- #36
TFM
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malawi_glenn said:no.
what is [tex] e^x e^x [/tex] ?
isn't it [tex] e^{x^2} [/tex]
?
malawi_glenn said:no.
what is [tex] e^x e^x [/tex] ?
why should i then ask?TFM said:isn't it [tex] e^{x^2} [/tex]
?
TFM said:I see now, when you multiply powers, they add up
So:
[tex] e^x * e^x = e^{2x} [/tex]
TFM said:OKay, so:
[tex] \psi (x) = B \sqrt{x}e^{-\beta x} [/tex]
[tex] \psi (x)^* = B \sqrt{x}e^{-\beta x} [/tex]
Thus:
[tex] P(x) = B \sqrt{x}e^{-\beta x}B sqrt{x}e^{-\beta x} [/tex]
This gives:
[tex] P(x) = B^2 xe^{-2\beta x} [/tex]
This is the right version as I have carefully copied it from the Question.
So now:
[tex] <x> = \int^{\infty}_{0} x P(x) dx [/tex]
[tex] <x> = \int^{\infty}_{0} x B^2 xe^{-2\beta x} dx [/tex]
[tex] <x> = B^2\int^{\infty}_{0} x^2 e^{-2\beta x} dx [/tex]
Now:
[tex] f(x) = x^2, f'(x) = 2x [/tex]
[tex] g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta} e^{-\beta x} [/tex]
Thus giving:
[tex] \frac{x^2}{\beta} - \int {-\frac{2x}{\beta}e^{-\beta x}} [/tex]
Okay so taking parts again:
[tex] f(x) = 2x, f'(x) = 2 [/tex]
[tex] g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta} e^{-\beta x} [/tex]
Giving:
[tex] \frac{2x}{\beta} - \int{-\frac{2}{\beta}e^{- \beta x}} [/tex]
Now then the integral now gives:
[tex] \frac{2}{\beta}\int{e^{-\beta x}} [/tex]
which is:
[tex] -\beta x e^{-\beta x} [/tex]
And put all together:
[tex] <x> = \frac{x^2}{\beta} + \frac{2}{\beta}[\frac{2x}{\beta} + \frac{2}{\beta}[-\beta e^{-\beta x}]] [/tex]
Does this look okay?
TFM said:Hmm, how do we get an [tex] e^2x [/tex]
?
TFM said:I cheated slightly and just stuck 2s in...
So properly this time:
[tex] P(x) = B^2 xe^{-2\beta x} [/tex]
[tex] B^2\int^{\infty}_{0} x^2 e^{-2\beta x} dx [/tex]
[tex]f(x) = x^2, f'(x) = 2x [/tex]
[tex] g'(x) = e^{-2\beta x}, g(x) = -\frac{1}{2\beta}e^{-2\beta x} [/tex]
Thus:
[tex] [-x^2\frac{1}{2\beta} - \int - 2x \frac{1}{2\beta}e^{-2\beta x}] [/tex]
[tex] [-x^2\frac{1}{2\beta} + \frac{1}{2\beta} \int 2xe^{-2\beta x}] [/tex]
Now:
[tex]f(x) = 2x, f'(x) = 2 [/tex]
[tex] g'(x) = e^{-2\beta x}, g(x) = -\frac{1}{2\beta}e^{-2\beta x} [/tex]
[tex] [-2x\frac{1}{2\beta}e^{-2\beta x} - \int -2\frac{1}{2\beta}e^{-2\beta x}] [/tex]
[tex] [-2x\frac{1}{2\beta}e^{-2\beta x} + \frac{2}{2 \beta} \int e^{-2\beta x}] [/tex]
And now:
[tex] \int e^{-2\beta x} = -2\beta e^{-2\beta x} [/tex]
Okay so far?
TFM said:[tex] B^2 \left[-\frac{x^2}{2 \beta}e^{-2\beta x} + \frac{1}{\beta} \left[-\frac{x}{2 \beta}e^{-2\beta x} -\frac{1}{4 \beta} e^{- 2 \beta x} \right] \right] [/tex]
[tex] B^2 \left[-\frac{x^2}{2\beta} - \frac{x}{2\beta^2}e^{-2\beta x} - \frac{1}{4\beta^3}e^{-2\beta x} \right] [/tex]
Okay?
TFM said:Okay so now we have found
[tex] <x> = B^2 \left[\frac{1}{4\beta^3}\right] [/tex]
This is the Average Position
So now I need to find the most probable I have to find where P(x) is a maximum... does this mean I have to differentiate:
[tex] P(x) = B^2 xe^{-2\beta x} [/tex]
TFM said:Okay so now we have found
[tex] <x> = B^2 \left[\frac{1}{4\beta^3}\right] [/tex]
This is the Average Position
malawi_glenn said:yeah that is correct (also B = -2'bets' is fine)