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Wave functions and probabilities

  • Thread starter kidsmoker
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  • #1
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Homework Statement



http://img200.imageshack.us/img200/9268/29360438.jpg [Broken]


Homework Equations



[tex]P=\left|\Psi \right| ^{2}dV[/tex]

The Attempt at a Solution



Okay, so [tex]r^{2} = x^{2}+y^{2}+z^{2}[/tex] and [tex]\left|\Psi \right| ^{2} = A^{2}e^{-2\alpha r^{2}}[/tex] .

The volume of the of the bit we're interested in should be

[tex]dV = 4\pi(r+dr)^{3} - 4\pi r^{3} \approx 12\pi r^{2}dr[/tex] if we ignore the [tex](dr)^{2}[/tex] and [tex](dr)^{3}[/tex] terms. Have I done something wrong here, as I was expecting to just end up with [tex]4\pi r^{2}dr[/tex]?

Assuming it's correct, the probability is then

[tex]P = A^{2}e^{-2 \alpha r^{2}}12\pi r^{2}dr[/tex] .

To find where this has a maximum value, would I set [tex]\frac{dP}{dr}=0[/tex] and then find the corresponding r values? But how do I take the derivative when there's a dr term in there? :confused:

Thanks for any help!
 
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Answers and Replies

  • #2
tiny-tim
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Hi kidsmoker! :smile:
The volume of the of the bit we're interested in should be

[tex]dV = 4\pi(r+dr)^{3} - 4\pi r^{3} \approx 12\pi r^{2}dr[/tex] if we ignore the [tex](dr)^{2}[/tex] and [tex](dr)^{3}[/tex] terms. Have I done something wrong here, as I was expecting to just end up with [tex]4\pi r^{2}dr[/tex]?
Yup! :biggrin:

4 is for areas

try 4/3 ! :wink:

(or just multiply the area by dr)
 
  • #3
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Hi kidsmoker! :smile:


Yup! :biggrin:

4 is for areas

try 4/3 ! :wink:

(or just multiply the area by dr)
Oh yeah, hahaha, i always get that wrong!

So we have [tex]P = A^{2}e^{-2 \alpha r^{2}}4\pi r^{2}dr[/tex] then what happens to the dr when i take the derivative? :s

Thanks.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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… what happens to the dr when i take the derivative?
Just ignore it :wink: … it's a constant

think of it as called something other than dr! :rolleyes:
 

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