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

## Homework Statement

part (a):

What is the average position of a particle

part (b)

What is the most likely position of the particle

N/A

## 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

Related Advanced Physics Homework Help News on Phys.org
malawi_glenn
Homework Helper

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 [Broken]

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 by a moderator:

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

malawi_glenn
Homework Helper

you have still not said anything about what course it is, and what probability density function you have. Also, show attempt to solution ....

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:

$$<x> = \int^{\infty}_{0} B^2 x e^{-\beta x}$$

Using integration by parts gives me:

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

malawi_glenn
Homework Helper

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:

$$<x> = \int^{\infty}_{0} x P(x) \, dx$$

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.

Last edited:

Isn't it normally though:

$$<x> = \int^\infty_\infty x P(x) dx$$

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

$$<x> = \int^{\infty}_{0} x P(x) dx$$

malawi_glenn
Homework Helper

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

and then you performed this integral:

$$<x> = \int^{\infty}_{0} B^2 x e^{-\beta x} dx$$

Which is just $$\int^{\infty}_{0} P(x) dx \neq <x>$$

So be careful.

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:

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

and how do you insert the limit of infinity?

malawi_glenn
Homework Helper

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

And if $$P(x) = \Psi (x)^* \, \Psi (x)$$

Then

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

Now I can give you a hint: $xe^{-x} \rightarrow 0$ when $x\rightarrow \infty$

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?

malawi_glenn
Homework Helper

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...

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?

malawi_glenn
Homework Helper

f 'prime' means derivative of f

that is standard notation

are you sure you have done integration by parts before?

Last edited:

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?

malawi_glenn
Homework Helper

Nope, what is derivative of x^{3/2} ?

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?

malawi_glenn
Homework Helper

yes, now the second integral you have to do integration by parts once more.

Okay, so:

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

So:

$$f(x) = x^{3/2}, f'(x) = \frac{1}{2}x^{-1/2}$$

$$g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta}e^{-\beta}$$

Giving:

$$-x^{3/2}\frac{1}{\beta}e^{-\beta} - \int{-\frac{1}{2}x^{-1/2}\frac{1}{\beta}e^{-\beta} }$$

$$-x^{3/2}\frac{1}{\beta}e^{-\beta} + \frac{1}{2\beta} \int{x^{-1/2}e^{-\beta} }$$

I still seem to get another integral that needs another int. by parts

malawi_glenn
Homework Helper

What are you doing???

you have to evaluate integral

$$\int x^{1/2}e^{-\beta x} \, dx$$

Why can't you do it?

Do we not need to use integration by parts then?

malawi_glenn
Homework Helper

$$-x^{3/2}\frac{1}{\beta} e^{-\beta x} + \frac{3}{2\beta} \int x^{1/2}e^{-\beta x} \, dx$$

now you have an integral

$$\int x^{1/2}e^{-\beta x} \, dx$$

which you have to integrate by parts

That's what you get when you copy stuff and don't bother to check properly.

So:

$$\int x^{1/2}e^{-\beta x} dx$$

Thus:

$$f(x) = x^{1/2}, f(x) = \frac{1}{2}x^{-1/2}$$

$$g'(x) = e^{-\beta x}, g(x) = -\frac{1}{\beta}e^{-\beta x}$$

Giving:

$$= -x^{1/2}\frac{1}{\beta}e^{-\beta x} - \int {\frac{1}{2}x^{-1/2} \frac{1}{\beta}e^{-\beta x}}$$

$$= -x^{1/2}\frac{1}{\beta}e^{-\beta x} + \frac{1}{2\beta}\int {x^{-1/2}e^{-\beta x}}$$

Is this better?

malawi_glenn
Homework Helper

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) ;-)

$$\int x^{1/2}e^{-\beta x} dx = \frac{\sqrt{\pi}\text{erf} (\sqrt{\beta x})}{2\beta ^{3/2}} - \dfrac{1}{\beta}\sqrt{x}e^{-\beta x}$$

who gave you this assignment?

http://en.wikipedia.org/wiki/Error_function
http://mathworld.wolfram.com/Erf.html

I am wondering if I made a mistake with the probability function, We was given the wave function:

$$\psi = B \sqrt{x}e^{-\beta x} for x \geq 0$$

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...