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

[TFM]

no.

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

isn't it $$e^{x^2}$$

?

 

[malawi_glenn]

isn't it $$e^{x^2}$$

?

why should i then ask?

with your logic:

$$x^1x^1 = x\cdot x = x^{1^1} = x$$

You should know this, algebra we learn when we are 15years old.

 

[TFM]

I see now, when you multiply powers, they add up

So:

$$e^x * e^x = e^{2x}$$

 

[malawi_glenn]

I see now, when you multiply powers, they add up

So:

$$e^x * e^x = e^{2x}$$

yup, and when you divide, you substract.

Now, after taking gabbagabbahey's remark into consideration, you can continue

 

[TFM]

OKay, so:

$$\psi (x) = B \sqrt{x}e^{-\beta x}$$

$$\psi (x)^* = B \sqrt{x}e^{-\beta x}$$

Thus:

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

This gives:

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

This is the right version as I have carefully copied it from the Question.

So now:

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

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

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

Now:

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

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

Thus giving:

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

Okay so taking parts again:

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

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

Giving:

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

Now then the integral now gives:

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

which is:

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

And put all together:

$$<x> = \frac{x^2}{\beta} + \frac{2}{\beta}[\frac{2x}{\beta} + \frac{2}{\beta}[-\beta e^{-\beta x}]]$$

Does this look okay?

 

[malawi_glenn]

OKay, so:

$$\psi (x) = B \sqrt{x}e^{-\beta x}$$

$$\psi (x)^* = B \sqrt{x}e^{-\beta x}$$

Thus:

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

This gives:

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

This is the right version as I have carefully copied it from the Question.

So now:

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

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

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

Now:

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

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

Thus giving:

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

Okay so taking parts again:

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

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

Giving:

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

Now then the integral now gives:

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

which is:

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

And put all together:

$$<x> = \frac{x^2}{\beta} + \frac{2}{\beta}[\frac{2x}{\beta} + \frac{2}{\beta}[-\beta e^{-\beta x}]]$$

Does this look okay?

No, you miss a factor e^-2x somewhere

and minus sign...

$$\frac{B^2}{4}(-2x(x+1)-1)e^{-2x}$$

You have worked hard, so I give you this one ;-)

I leave for you to insert the "betas" att correct places

 

[TFM]

Hmm, how do we get an $$e^2x$$

?

 

[malawi_glenn]

Hmm, how do we get an $$e^2x$$

?

you interprent me wrong

it should be beta included

 

[TFM]

Okay, I see what has happened, I forgot the 2 part iof teh exponential, and the B^2 taken out, so:

$$= B^2(\frac{x^2}{2\beta} + \frac{2}{1\beta}[\frac{2x}{1\beta} + \frac{2}{2\beta}[-2\beta e^{-2\beta x}]])$$

Does this look better?

 

[malawi_glenn]

no ...

wrong wrong, start over again.

It is integration by parts two times, you could do it earlier, why not now?

 

[TFM]

I cheated slightly and just stuck 2s in...

So properly this time:

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

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

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

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

Thus:

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

$$[-x^2\frac{1}{2\beta} + \frac{1}{2\beta} \int 2xe^{-2\beta x}]$$

Now:

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

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

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

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

And now:

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

Okay so far?

 

[malawi_glenn]

I cheated slightly and just stuck 2s in...

So properly this time:

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

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

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

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

Thus:

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

$$[-x^2\frac{1}{2\beta} + \frac{1}{2\beta} \int 2xe^{-2\beta x}]$$

Now:

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

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

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

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

And now:

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

Okay so far?

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

is wrong

And also this one, why are you doing this wrong?

$$\int^{\infty}_{0} x^2 e^{-2\beta x} dx = [-x^2\frac{1}{2\beta} ]- \int - 2x \frac{1}{2\beta}e^{-2\beta x}$$

You are forgetting the $$e^{-2\beta x}$$ on the first term.

 

[TFM]

OKay, let's try this step by step...

$$B^2 \int x^2 e^{-\beta x}$$

Using:

$$f(x) = x^2 and f'(x) = 2x$$

and

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

So:

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

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

Okay so far?

 

[malawi_glenn]

No, the last line is wrong, 'x' should be inside the integral.

 

[TFM]

Ah yes, I see that:

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

Okay so now, we have:

$$\int xe^{-\beta x}$$

So:

$$f(x) = x, f'(x) = 1$$

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

So:

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

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

Is this okay?

 

[malawi_glenn]

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

 

[TFM]

Not again!

Okay so:

First One:

$$f(x) = x^2 and f'(x) = 2x$$

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

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:

$$f(x) = x, f'(x) = 1$$

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

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]

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

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)

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

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

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:

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

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

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