Normalize a wave function

  • Thread starter vorcil
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  • #1
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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
[/tex]


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?
 

Answers and Replies

  • #2
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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!
 
  • #3
395
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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
 
  • #4
395
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the same with the middle equation,

[tex]
\Psi(x,0)^2 =(A\frac{(b-x}{b-a})^2) =1
[/tex]
 
  • #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.
 
  • #7
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When I am normalizing the wave function, do I add the equations from all the parts of the wave function?

i.e

[tex] \int_{-\infty}^{0} +\int_{0}^{a}+\int_{a}^{b}+\int_{b}^\infty [/tex]
 
  • #8
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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]
 
  • #9
395
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[tex]
= |\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
[/tex]



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:
  • #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.
 
  • #11
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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]
 
  • #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.
 
  • #13
395
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[tex]\int (b^2 - 2bx + x^2) dx = b^2x - \frac{2bx^2}{2} + \frac{x^3}{3} [/tex]

[tex]

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

[/tex]

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???
from

[tex] \int_a^b (b-x)^2 dx [/tex]
to
[tex] \left -\frac{(b-x)^3}{3} \right|_a^b [/tex]
 
  • #14
124
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[tex]\int (b^2 - 2bx + x^2) dx = b^2x - \frac{2bx^2}{2} + \frac{x^3}{3} [/tex]

[tex]

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

[/tex]

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???
from

[tex] \int_a^b (b-x)^2 dx [/tex]
to
[tex] \left -\frac{(b-x)^3}{3} \right|_a^b [/tex]

They used u=(b-x) => du=-dx

You are on a roll and almost there -- keep up the good work!
 
  • #15
395
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embarassing

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]

:(
 
  • #16
124
0
embarassing

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]

:(

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.
 
  • #17
395
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[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} \\
[/tex]

i Don't see how that equation,

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

[tex] \frac{(b-x)^3}{3} [/tex]

can someone just explain please
 
  • #18
124
0
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.
 

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