(ell)/4) and A = root(2/a)
    am i correct?:)
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