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

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  • #76
gabbagabbahey
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[itex]x \to \infty[/itex] is one extremum, but certainly not a maximum...[itex]P(x \to \infty)=0[/itex]....There is one more solution to that equation which does correspond to a maximum.
 
  • #77
TFM
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Well, you want x on one side, so:

[tex] 0 = -8x\beta^3 e^{-2\beta x} + 4\beta^2e^{-2\beta x} [/tex]

[tex] 8x\beta^3 e^{-2\beta x} = 4\beta^2e^{-2\beta x} [/tex]

[tex] 2x\beta^3 e^{-2\beta x} = \beta^2e^{-2\beta x} [/tex]

[tex] 2x e^{-2\beta x} = \frac{\beta^2}{\beta^3}e^{-2\beta x} [/tex]

[tex] 2x e^{-2\beta x} = \beta^{-2}e^{-2\beta x} [/tex]

[tex] 2x = \beta^{-2}\frac{e^{-2\beta x}}{e^{-2\beta x}} [/tex]

[tex] 2x = \beta^{-2} [/tex]

[tex] x = \frac{\beta^{-2}}{2} [/tex]

How is this?
 
  • #78
malawi_glenn
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[tex]
2x e^{-2\beta x} = \frac{\beta^2}{\beta^3}e^{-2\beta x}
[/tex]

is not

[tex]
2x e^{-2\beta x} = \beta^{-2}e^{-2\beta x}
[/tex]

(I am faster than gabbagabbahey ! :-D )
 
  • #79
gabbagabbahey
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[tex]\frac{\beta^2}{\beta^3}=\frac{1}{\beta}\neq\beta^{-2}[/tex] :wink:
 
  • #80
TFM
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Hmm, why did I do that???

[tex] 2x e^{-2\beta x} = \beta^{-1}e^{-2\beta x} [/tex]

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

[tex] 2x = \beta^{-1} [/tex]

[tex] x = \frac{\beta^{-1}}{2} [/tex]

Is this better?
 
  • #81
malawi_glenn
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yes! :-)

or nicer [tex]
x = \frac{1}{2\beta}
[/tex]
 
  • #82
gabbagabbahey
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Don't forget to express [itex]\langle x \rangle[/itex] in terms of [itex]\beta[/itex] in order to properly compare the average value to the expected value :smile:
 
  • #83
TFM
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Okay so:

Average Position:

[tex] \frac{4\beta^2}{4\beta^3}\right] [/tex]

[tex] \frac{1}{\beta}\right] [/tex]

And the Most probable position:

[tex] \frac{1}{2\beta} [/tex]
 
  • #84
malawi_glenn
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yup, quite cool isn't it?

the reason for this is that the wave function has a long 'tail' which makes more 'probability' lie to the right of its maximum (c.f Maxwell speed distribution)
 
  • #85
TFM
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Indeed, not much difference, just *half more

Would the Standard deviation for this be the difference, or will I need to use:

[tex] |Delta x = \sqrt{<x^2> - <x>^2} [\tex]
 
  • #86
malawi_glenn
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you have to use it
 
  • #87
TFM
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Okay so:

[tex] <x^2> = \int^{\infty}_0 x^2 P(x) [/tex]

P(x) = B^2 x e^{-2\beta x}

[tex] <x^2> = \int^{\infty}_0 x^2B^2 x e^{-2\beta x} [/tex]

[tex] <x^2> = \int^{\infty}_0 x^3B^2 e^{-2\beta x} [/tex]

[tex] <x^2> = B^2 \int^{\infty}_0 x^3e^{-2\beta x} [/tex]

I have a feeling that this integral will have to be done three times...

Okay so first time:

[tex] \left[\int^{\infty}_0 x^3e^{-2\beta x}\right] [/tex]

[tex] f(x) = x^3, f'(x) = 3x^2 [/tex]

[tex] g'(x) = e^{-2\beta x}, g(x) = -\frac{1}{2\beta}e^{-2\beta x} [/tex]

Thus:

[tex] \left[ -x^3\frac{1}{2\beta}e^{-2\beta x} - \int -3x^2\frac{1}{2\beta}e^{-2\beta x} \right] [/tex]

[tex] \left[ -x^3\frac{1}{2\beta}e^{-2\beta x} + \frac{3}{2\beta}\int x^2e^{-2\beta x} \right] [/tex]

Okay so far?
 
  • #88
malawi_glenn
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so now you added a question to the original one, how rude!!

and I/we don't have to feed you like a baby, have some confident
[tex]
\int x^2e^{-2\beta x} \, dx
[/tex]
now you know what the integral is, you did earlier
 
  • #89
TFM
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Sorry
 
  • #90
malawi_glenn
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how's it going?
 
  • #91
TFM
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OKay, well I have:

[tex] B^2\left[ -\frac{1}{2\beta}e^{-2\beta x} + frac{3}{2 \beta\left[ -frac{-x^2}{2\beta}e^{-2 \beta x} + \frac{1}{\beta} \left[ \frac{x}{2\beta}e^{-2\beta x} + 1\frac{1}{2\beta} \left[ -\frac{1}{2\beta} e^{-2\beta x} \right]\right] /right]} \right] [/tex]

Which I have reduced to:

[tex] B^2\left[-\frac{x^3}{2\beta}e^{-2\beta x} + \frac(-\frac{3x^2}{4\beta^2}e^{-2\beta x} + \frac{3x}{4\beta^3}e^{-2\beta x} - \frac{3}{8\beta^4}e^{-2 \beta x} \right] [/tex]

[tex] B^2 = 4\beta^2[/tex]

[tex] \left[-\frac{B^2x^3}{2\beta}e^{-2\beta x} - \frac{B^23x^2}{4\beta^2}e^{-2\beta x} + \frac{B^23x}{4\beta^3}e^{-2\beta x} - \frac{3B^2}{8\beta^4}e^{-2 \beta x} \right] [/tex]

[tex] \left[-\frac{4\beta x^3}{2}e^{-2\beta x} - 3x^2e^{-2\beta x} + \frac{3x}{4\beta}e^{-2\beta x} - \frac{3}{2\beta^2}e^{-2 \beta x} \right] [/tex]
 

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