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Wave function of a particle in a infinite well.

  1. Jul 15, 2003 #1
    You're not understading:

    Let me give you all my work to alleaviate any confusion.

    Show that A = (2/L)1/2

    &psi(x) = A Sin(&pi x/L)

    &psi2(x) = A2 Sin2(&pi x/L)

    [inte]0L &psi2dx = 1

    A2[inte]0L Sin2(&pi x/L) dx = 1

    Actually...


    I forgot to resubsitute...

    BTW: I only use a calculator/computer to check my work.

    Thanks...
     
    Last edited: Jul 15, 2003
  2. jcsd
  3. Jul 15, 2003 #2

    Tom Mattson

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    First, you must have done something wrong because you should get:

    1=A2(L/2)

    (I know for a fact that A=(2/L)1/2).

    Ordinary algebra. Once you do the integral correctly and get

    A2(L/2)=1

    just multiply by 2/L and take the square root.
     
  4. Jul 15, 2003 #3
    Yeah, that's what my fancy calculator told me, so what the heck did I do wrong?!
     
  5. Jul 15, 2003 #4

    Tom Mattson

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    What you did wrong was take the answer off a calculator.

    Do a u-substitution:

    u=(π/L)x
    du=(π/L)dx

    then look up ∫sin2(u)du in an integral table. Don't forget to evaluate it at each limit.
     
  6. Jul 15, 2003 #5
    See my original post: I'm [zz)]
     
  7. Jul 15, 2003 #6

    Tom Mattson

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    Pardon, but I think I am. :)

    Yes, I assumed all of that. The problem is that the answer you posted:

    1 = A2[1/2L - 1/4Sin[2L]]

    is not right, even prior to substitution. Like I said, try it with a u-substitution and an integral table.

    edit: fixed quote bracket
     
  8. Jul 15, 2003 #7
    I worked out all of the kinks. I was wrong.

    I found:

    1 = L/&pi A2[1/2&pi]

    The integral was more challenging than I thought.

    Thanks for your help.

    I would of never thought of the u substitution, I haven't done that sort of thing since Calculus II.
     
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