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

by TFM
Tags: average, difference, particle, position
 P: 1,031 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
 HW Helper Sci Advisor P: 4,739 They are not a priori the same, consider as an analogy the maxwell veolcity distribution: http://www.ac.wwu.edu/~vawter/Physic...heoryGas05.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)
 P: 1,031 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 $$\int^\infty_\infty x P(x) dx$$, and for (b), locate its position on the graph. But how would you work out it's function? TFM
HW Helper
P: 4,739

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

you have still not said anything about what course it is, and what probability density function you have. Also, show attempt to solution ....
 P: 1,031 Ops, Sorry, Okay so, I have the P(x) to be $$B^2 x e^{-\beta x}$$ 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: $$= \int^{\infty}_{0} B^2 x e^{-\beta x}$$ Using integration by parts gives me: $$= [-\beta x e^{-\beta x} - \beta^2 e^{-\beta x}]^{\infty}_0$$
 HW Helper Sci Advisor P: 4,739 ok, now you are not quite correct, if the probability function $B^2 x e^{-\beta x}$ is the wave function modulus square, then this is the mean value: $$= \int^{\infty}_{0} x P(x) \, dx$$ 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.
 P: 1,031 Isn't it normally though: $$= \int^\infty_\infty x P(x) dx$$ But changes sometimes due to limits impsoed on the system, eg with this specific function it goes to: $$= \int^{\infty}_{0} x P(x) dx$$
 HW Helper Sci Advisor P: 4,739 Yes, but you said that $$P(x) = B^2 x e^{-\beta x}$$ and then you performed this integral: $$= \int^{\infty}_{0} B^2 x e^{-\beta x} dx$$ Which is just $$\int^{\infty}_{0} P(x) dx \neq$$ So be careful.
 P: 1,031 Okay fair, enough, I can also see what I have done... P(x) actually is $$B^2 \sqrt{x} e^{-\beta x}$$ But the version I copied from my workings I had already inserted the extra x, making $$B^2 x e^{-\beta x}$$ So does my integral look right now: $$= [-\beta x e^{-\beta x} - \beta^2 e^{-\beta x}]^{\infty}_0$$ and how do you insert the limit of infinity?
 HW Helper Sci Advisor P: 4,739 if $$P(x) = B^2 \sqrt{x} e^{-\beta x}$$ And if $$P(x) = \Psi (x)^* \, \Psi (x)$$ Then $$= \int _0^{\infty} x\,P(x) \, dx = B^2\int _0^{\infty} x\, \sqrt{x} e^{-\beta x} \, dx$$ Now I can give you a hint: $xe^{-x} \rightarrow 0$ when $x\rightarrow \infty$
 P: 1,031 I feel silly, I should have know that $$e^{-\infty} \approx 0$$ $$= \int _0^{\infty} x\,P(x) \, dx = B^2\int _0^{\infty} x\, \sqrt{x} e^{-\beta x}$$ Okay, so now: $$B^2\int _0^{\infty} x\, x^{1/2} e^{-\beta x} = B^2\int _0^{\infty} x^{3/2} e^{-\beta x}$$ Integration by parts: $$u = x^{3/2}, du = (x/2)^{3/2} dv = e^{-\beta x}, v = -\beta e^{-\beta x}$$ This gives: $$-\beta x^{3/2} e^{-\beta x} - \int {(x/2)^{3/2} e^{-\beta x}}$$ 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?
 HW Helper Sci Advisor P: 4,739 you might want to do the integration starting all over again... Hint: consider $$\int f g^{'} dx = fg - \int f^{'} g dx$$ with $$f(x) = x^{3/2}$$ $$g^{'} (x) = e^{-\beta x}$$ you have done the derivative of x^{3/2} wrong...
 P: 1,031 Okay, so: $$f(x) = x^{3/2}, f'(x) = \frac{2}{5}x^{5/2}$$ $$g'(x) = e^{-\beta x}, g(x) = -\beta e^{-\beta x}$$ Thus: $$-x^{3/2}\beta e^{-\beta x} - \int{-\frac{2}{5}x^{5/2}\beta e^{-\beta x} }$$ Okay so far?
 HW Helper Sci Advisor P: 4,739 f 'prime' means derivative of f that is standard notation are you sure you have done integration by parts before?
 P: 1,031 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: $$f(x) = x^{3/2}, f'(x) = \frac{1}{2}x^{1/2}$$ $$g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta} e^{-\beta x}$$ So: $$= -x^{3/2}\frac{1}{\beta} e^{-\beta x} - \int{-\frac{1}{2}x^{1/2}\frac{1}{\beta} e^{-\beta x} }$$ $$= -x^{3/2}\frac{1}{\beta} e^{-\beta x} - -\frac{1}{2\beta} \int{x^{1/2}e^{-\beta x} }$$ Is this okay now?
 HW Helper Sci Advisor P: 4,739 Nope, what is derivative of x^{3/2} ?
 P: 1,031 Okay: $$f(x) = x^{3/2}, f'(x) = \frac{3}{2}x^{1/2}$$ Thus: $$= -x^{3/2}\frac{1}{\beta} e^{-\beta x} + \frac{3}{2\beta} \int{x^{1/2}e^{-\beta x} }$$ Better?
 HW Helper Sci Advisor P: 4,739 yes, now the second integral you have to do integration by parts once more.

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