Probability distribution momentum for particle

• renec112
In summary, the probability distribution for the momentum of a particle with mass m is given by: ## \Phi (p) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \Psi(x,0) \cdot e^{-ipx} dx##f

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.

## = \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.

## = \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##.
Oh off course not.. My blunder..
t is time - are you thinking about finding an expression for t and substituting it?

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##

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.

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 :)