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Probability of finding a particle?

  • Thread starter iafatel
  • Start date
4
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1. Homework Statement

For a particle in the ground state of a rigid box, calculate the probability of finding it between x=0 and x=[tex]\frac{L}{3}[/tex]

2. Homework Equations

[tex]\left|\psi^{2}\right| = \frac{2}{L}Sin^{2}\left(\frac{nx\pi}{L}dx\right)[/tex]

3. The Attempt at a Solution

[tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} Sin^{2}\left(\frac{x\pi}{L}\right)[/tex]
[tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} \frac{1-Cos \left(\frac{2x\pi}{L}\right)}{2}[/tex]

[tex]\frac{1}{L} \int^{\frac{L}{3}}_{0} 1 - \int^{0}_{\frac{L}{3}} Cos \left(\frac{2x\pi}{L}\right)}[/tex]

[tex]\frac{1}{L} \left[ \left|x\right|^{\frac{L}{3}}_{0} - \left|\frac{L}{2\pi}Sin\left(\frac{2x\pi}{L}\right)\right|^{\frac{L}{3}}_{0} [/tex]


[tex]\frac{1}{L} \left[ \frac{L}{3}} - \frac{L}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

factor out the L

[tex]\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

stuck here...when I work it out I get a negative number...am I mising something?
 
Last edited:

Answers and Replies

Mentz114
Gold Member
5,424
290
[tex]\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]

I get a positive number -

sin(2pi/3) = 0.8660

0.8660/(2*pi ) = 0.1378

1/3 - 0.1378 = 0.1955
 
4
0
oh wow...stupid math mistake....forgot the parenthesis on my calc =(

thank you!
 
2
0
what would the probability of the particle be if x=1.95 and 2.05?
 
2
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the answer is supposed to be .007, but I keep getting -.2921
 
I think you should be using cos instead of sin for the ground state.
 

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