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

[malawi_glenn]

yeah, but you have forgotten that the original exponential function was <br /> e^{-2\beta x}<br />

 

[TFM]

Not again!

Okay so:

First One:

f(x) = x^2 and f&#039;(x) = 2x

g&#039;(x) = e^{-2\beta x} and g(x) = -\frac{1}{2 \beta}e^{-2\beta x}

Giving:

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

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

So now:

f(x) = x, f&#039;(x) = 1

g&#039;(x) = e^{-2\beta x} and g(x) = -\frac{1}{2 \beta}e^{-2\beta x}

Giving:

\left[-x\frac{1}{2 \beta}e^{-2\beta x} - \int \frac{1}{2 \beta}e^{-2\beta x}

\left[-\frac{x}{2 \beta}e^{-2\beta x} + \frac{1}{2 \beta} \int e^{-2\beta x}

Does this look better?

 

[malawi_glenn]

yes, now just do the final integral, and but everything together and perform the limits.
Remember to keep track on those 'beta's, be careful.

 

[TFM]

Okay so the integral of:

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

So if we put together, from the top:

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

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

\left[ -2 \beta e^{- 2 \beta x} \right]

So:

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

And:

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

One step at a time:

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


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

And:

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

Okay so far?

 

[malawi_glenn]

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

WRONG

it is not the first time you do this mistake

 

[TFM]

Hmm, if we integrate e^kx, it becomes 1/k e^kx

So:

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

So:

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

goes to:

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

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

Okay?

 

[malawi_glenn]

bravo!

now, insert the limits

 

[TFM]

Okay, so the limits were between infinity and 0, so:

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

so using the approximation e^-infinty = 0:

B^2 \left[\left(-\frac{\infty^2}{2\beta} - \frac{\infty}{2\beta^2}e^{-2\beta \infty} - \frac{1}{4\beta^3}e^{-2\beta \infty} \right) - \left( -\frac{0^2}{2\beta} - \frac{0}{2\beta^2}e^{-2\beta 0} - \frac{1}{4\beta^3}e^{-2\beta 0} \right) \right]

B^2 \left[\left(-\frac{\infty^2}{2\beta} \right) - \left(-\frac{1}{4\beta^3}e^{-2\beta 0} \right) \right]

e^0 = 1

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

So what do I do about the first fraction, with infinity squared on top?

 

[gabbagabbahey]

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

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

Okay?

You've dropped a factor of e^{-2\beta x} from the first term in going from one line to the other. This is why you are getting infinity when you plug in the limits.

 

[malawi_glenn]

oh I didn't see that, you made that mistake AGAIN! What is 'wrong'? You still had it in the next last line in post #56, why did you remove it?

And one more thing e^x x-> infty IS 0, not just approximate.This guy is correct (next last line in post #56)

<br /> B^2 \left[-\frac{x^2}{2 \beta}e^{-2\beta x} + \frac{1}{\beta} \left[-\frac{x}{2 \beta}e^{-2\beta x} -\frac{1}{4 \beta} e^{- 2 \beta x} \right] \right] <br />

but not this one:
<br /> B^2 \left[-\frac{x^2}{2\beta} - \frac{x}{2\beta^2}e^{-2\beta x} - \frac{1}{4\beta^3}e^{-2\beta x} \right] <br />

Be more careful in the future...

 

[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]

&lt;x&gt; = B^2 \left[\frac{1}{4\beta^3}\right]

Is this okay now?

 

[malawi_glenn]

yes .

 

[TFM]

Okay so now we have found

&lt;x&gt; = 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

&lt;x&gt; = 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

&lt;x&gt; = 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?

 

[TFM]

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]

 

[malawi_glenn]

if you want to check your integrals:

http://integrals.wolfram.com/index.jsp

;-)