# Difference between the average position and the most likely position of a particle

by TFM
Tags: average, difference, particle, position
 P: 1,031 Okay, so: $$= -x^{3/2}\frac{1}{\beta} e^{-\beta x} + \frac{3}{2\beta} \int{x^{1/2}e^{-\beta x} }$$ So: $$f(x) = x^{3/2}, f'(x) = \frac{1}{2}x^{-1/2}$$ $$g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta}e^{-\beta}$$ Giving: $$-x^{3/2}\frac{1}{\beta}e^{-\beta} - \int{-\frac{1}{2}x^{-1/2}\frac{1}{\beta}e^{-\beta} }$$ $$-x^{3/2}\frac{1}{\beta}e^{-\beta} + \frac{1}{2\beta} \int{x^{-1/2}e^{-\beta} }$$ I still seem to get another integral that needs another int. by parts
 HW Helper Sci Advisor P: 4,739 What are you doing??? you have to evaluate integral $$\int x^{1/2}e^{-\beta x} \, dx$$ Why can't you do it?
 P: 1,031 Do we not need to use integration by parts then?
 HW Helper Sci Advisor P: 4,739 your last post is non-sense. you had $$-x^{3/2}\frac{1}{\beta} e^{-\beta x} + \frac{3}{2\beta} \int x^{1/2}e^{-\beta x} \, dx$$ now you have an integral $$\int x^{1/2}e^{-\beta x} \, dx$$ which you have to integrate by parts
 P: 1,031 That's what you get when you copy stuff and don't bother to check properly. So: $$\int x^{1/2}e^{-\beta x} dx$$ Thus: $$f(x) = x^{1/2}, f(x) = \frac{1}{2}x^{-1/2}$$ $$g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta}e^{-\beta x}$$ Giving: $$= -x^{1/2}\frac{1}{\beta}e^{-\beta x} - \int {\frac{1}{2}x^{-1/2} \frac{1}{\beta}e^{-\beta x}}$$ $$= -x^{1/2}\frac{1}{\beta}e^{-\beta x} + \frac{1}{2\beta}\int {x^{-1/2}e^{-\beta x}}$$ Is this better?
 HW Helper Sci Advisor P: 4,739 oh crap, you can't solve this since these are not standard integrals. Sorry, I only recognize them when there is x^1/2.. I must see it as Sqrt(x) ;-) The answer is $$\int x^{1/2}e^{-\beta x} dx = \frac{\sqrt{\pi}\text{erf} (\sqrt{\beta x})}{2\beta ^{3/2}} - \dfrac{1}{\beta}\sqrt{x}e^{-\beta x}$$ who gave you this assignment? http://en.wikipedia.org/wiki/Error_function http://mathworld.wolfram.com/Erf.html
 P: 1,031 I am wondering if I made a mistake with the probability function, We was given the wave function: $$\psi = B \sqrt{x}e^{-\beta x} for x \geq 0$$ Now I looked through my calculations, and I had a graph from this, and I assumed that this was also the probabiltiy function, so I assume I may have made an error somewhere...
HW Helper
P: 5,004
 Quote by malawi_glenn oh crap, you can't solve this since these are not standard integrals. Sorry, I only recognize them when there is x^1/2.. I must see it as Sqrt(x) ;-) The answer is $$\int x^{1/2}e^{-\beta x} dx = \frac{\sqrt{\pi}\text{erf} (\sqrt{\beta x})}{2\beta ^{3/2}} - \dfrac{1}{\beta}\sqrt{x}e^{-\beta x}$$ who gave you this assignment? http://en.wikipedia.org/wiki/Error_function http://mathworld.wolfram.com/Erf.html
The integral goes from 0 to infinity correct?...If so a simple substitution $u=\beta\sqrt{x}$ gives

$$\int_0^{\infty} x^{-1/2}e^{-\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{-u^2} du$$

For which the solution is well known and easily derivable.
 HW Helper Sci Advisor P: 4,739 but oh my godness... wave function is NOT probability function, look at post #10. why didn't you confirm this to me?
HW Helper
P: 4,739
 Quote by gabbagabbahey The integral goes from 0 to infinity correct?...If so a simple substitution $u=\beta\sqrt{x}$ gives $$\int_0^{\infty} x^{-1/2}e^{-\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{-\beta x} du$$ For which the solution is well known and easily derivable.
yeah, that is true ;-)

but you still have x, but you integrate over u, that is strange
HW Helper
P: 5,004
 Quote by malawi_glenn yeah, that is true ;-) but you still have x, but you integrate over u, that is strange
Take a look at the edited version of my post
HW Helper
P: 4,739
 Quote by gabbagabbahey Take a look at the edited version of my post
I saw, just want to be a bit rude since I did't remember that trick!

Anyway, OP do not need this integral anymore since we found a mistake on the way
HW Helper
P: 5,004
 Quote by malawi_glenn Anyway, OP do not need this integral anymore since we found a mistake on the way
Indeed! There is a big difference between a wavefunction and its probability distribution!
 P: 1,031 Okay, so: $$\psi(x) = Bxe^{-\beta x}, \psi(x)^* = Bxe^{-\beta x}$$ Thus: $$P(x) = B^2 x^2 e^{(-\beta x)^2}$$ Does this look better now?
 HW Helper Sci Advisor P: 4,739 no. what is $$e^x e^x$$ ?
HW Helper
P: 5,004
 Quote by TFM Okay, so: $$\psi(x) = Bxe^{-\beta x}, \psi(x)^* = Bxe^{-\beta x}$$ Thus: $$P(x) = B^2 x^2 e^{(-\beta x)^2}$$ Does this look better now?
I thought $\psi(x)$ was:

 Quote by TFM I am wondering if I made a mistake with the probability function, We was given the wave function: $$\psi = B \sqrt{x}e^{-\beta x} for x \geq 0$$
Which is it?
HW Helper
 Quote by gabbagabbahey I thought $\psi(x)$ was: Which is it?
 Quote by malawi_glenn no. what is $$e^x e^x$$ ?
isn't it $$e^{x^2}$$