Difference between the average position and the most likely position of a particleby TFM Tags: average, difference, particle, position 

#19
Feb809, 01:33 PM

P: 1,031

Okay, so:
[tex] = x^{3/2}\frac{1}{\beta} e^{\beta x} + \frac{3}{2\beta} \int{x^{1/2}e^{\beta x} } [/tex] So: [tex] f(x) = x^{3/2}, f'(x) = \frac{1}{2}x^{1/2} [/tex] [tex] g'(x) = e^{\beta x}, g(x) = \frac{1}{\beta}e^{\beta} [/tex] Giving: [tex] x^{3/2}\frac{1}{\beta}e^{\beta}  \int{\frac{1}{2}x^{1/2}\frac{1}{\beta}e^{\beta} } [/tex] [tex] x^{3/2}\frac{1}{\beta}e^{\beta} + \frac{1}{2\beta} \int{x^{1/2}e^{\beta} } [/tex] I still seem to get another integral that needs another int. by parts 



#20
Feb809, 01:36 PM

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What are you doing???
you have to evaluate integral [tex]\int x^{1/2}e^{\beta x} \, dx [/tex] Why can't you do it? 



#21
Feb809, 01:38 PM

P: 1,031

Do we not need to use integration by parts then?




#22
Feb809, 01:42 PM

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your last post is nonsense.
you had [tex] x^{3/2}\frac{1}{\beta} e^{\beta x} + \frac{3}{2\beta} \int x^{1/2}e^{\beta x} \, dx [/tex] now you have an integral [tex] \int x^{1/2}e^{\beta x} \, dx [/tex] which you have to integrate by parts 



#23
Feb809, 01:49 PM

P: 1,031

That's what you get when you copy stuff and don't bother to check properly.
So: [tex] \int x^{1/2}e^{\beta x} dx [/tex] Thus: [tex] f(x) = x^{1/2}, f(x) = \frac{1}{2}x^{1/2} [/tex] [tex] g'(x) = e^{\beta x}, g(x) = \frac{1}{\beta}e^{\beta x} [/tex] Giving: [tex] = x^{1/2}\frac{1}{\beta}e^{\beta x}  \int {\frac{1}{2}x^{1/2} \frac{1}{\beta}e^{\beta x}} [/tex] [tex] = x^{1/2}\frac{1}{\beta}e^{\beta x} + \frac{1}{2\beta}\int {x^{1/2}e^{\beta x}} [/tex] Is this better? 



#24
Feb809, 02:16 PM

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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 [tex] \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} [/tex] who gave you this assignment? http://en.wikipedia.org/wiki/Error_function http://mathworld.wolfram.com/Erf.html 



#25
Feb809, 02:26 PM

P: 1,031

I am wondering if I made a mistake with the probability function, We was given the wave function:
[tex] \psi = B \sqrt{x}e^{\beta x} for x \geq 0 [/tex] 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... 



#26
Feb809, 02:28 PM

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P: 5,004

[tex]\int_0^{\infty} x^{1/2}e^{\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{u^2} du[/tex] For which the solution is well known and easily derivable. 



#27
Feb809, 02:28 PM

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but oh my godness... wave function is NOT probability function, look at post #10.
why didn't you confirm this to me? 



#28
Feb809, 02:30 PM

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but you still have x, but you integrate over u, that is strange 



#30
Feb809, 02:37 PM

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Anyway, OP do not need this integral anymore since we found a mistake on the way 



#31
Feb809, 02:40 PM

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#32
Feb809, 02:45 PM

P: 1,031

Okay, so:
[tex] \psi(x) = Bxe^{\beta x}, \psi(x)^* = Bxe^{\beta x} [/tex] Thus: [tex] P(x) = B^2 x^2 e^{(\beta x)^2} [/tex] Does this look better now? 



#36
Feb809, 02:56 PM

P: 1,031

? 


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