What distinguishes the average and most likely position of a particle?

[TFM]

Okay, so should have been:

B^2 \left[-\frac{x^2}{2\beta}e^{-2\beta x} - \frac{x}{2\beta^2}e^{-2\beta x} - \frac{1}{4\beta^3}e^{-2\beta x} \right]^{\infty}_0

Which means once all the values are put in, we get:


B^2 \left[\left(-\frac{\infty^2}{2\beta}e^{-2 \beta \infty} \right) - \left(-\frac{1}{4\beta^3} \right) \right]

Thus:

B^2 \left[-\left(-\frac{1}{4\beta^3} \right) \right]

<x> = B^2 \left[\frac{1}{4\beta^3}\right]

Is this okay now?

 

[malawi_glenn]

yes .

 

[TFM]

Okay so now we have found

<x> = B^2 \left[\frac{1}{4\beta^3}\right]

This is the Average Position

So now I need to find the most probable I have to find where P(x) is a maximum... does this mean I have to differentiate:

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

 

[malawi_glenn]

Okay so now we have found

<x> = B^2 \left[\frac{1}{4\beta^3}\right]

This is the Average Position

So now I need to find the most probable I have to find where P(x) is a maximum... does this mean I have to differentiate:

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

yeah, you have to find where maximum is.

 

[TFM]

Okay So I am assuming I need to use the product rule, so:

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

Product rule:

uv = udv + vdu

u = B^2 x, du = B^2

v = e^{-2\beta x}, dv = -2 \beta e^{-2\beta x}

Thus:

= -B^2x2 \beta e^{-2\beta x} + B^2e^{-2\beta x}

Does this look okay?

 

[gabbagabbahey]

Okay so now we have found

<x> = B^2 \left[\frac{1}{4\beta^3}\right]

This is the Average Position

Assuming that the wavefunction is normalized such that

\int_0^{\infty}P(x)dx=1

You can actually compute what the value of B must be in terms of\beta and use that to express the average position in terms of \beta only. That should make it much easier to compare the average value to the expected value.

 

[malawi_glenn]

yes your derivative is ok.

I think he knows what B is, he said it was a function of 'beta'

 

[TFM]

Okay, now I have this in my workings out...

I have B = 2\beta

Does this seem about right?

 

[malawi_glenn]

yeah that is correct (also B = -2'bets' is fine)

 

[gabbagabbahey]

yeah that is correct (also B = -2'bets' is fine)

So is B=\pm 2\beta i; all that really matters is that |B|^2=4\beta^2.

 

[malawi_glenn]

yes that is true, the wave functions are, in general, complex valued. However, their modulus square is real :-)

 

[TFM]

Okay so if we insert in B:

= -(2\beta)^2x2 \beta e^{-2\beta x} + (2\beta)^2e^{-2\beta x}

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

So does this look okay?

 

[malawi_glenn]

yes, then what is the value of x which gives the derivative equal to zero?

 

[TFM]

It is going to be infinity, because both sides have e^-x, and only one has x before hand, so we need the e^-x = 0, which means x must be infinity

 

[malawi_glenn]

no no no no no

<br /> 0 = -8x\beta^3 e^{-2\beta x} + 4\beta^2e^{-2\beta x} <br />

Show me how you solve it.

 

[gabbagabbahey]

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.

 

[TFM]

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]

<br /> 2x e^{-2\beta x} = \frac{\beta^2}{\beta^3}e^{-2\beta x} <br />

is not

<br /> 2x e^{-2\beta x} = \beta^{-2}e^{-2\beta x} <br />

(I am faster than gabbagabbahey ! :-D )

 

[gabbagabbahey]

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

 

[TFM]

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]

yes! :-)

or nicer <br /> x = \frac{1}{2\beta} <br />

 

[gabbagabbahey]

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

 

[TFM]

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]

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)

 

[TFM]

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{&lt;x^2&gt; - &lt;x&gt;^2} [\tex]

 

[malawi_glenn]

you have to use it

 

[TFM]

Okay so:

&lt;x^2&gt; = \int^{\infty}_0 x^2 P(x)

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

&lt;x^2&gt; = \int^{\infty}_0 x^2B^2 x e^{-2\beta x}

&lt;x^2&gt; = \int^{\infty}_0 x^3B^2 e^{-2\beta x}

&lt;x^2&gt; = 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&#039;(x) = 3x^2

g&#039;(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]

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
<br /> \int x^2e^{-2\beta x} \, dx<br />
now you know what the integral is, you did earlier

 

[TFM]

Sorry

 

[malawi_glenn]

how's it going?