Probability of finding a particle?

1. Mar 27, 2008

iafatel

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$\left|\psi^{2}\right| = \frac{2}{L}Sin^{2}\left(\frac{nx\pi}{L}dx\right)$$

3. The attempt at a solution

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

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

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

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

factor out the L

$$\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right]$$

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

Last edited: Mar 27, 2008
2. Mar 27, 2008

Mentz114

$$\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right]$$

I get a positive number -

sin(2pi/3) = 0.8660

0.8660/(2*pi ) = 0.1378

1/3 - 0.1378 = 0.1955

3. Mar 27, 2008

iafatel

oh wow...stupid math mistake....forgot the parenthesis on my calc =(

thank you!

4. Jan 23, 2010

tazgurl78

what would the probability of the particle be if x=1.95 and 2.05?

5. Jan 23, 2010

tazgurl78

the answer is supposed to be .007, but I keep getting -.2921

6. Apr 19, 2011

sergioeldiego

I think you should be using cos instead of sin for the ground state.