Probability distribution momentum for particle

[renec112]

Homework Statement

A particle with mass m is moving on the x-axis and is described by
$\psi_b = \sqrt{b} \cdot e^{-b |x|}$
Find the probability distribution for the particles momentum

Homework Equations

$\Phi (p)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \Psi(x,0) \cdot e^{-ipx} dx$

The Attempt at a Solution

I just inserted $\Psi(x,0) \$ and had a go
$= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \sqrt{b} \cdot e^{-b |x|} \cdot e^{-ipx} dx$
move constants out
$= \sqrt{\frac{b}{2 \pi}}\int_{-\infty}^\infty e^{-b |x|} \cdot e^{-ipx} dx$
combine $e$
$= \sqrt{\frac{b}{2 \pi}}\int_{-\infty}^\infty e^{-b |x| -ipx} dx$
split integral by
$| x| = \begin{cases} \mbox{x,} & \mbox{if } x>0 \\ \mbox{-x,} & \mbox{if } x <0 \end{cases}$
so we have
$= \sqrt{\frac{b}{2 \pi}} (\int_{-\infty}^0e^{b x -ipx} dx + \int_{0}^\infty e^{-b x -ipx} dx)$
perform integration
$= \sqrt{\frac{b}{2 \pi}} ([\frac{e^{b x -ipx}}{ip/t - b}]_{-\infty}^0 + [\frac{e^{-b x -ipx}}{ip/t-b}]_0^{\infty})$
evaluating these integrals fails i get
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{ip/t - b} - \frac{1}{ip/t - b}) = 0$

Can you spot my mistakes? would love some input.

 

[DrClaude]

$= \sqrt{\frac{b}{2 \pi}} (\int_{-\infty}^0e^{b x -ipx} dx + \int_{0}^\infty e^{-b x -ipx} dx)$
perform integration
$= \sqrt{\frac{b}{2 \pi}} ([\frac{e^{b x -ipx}}{ip/t - b}]_{-\infty}^0 + [\frac{e^{-b x -ipx}}{ip/t-b}]_0^{\infty})$

Check the denominators.

 

[renec112]

Check the denominators.

Thanks!
I just had a look and i see it should be:
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/t + b} - \frac{1}{-ip/t - b})$
Giving me
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/t} + \frac{1}{-ip/t } + \frac{1 }{b}-\frac{1}{b})$
$= \sqrt{\frac{b}{2 \pi}} (-2\frac{1}{-ip/t} )$
$= -2 \sqrt{\frac{b}{2 \pi}} \frac{t}{ip}$
Taking the norm squared to get the probability distribution:
$= 4 \frac{b}{2 \pi} \frac{t^2}{p^2}$

Seems legit to me, but i am not sure.

 

[DrClaude]

$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/t + b} - \frac{1}{-ip/t - b})$
Giving me
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/t} + \frac{1}{-ip/t } + \frac{1 }{b}-\frac{1}{b})$

What is $t$? And $1/(a+b) \neq 1/a + 1/b$.

 

[renec112]

What is $t$? And $1/(a+b) \neq 1/a + 1/b$.

Thanks for the reply!
Oh off course not.. My blunder..
t is time - are you thinking about finding an expression for t and substituting it?

 

[DrClaude]

t is time - are you thinking about finding an expression for t and substituting it?

No, I just don't understand why are introducing time in the picture. You are Fourier transforming a function from $x$ to $p$, that's it.

 

[renec112]

No, I just don't understand why are introducing time in the picture. You are Fourier transforming a function from $x$ to $p$, that's it.

Okay this is embarrassing. It was suppose to be a $\hbar$, but when i wrote from my notes to latex i thought it was a $t$

 

[DrClaude]

Okay this is embarrassing. It was suppose to be a $\hbar$, but when i wrote from my notes to latex i thought it was a $t$

Then it should also appear in the exponential: $e^{-i p x / \hbar}$.

 

[renec112]

Right, so i have:
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/\hbar + b} - \frac{1}{-ip/\hbar - b})$

Guess there's not much to do - you think taking the norm squared here is a reasonable idea?

Thanks for helping me out.

 

[DrClaude]

Right, so i have:
$= \sqrt{\frac{b}{2 \pi}} (\frac{1}{-ip/\hbar + b} - \frac{1}{-ip/\hbar - b})$

Guess there's not much to do

Find the common denominator and add the two terms together.

 

[renec112]

Find the common denominator and add the two terms together.

Off course! Thanks for being so patient with me. Now it works out :)