The wave fuction for particle in a box

  • Thread starter VHAHAHA
  • Start date
  • #1
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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
and
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?
 

Answers and Replies

  • #2
Meir Achuz
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Change the variable to u=x+1/4.
 
  • #3
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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?
 
  • #4
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sorry
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
 
  • #5
HallsofIvy
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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.
 
  • #6
Bill_K
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(it's a little difficult to distinguish between "l" ('ell') and "1" ('one') depending upon which font you use.)
Script ell is easier to read, one reason I use UTF.
 
  • #7
58
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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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