(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that for a particle in a one dimensional box of length L, the probability of observing a value of [tex]p_x[/tex] (recall [tex]\hat{p}_x[/tex] is Hermitian and that [tex]\Psi[/tex] is not an eigenfunction of [tex]\hat{p}_x[/tex]) between p and dp is:

[tex]\frac{4|N|^2 s^2}{L(s^2-b^2)^2}[1-(-1)^n \cos(bL)]dp[/tex]

where [tex]s=n\pi L^{-1}[/tex] and [tex]b=p\bar{h}^{-1}[/tex] and the constant N so that the integral of this result from - infinity to + infinity is one.

Evaluate this result from

[tex]p=\pm \frac{nh}{2L}[/tex]

What is the significance of this choice for p?

2. Relevant equations

[tex]p[/tex]

[tex]p[/tex]

[tex]\Psi = \left(\frac{2}{L}\right)^{1/2} \sin \frac{n\pi x}{L}[/tex]

3. The attempt at a solution

I tried

[tex]

<p_x> = \int_p^{p+dp} \Psi^* \frac{\bar{h}}{i} \frac{d}{dx}\Psi

= \frac{4\pi\bar{h}}{i L^2} \int_p^{p+dp} \sin \frac{n\pi x}{L} \cos \frac{n\pi x}{L} [/tex]

but it doesn't seem to be taking me where I want to go, and I wasn't sure whether integrating from p to p+dp was kosher. The source of the (-1)^n term in the final result is a bit baffling.

Any suggestions?

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# Homework Help: Particle in a one dimensional box: probability of observing px between p and dp

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