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Expectation Value

  • Thread starter roeb
  • Start date
  • #1
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Homework Statement


Find the expectation value <x> if:
from 0 <= x <= a, psi = A x/a
from a <= x <= b, psi = A(b-x)/(b-a)

Normalizing gives me that A = sqrt(3/b) (verified correct)


Homework Equations





The Attempt at a Solution


[tex]<x> = \int_0^b x \psi^2 dx = \int_0^a \frac{A^2 x^3}{a^2}dx + \int_a^b A^2 \frac{x (b-x)^2 }{(b-a)^2} dx[/tex]


[tex]\int_0^a A^2 \frac{ x^3 }{ a^2} = A^2 \frac{ a^2 }{4} = 3/4 a^2/b[/tex]

[tex]A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = -A^2 \frac{x (b-x)^3}{(b-a)^2} - A^2 \frac{(b-x)^4}{12(b-a)^2}[/tex] eval from b to a

[tex]<x> = \frac{3a^2}{4b } + \frac{+ 3(b-a)a}{b} + \frac{3(b-a)^2}{12b} = \frac{-8a^2 + b^2 + 10ab}{4b}[/tex]

The answer is supposed to be (2a+b)/4... Does anyone see where I've gone wrong? I have double checked the antideriv. for the x(b-x)^2 integral and it seems to work.
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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The antiderivative of the x(b-x)^2 does NOT work. Try it again. How did you get that form anyway?
 
  • #3
107
1
Thanks for your reply

I used integration by parts,

u = x
du = dx

dv = (b-x)^2
v = -(b-x)^3/3

vu - int(v du)

[tex]\frac{-(b-x)^3 x}{3} - \frac{(b-x)^4}{12}[/tex]

so
[tex]A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = \frac{A^2}{(b-a)^2} * (\frac{-x(b-x)^3}{4} - \frac{(b-x)^4}{12})[/tex]
I ran through it again but I can't seem to see what I am doing incorrectly.
 
Last edited:
  • #4
25
0
The 0 to a integration seems to be correct.

The a to b integration seems little bit weird. The approach that you did by using the integration by part can lead you to the answer.

[tex]A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = \frac{A^2}{(b-a)^2} * (\frac{-x(b-x)^3}{3} - \frac{(b-x)^4}{12})[/tex]

[tex] = \frac{3}{b} (\frac{a(b-a)}{3}+\frac{(b-a)^2}{12})[/tex]

Now you can get <x> value by doing:

[tex]<x>=\frac{3a^2}{4b}+\frac{4a(b-a)}{4b}+\frac{(b-a)^2}{4b}=\frac{2a+b}{4}[/tex]

Pretty much a straight forward process.

Hope it helped =)
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
In the original post you were missing the denominator 3 on the (b-x)^3 part. In post 3 you had it except in the last line you replaced it with a 4. You are doing everything right. I think you are just making a bookkeeping error. You could also integrate x*(b-x)^2 by just expanding it into xb^2-2bx^2-x^3. That's what I did.
 

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