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

[gabbagabbahey]

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!

 

[TFM]

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?

 

[malawi_glenn]

no.

what is e^x e^x ?

 

[gabbagabbahey]

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:

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?

 

[malawi_glenn]

I thought \psi(x) was:



Which is it?

Yeah, that is one more fundamental remark

 

[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 the 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?

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

is wrong

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

<br /> \int^{\infty}_{0} x^2 e^{-2\beta x} dx <br /> =<br /> [-x^2\frac{1}{2\beta} ]- \int - 2x \frac{1}{2\beta}e^{-2\beta x}<br />

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&#039;(x) = 2x

and

g&#039;(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&#039;(x) = 1

g&#039;(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 <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...