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

[Dunhausen]

Homework Statement

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

\frac{4|N|^2 s^2}{L(s^2-b^2)^2}[1-(-1)^n \cos(bL)]dp
where s=n\pi L^{-1} and b=p\bar{h}^{-1} and the constant N so that the integral of this result from - infinity to + infinity is one.

Evaluate this result from
p=\pm \frac{nh}{2L}

What is the significance of this choice for p?

Homework Equations

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

The Attempt at a Solution

I tried
<br /> &lt;p_x&gt; = \int_p^{p+dp} \Psi^* \frac{\bar{h}}{i} \frac{d}{dx}\Psi<br /> = \frac{4\pi\bar{h}}{i L^2} \int_p^{p+dp} \sin \frac{n\pi x}{L} \cos \frac{n\pi x}{L}
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?

 

[vela]

Are you familiar with bra-ket notation, aka Dirac notation?

 

[Dunhausen]

My course doesn't use it (so far) but if you prefer to write that way I can try to muscle through it with reference to the wikipedia article. :p

 

[vela]

No, that's okay. It's just that the notation suggests a certain way to explain things.

First, you should recall that the probability of finding a particle between x and x+dx is given by

P(x \le X \le x+dx) = \psi^*(x)\psi(x) dx

Note that you're not doing an integral and you're not throwing the operator \hat{x} in there anywhere. Remember that the complex amplitude of finding the particle at x is just the wavefunction evaluated at x. The probability (density) is then the square of the absolute value of the amplitude.

Now you're being asked to find the probability that the momentum of the particle is between p and p+dp. It's the same problem except this time you need the complex amplitude of finding the particle with momentum p, i.e.

P(p \le p_x \le p+dp) = \psi_p^*(p)\psi_p(p) dp

In other words, you need to find the momentum representation \psi_p(p) of the particle's state. How to do this should be in your notes and textbook.

 

[Dunhausen]

Thanks vella! That makes a lot of sense to me. :)

In other words, you need to find the momentum representation LaTeX Code: \\psi_p(p) of the particle's state. How to do this should be in your notes and textbook.

You would think that! But it really isn't. (fyi this is a physical chemistry course so it doesn't quite fit the normal physics curriculum) If I know my professor, there is another way to do this problem which has something to do with properties of the Hermitian, but I'm just happy to solve it by any means necessary.

Anyway I found http://www.google.com/url?sa=t&source=web&ct=res&cd=3&ved=0CBgQFjAC&url=http%3A%2F%2Fwww.ecse.rpi.edu%2F~schubert%2FCourse-ECSE-6968%2520Quantum%2520mechanics%2FCh03%2520Position%26momentum%2520space.pdf&ei=ajiCS_CxLYL0sgPD-OX6Aw&usg=AFQjCNH5tdKEyUg9XDGzKPunWoPaaBM83g&sig2=9CzZhXrFYsgNpO2puHyOuQ article which described how to go into "momentum space" using a Fourier transform:

<br /> \Phi(p) = \int_\infty^\infty \Psi(x) e^{-ipx/h} dx<br />

So I put that into Wolfram|Alpha (integrating from 0 to q, where q is the length of the square well) and got this

http://img199.imageshack.us/img199/7875/63803268.gif

I want to say it seems it will give me the right (or very similar) coefficients to that shown in my topic post, although I'm still not sure where that (-1)^n term comes from unless I have to write things out as their infinite series.

Anyway, I will hack away at it, but if anyone has any other insights, they are more than appreciated. :)

 

[vela]

That's exactly what you want to do. Remember that Wolfram Alpha doesn't know that n is an integer, so it doesn't simplify the result as much as it could, in particular the sine and cosine terms.

 

[Dunhausen]

Remember that Wolfram Alpha doesn't know that n is an integer, so it doesn't simplify the result as much as it could, in particular the sine and cosine terms.

Ah, yes! That was the clue I needed to finish this! The LaTeX here is being pretty buggy (and it's a long problem) so I'll post a scan of my solution later.

Thank you again for your help! I would have spent many fruitless hours doing the wrong thing otherwise.