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Probability for a momentum value for particle in a box

  • Thread starter pedroobv
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  • #1
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[SOLVED] Probability for a momentum value for particle in a box

Homework Statement



This is the problem 7.37 from Levine's Quantum Chemistry Textbook.

Show that, for a particle in a one-dimensional box of length [tex]l[/tex], the probability of observing a value of [tex]p_{x}[/tex] between [tex]p[/tex] and [tex]p+dp[/tex] is:

[tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​

Where [tex]s\equiv n\pi l^{-1}[/tex] and [tex]b\equiv p\hbar^{-1} [/tex]. The constant [tex]N[/tex] is to be chosen so that the integral from minus infinity to infinity of the previous equation is unity.

Homework Equations



Since [tex]p[/tex] has a continue range of eigenvalues, the probability of finding a value of [tex]p[/tex] beetween [tex]p[/tex] and [tex]p+dp[/tex] for a system in the state [tex]\Psi[/tex] is

[tex]|\left\langle g_{p}(x)|\Psi (x,t) \right\rangle |^{2}dp [/tex]​

where [tex]g_{p}[/tex] are the eigenfunctions of [tex]\hat{p}[/tex].

For a particle in a box:

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

and the eigenfunctions of the momentum operator are:

[tex]g=Ne^{ipx/\hbar}[/tex]​

The Attempt at a Solution



I made the corresponding substitutions to obtain the integral:

[tex]\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx[/tex]​

Which after computing and using some identities gives:

[tex]\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx = \frac{\sqrt{2}Ns\left[1-(-1)^{n}e^{-ibl}\right] }{\sqrt{l}\left(s^{2}-b^{2}\right)} [/tex]​

Which looks a lot like the square root of the expression that I seek, however I'm stuck at this point. Any help will be appreciated, if more informations is needed please tell me.

Thanks in advance
 

Answers and Replies

  • #2
9
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Sorry, I made a mistake in the first equation. Instead of

[tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​

Must be:

[tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b^{2})^{2}}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​
 
  • #3
9
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Ok thanks anyway, I already solved it. I forgot to take the absolute value.
 

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