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

  1. 1. The problem statement, all variables and given/known data

    part (a):

    What is the average position of a particle

    part (b)

    What is the most likely position of the particle

    2. Relevant equations

    N/A

    3. The attempt at a solution

    What is the difference between these two questions, because on first glance, I would have thought they were the same?

    TFM
     
  2. jcsd
  3. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    They are not a priori the same, consider as an analogy the maxwell veolcity distribution:

    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Thermal/gifs/KinTheoryGas05.gif

    The peak is most probable-speed and the location on the x-axis where you have equal amount of probability to the left as to the right of that point is the mean-speed.

    This is general for all probability distributions, e.g. wave functions in quantum mechanics. The most probable value and mean value only coincide if the distribution is symmetric around its maximum.

    As an example, consider my Cheat - Dice:

    It has 6sides, 4 of them has the value 1, and the rest of them has the value 6. What is the most probable outcome? And what is the mean-value of my dice? (a normal dice has 3.5)
     
    Last edited: Feb 8, 2009
  4. Re: Difference between the average position and the most likely position of a particl

    Okay, that makes sense, since the two questions are:

    (a)

    Calculate the average position of the particle on the x axis, as a function of β. Mark the average position in your sketch.

    (b)

    Mark the most likely position of the particle in your sketch and calculate it as a function of β.

    So for this, you will have to work out (a), using [tex] \int^\infty_\infty x P(x) dx [/tex], and for (b), locate its position on the graph. But how would you work out it's function?

    TFM
     
  5. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    you have still not said anything about what course it is, and what probability density function you have. Also, show attempt to solution ....
     
  6. Re: Difference between the average position and the most likely position of a particl

    Ops, Sorry,

    Okay so, I have the P(x) to be [tex] B^2 x e^{-\beta x} [/tex]

    Where B is a function of beta

    We also know from the wave function that the integrals are actually between 0 and infinity, because the wave function is 0, thus:

    [tex] <x> = \int^{\infty}_{0} B^2 x e^{-\beta x} [/tex]

    Using integration by parts gives me:

    [tex] <x> = [-\beta x e^{-\beta x} - \beta^2 e^{-\beta x}]^{\infty}_0 [/tex]
     
  7. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    ok, now you are not quite correct, if the probability function [itex]
    B^2 x e^{-\beta x}
    [/itex] is the wave function modulus square, then this is the mean value:

    [tex] <x>
    = \int^{\infty}_{0} x P(x) \, dx
    [/tex]

    So please make sure you have done this, you have written something else in post #3 ...

    Then most probable, it is just to find where maximum of P(x) is located.
     
    Last edited: Feb 8, 2009
  8. Re: Difference between the average position and the most likely position of a particl

    Isn't it normally though:

    [tex] <x> = \int^\infty_\infty x P(x) dx [/tex]

    But changes sometimes due to limits impsoed on the system, eg with this specific function it goes to:

    [tex] <x> = \int^{\infty}_{0} x P(x) dx [/tex]
     
  9. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    Yes, but you said that [tex]
    P(x) = B^2 x e^{-\beta x}
    [/tex]

    and then you performed this integral:

    [tex]
    <x> = \int^{\infty}_{0} B^2 x e^{-\beta x} dx
    [/tex]

    Which is just [tex]
    \int^{\infty}_{0} P(x) dx \neq <x>
    [/tex]

    So be careful.
     
  10. Re: Difference between the average position and the most likely position of a particl

    Okay fair, enough, I can also see what I have done...

    P(x) actually is
    [tex] B^2 \sqrt{x} e^{-\beta x} [/tex]

    But the version I copied from my workings I had already inserted the extra x, making

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

    So does my integral look right now:

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

    and how do you insert the limit of infinity?
     
  11. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    if [tex]

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

    [/tex]

    And if [tex]P(x) = \Psi (x)^* \, \Psi (x) [/tex]

    Then

    [tex] <x> = \int _0^{\infty} x\,P(x) \, dx = B^2\int _0^{\infty} x\, \sqrt{x} e^{-\beta x} \, dx[/tex]

    Now I can give you a hint: [itex]xe^{-x} \rightarrow 0[/itex] when [itex]x\rightarrow \infty [/itex]
     
  12. Re: Difference between the average position and the most likely position of a particl

    I feel silly, I should have know that [tex] e^{-\infty} \approx 0 [/tex]

    [tex] = \int _0^{\infty} x\,P(x) \, dx = B^2\int _0^{\infty} x\, \sqrt{x} e^{-\beta x} [/tex]

    Okay, so now:

    [tex] B^2\int _0^{\infty} x\, x^{1/2} e^{-\beta x} = B^2\int _0^{\infty} x^{3/2} e^{-\beta x} [/tex]

    Integration by parts:

    [tex] u = x^{3/2}, du = (x/2)^{3/2} dv = e^{-\beta x}, v = -\beta e^{-\beta x} [/tex]

    This gives:

    [tex] -\beta x^{3/2} e^{-\beta x} - \int {(x/2)^{3/2} e^{-\beta x}}[/tex]

    Hmm, I need to use the integration by parts again, but if I do, I will still be left with another integral that needs int by parts again... Have I done something wrong?
     
  13. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    you might want to do the integration starting all over again...

    Hint: consider [tex]\int f g^{'} dx = fg - \int f^{'} g dx[/tex]
    with
    [tex] f(x) = x^{3/2} [/tex]
    [tex] g^{'} (x) = e^{-\beta x} [/tex]

    you have done the derivative of x^{3/2} wrong...
     
  14. Re: Difference between the average position and the most likely position of a particl

    Okay, so:

    [tex] f(x) = x^{3/2}, f'(x) = \frac{2}{5}x^{5/2} [/tex]

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

    Thus:

    [tex] -x^{3/2}\beta e^{-\beta x} - \int{-\frac{2}{5}x^{5/2}\beta e^{-\beta x} } [/tex]

    Okay so far?
     
  15. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    f 'prime' means derivative of f

    that is standard notation

    are you sure you have done integration by parts before?
     
    Last edited: Feb 8, 2009
  16. Re: Difference between the average position and the most likely position of a particl

    Yeah, I did it for A Level a few years back, although, we used u/v and du/dv. I can do most of them okay...

    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 x} [/tex]

    So:

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

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

    Is this okay now?
     
  17. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    Nope, what is derivative of x^{3/2} ?
     
  18. Re: Difference between the average position and the most likely position of a particl

    Okay:

    [tex] f(x) = x^{3/2}, f'(x) = \frac{3}{2}x^{1/2} [/tex]

    Thus:

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

    Better?
     
  19. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    yes, now the second integral you have to do integration by parts once more.
     
  20. Re: Difference between the average position and the most likely position of a particl

    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
     
  21. malawi_glenn

    malawi_glenn 4,726
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    Re: Difference between the average position and the most likely position of a particl

    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?
     
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