Wave function of a particle in a infinite well.

  • Thread starter frankR
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  • #1
frankR
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Originally posted by Tom
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.

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...
 
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Answers and Replies

  • #2
quantumdude
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Originally posted by frankR
1 = A2[1/2L - 1/4Sin[2L]]

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).

How the heck do I find A?

Ordinary algebra. Once you do the integral correctly and get

A2(L/2)=1

just multiply by 2/L and take the square root.
 
  • #3
frankR
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Yeah, that's what my fancy calculator told me, so what the heck did I do wrong?!
 
  • #4
quantumdude
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Originally posted by frankR
Yeah, that's what my fancy calculator told me, so what the heck did I do wrong?!

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.
 
  • #5
frankR
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See my original post: I'm [zz)]
 
  • #6
quantumdude
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Originally posted by frankR
You're not understading:

Pardon, but I think I am. :)

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

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
 
  • #7
frankR
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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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