Wave functions and probabilities

[kidsmoker]

Homework Statement

http://img200.imageshack.us/img200/9268/29360438.jpg

Homework Equations

P=\left|\Psi \right| ^{2}dV

The Attempt at a Solution

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

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

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

Assuming it's correct, the probability is then

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

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

Thanks for any help!

 

[tiny-tim]

Hi kidsmoker!

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

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

Yup!

4 is for areas …

try 4/3 ! ​

(or just multiply the area by dr)

 

[kidsmoker]

Hi kidsmoker!


Yup!

4 is for areas …

try 4/3 ! ​

(or just multiply the area by dr)

Oh yeah, hahaha, i always get that wrong!

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

Thanks.

 

[tiny-tim]

… what happens to the dr when i take the derivative?

Just ignore it … it's a constant …

think of it as called something other than dr!