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Probability distribution momentum for particle

  • #1
35
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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.
 

Answers and Replies

  • #2
DrClaude
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## = \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.
 
  • #3
35
4
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.
 
  • #4
DrClaude
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7,163
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## = \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##.
 
  • #5
35
4
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?
 
  • #6
DrClaude
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7,163
3,306
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.
 
  • #7
35
4
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##
 
  • #8
DrClaude
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7,163
3,306
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}##.
 
  • #9
35
4
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.
 
  • #10
DrClaude
Mentor
7,163
3,306
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.
 
  • #11
35
4
Find the common denominator and add the two terms together.
Off course! Thanks for being so patient with me. Now it works out :)
 

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