# Wave functions and probabilities

• kidsmoker
In summary, the conversation discusses the calculation of volume and probability in relation to the formula P=\left|\Psi \right| ^{2}dV. It is noted that the volume should be multiplied by 4/3 or 12\pi r^{2}dr, rather than 4, and the derivative of P should be taken without the dr term.

## Homework Statement

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

## 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!

Last edited by a moderator:
Hi kidsmoker! kidsmoker said:
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)

tiny-tim said:
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.

kidsmoker said:
… 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! 