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

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TFM
#19
Feb8-09, 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
malawi_glenn
#20
Feb8-09, 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?
TFM
#21
Feb8-09, 01:38 PM
P: 1,031
Do we not need to use integration by parts then?
malawi_glenn
#22
Feb8-09, 01:42 PM
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your last post is non-sense.

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
TFM
#23
Feb8-09, 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?
malawi_glenn
#24
Feb8-09, 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
TFM
#25
Feb8-09, 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...
gabbagabbahey
#26
Feb8-09, 02:28 PM
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Quote Quote by malawi_glenn View Post
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
The integral goes from 0 to infinity correct?...If so a simple substitution [itex]u=\beta\sqrt{x}[/itex] gives

[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.
malawi_glenn
#27
Feb8-09, 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?
malawi_glenn
#28
Feb8-09, 02:30 PM
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Quote Quote by gabbagabbahey View Post
The integral goes from 0 to infinity correct?...If so a simple substitution [itex]u=\beta\sqrt{x}[/itex] gives

[tex]\int_0^{\infty} x^{-1/2}e^{-\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{-\beta x} du[/tex]

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
gabbagabbahey
#29
Feb8-09, 02:32 PM
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Quote Quote by malawi_glenn View Post
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
malawi_glenn
#30
Feb8-09, 02:37 PM
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Quote Quote by gabbagabbahey View Post
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
gabbagabbahey
#31
Feb8-09, 02:40 PM
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Quote Quote by malawi_glenn View Post
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!
TFM
#32
Feb8-09, 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?
malawi_glenn
#33
Feb8-09, 02:50 PM
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no.

what is [tex] e^x e^x [/tex] ?
gabbagabbahey
#34
Feb8-09, 02:53 PM
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Quote Quote by TFM View Post
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?
I thought [itex]\psi(x)[/itex] was:

Quote Quote by TFM View Post
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]
Which is it?
malawi_glenn
#35
Feb8-09, 02:54 PM
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Quote Quote by gabbagabbahey View Post
I thought [itex]\psi(x)[/itex] was:



Which is it?
Yeah, that is one more fundamental remark
TFM
#36
Feb8-09, 02:56 PM
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Quote Quote by malawi_glenn View Post
no.

what is [tex] e^x e^x [/tex] ?
isn't it [tex] e^{x^2} [/tex]

?


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