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

• TFM
In summary, The two parts (a) and (b) are not the same, as they are asking for different values of the particle's position. Part (a) is asking for the average position, which can be calculated by finding the area under the probability distribution curve. Part (b) is asking for the most likely position, which can be found by locating the maximum of the probability distribution curve. In order to solve these problems, integration by parts can be used to evaluate the integrals.
TFM

## 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

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)

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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

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:

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

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.

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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:

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

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?

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?

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?

f 'prime' means derivative of f

that is standard notation

are you sure you have done integration by parts before?

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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?

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?

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

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?

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

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?

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...

malawi_glenn said:
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

The integral goes from 0 to infinity correct?...If so a simple substitution $u=\beta\sqrt{x}$ gives

$$\int_0^{\infty} x^{-1/2}e^{-\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{-u^2} du$$

For which the solution is well known and easily derivable.

but oh my godness... wave function is NOT probability function, look at post #10.

why didn't you confirm this to me?

gabbagabbahey said:
The integral goes from 0 to infinity correct?...If so a simple substitution $u=\beta\sqrt{x}$ gives

$$\int_0^{\infty} x^{-1/2}e^{-\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{-\beta x} du$$

For which the solution is well known and easily derivable.

yeah, that is true ;-)

but you still have x, but you integrate over u, that is strange

malawi_glenn said:
yeah, that is true ;-)

but you still have x, but you integrate over u, that is strange

Take a look at the edited version of my post

gabbagabbahey said:
Take a look at the edited version of my post

I saw, just want to be a bit rude since I did't remember that trick!

Anyway, OP do not need this integral anymore since we found a mistake on the way

malawi_glenn said:
Anyway, OP do not need this integral anymore since we found a mistake on the way

Indeed! There is a big difference between a wavefunction and its probability distribution!

Okay, so:

$$\psi(x) = Bxe^{-\beta x}, \psi(x)^* = Bxe^{-\beta x}$$

Thus:

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

Does this look better now?

no.

what is $$e^x e^x$$ ?

TFM said:
Okay, so:

$$\psi(x) = Bxe^{-\beta x}, \psi(x)^* = Bxe^{-\beta x}$$

Thus:

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

Does this look better now?

I thought $\psi(x)$ was:

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

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

Which is it?

gabbagabbahey said:
I thought $\psi(x)$ was:

Which is it?

Yeah, that is one more fundamental remark

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