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

gabbagabbahey
Homework Helper
Gold Member

$x \to \infty$ is one extremum, but certainly not a maximum...$P(x \to \infty)=0$....There is one more solution to that equation which does correspond to a maximum.

Well, you want x on one side, so:

$$0 = -8x\beta^3 e^{-2\beta x} + 4\beta^2e^{-2\beta x}$$

$$8x\beta^3 e^{-2\beta x} = 4\beta^2e^{-2\beta x}$$

$$2x\beta^3 e^{-2\beta x} = \beta^2e^{-2\beta x}$$

$$2x e^{-2\beta x} = \frac{\beta^2}{\beta^3}e^{-2\beta x}$$

$$2x e^{-2\beta x} = \beta^{-2}e^{-2\beta x}$$

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

$$2x = \beta^{-2}$$

$$x = \frac{\beta^{-2}}{2}$$

How is this?

malawi_glenn
Homework Helper

$$2x e^{-2\beta x} = \frac{\beta^2}{\beta^3}e^{-2\beta x}$$

is not

$$2x e^{-2\beta x} = \beta^{-2}e^{-2\beta x}$$

(I am faster than gabbagabbahey ! :-D )

gabbagabbahey
Homework Helper
Gold Member

$$\frac{\beta^2}{\beta^3}=\frac{1}{\beta}\neq\beta^{-2}$$

Hmm, why did I do that???

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

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

$$2x = \beta^{-1}$$

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

Is this better?

malawi_glenn
Homework Helper

yes! :-)

or nicer $$x = \frac{1}{2\beta}$$

gabbagabbahey
Homework Helper
Gold Member

Don't forget to express $\langle x \rangle$ in terms of $\beta$ in order to properly compare the average value to the expected value

Okay so:

Average Position:

$$\frac{4\beta^2}{4\beta^3}\right]$$

$$\frac{1}{\beta}\right]$$

And the Most probable position:

$$\frac{1}{2\beta}$$

malawi_glenn
Homework Helper

yup, quite cool isn't it?

the reason for this is that the wave function has a long 'tail' which makes more 'probability' lie to the right of its maximum (c.f Maxwell speed distribution)

Indeed, not much difference, just *half more

Would the Standard deviation for this be the difference, or will I need to use:

$$|Delta x = \sqrt{<x^2> - <x>^2} [\tex] malawi_glenn Science Advisor Homework Helper you have to use it Okay so: [tex] <x^2> = \int^{\infty}_0 x^2 P(x)$$

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

$$<x^2> = \int^{\infty}_0 x^2B^2 x e^{-2\beta x}$$

$$<x^2> = \int^{\infty}_0 x^3B^2 e^{-2\beta x}$$

$$<x^2> = B^2 \int^{\infty}_0 x^3e^{-2\beta x}$$

I have a feeling that this integral will have to be done three times...

Okay so first time:

$$\left[\int^{\infty}_0 x^3e^{-2\beta x}\right]$$

$$f(x) = x^3, f'(x) = 3x^2$$

$$g'(x) = e^{-2\beta x}, g(x) = -\frac{1}{2\beta}e^{-2\beta x}$$

Thus:

$$\left[ -x^3\frac{1}{2\beta}e^{-2\beta x} - \int -3x^2\frac{1}{2\beta}e^{-2\beta x} \right]$$

$$\left[ -x^3\frac{1}{2\beta}e^{-2\beta x} + \frac{3}{2\beta}\int x^2e^{-2\beta x} \right]$$

Okay so far?

malawi_glenn
Homework Helper

so now you added a question to the original one, how rude!!

and I/we don't have to feed you like a baby, have some confident
$$\int x^2e^{-2\beta x} \, dx$$
now you know what the integral is, you did earlier

Sorry

malawi_glenn
Homework Helper

how's it going?

OKay, well I have:

$$B^2\left[ -\frac{1}{2\beta}e^{-2\beta x} + frac{3}{2 \beta\left[ -frac{-x^2}{2\beta}e^{-2 \beta x} + \frac{1}{\beta} \left[ \frac{x}{2\beta}e^{-2\beta x} + 1\frac{1}{2\beta} \left[ -\frac{1}{2\beta} e^{-2\beta x} \right]\right] /right]} \right]$$

Which I have reduced to:

$$B^2\left[-\frac{x^3}{2\beta}e^{-2\beta x} + \frac(-\frac{3x^2}{4\beta^2}e^{-2\beta x} + \frac{3x}{4\beta^3}e^{-2\beta x} - \frac{3}{8\beta^4}e^{-2 \beta x} \right]$$

$$B^2 = 4\beta^2$$

$$\left[-\frac{B^2x^3}{2\beta}e^{-2\beta x} - \frac{B^23x^2}{4\beta^2}e^{-2\beta x} + \frac{B^23x}{4\beta^3}e^{-2\beta x} - \frac{3B^2}{8\beta^4}e^{-2 \beta x} \right]$$

$$\left[-\frac{4\beta x^3}{2}e^{-2\beta x} - 3x^2e^{-2\beta x} + \frac{3x}{4\beta}e^{-2\beta x} - \frac{3}{2\beta^2}e^{-2 \beta x} \right]$$