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Quantum Question!

  1. Dec 12, 2004 #1
    I have a question regarding the expectation value of particle:

    "Obtain an expression for the expectation value [tex]<p_{x}^n>[/tex] where n is 1,2,3... of a particle in an infinite box (U = infinity for x<0 and x>L; U=0 for 0<x<L) which is in an eigenstate of the energy."

    That is the question word for word.
    Now, I have basically no idea where to even begin with this. The Prof zipped through it just so he can say he covered the material during the final class. Now we have to answer questions and I am lost.

    I know no one likes to give answers without seeing work, but I simply don't know what to do.
    Any help with a push in the right direction would be great.

    Thanks.
     
  2. jcsd
  3. Dec 12, 2004 #2

    dextercioby

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    Homework Helper

    U mean finite box... :rolleyes: If then,use normalized wave functions of the system and compute that average.I'm afraid you'll have to know how to differentiate "n" times a trigonometric function.That's not so hard.It requires some attention with the phases.
    I hope u're aware of the definiton for an expectation value for an operator,call it [itex] \hat{A} [/itex],in the
    [tex] <\hat{A}>_{\psi(\vec{r})}=\int\int\int_{R^3} \psi^*(\vec{r}) A(\vec{r}) \psi(\vec{r}) [/tex]
    ,where of course,the operator must written in the coordinate representation.In your case,it's simple,since the problem is unidimentional.

    Daniel.
     
    Last edited: Dec 12, 2004
  4. Dec 12, 2004 #3
    dextercioby, thanks for the reply.

    As for the question, it definitely states an infinite box. Do you think this could be an error in the question?

    I know what you mean by the formula you provided. We did go over that in class. I also follow you on the fact that this is one dimension.
    I think I am getting lost now on what you mentioned about taking the derivative of a trig function to the n power.

    This is what I have so far...

    [tex] <p_{x}^n>=\int\psi^*(x) p(x)^n \psi(x) dx[/tex]

    where
    [tex] \psi^*=\psi_{n}=\sqrt{\frac{2}{L}} sin \left( \frac{n\pi{x}} {L} \right) [/tex]

    and
    [tex]\frac{d\psi}{dx}=\sqrt{\frac{2}{L}} \left( \frac{n\pi}{L} \right) cos \left( \frac{n\pi{x}}{L} \right) [/tex]

    I get
    [tex] <p_{x}^n>=\int\psi^* \left( \frac{\hbar}{i}\frac{d}{dx} \right)^n \psi dx[/tex]

    [tex] <p_{x}^n>=\left( \frac{\hbar}{i} \right)^n \left( \frac{2}{L} \right)^n \left( \frac{n\pi}{L} \right)^n \int_{0}^{L}sin \left( \frac{n\pi{x}}{L} \right) \left( cos\left(\frac{n\pi{x}}{L} \right) \right)^ndx[/tex]

    Whew! That was a lot of LaTex to write.
    Assuming I am correct to this point, what comes next? The cos function raised to the n power is really killing me here. Or, is this my expression, as the question calls for?

    Finally, I have read that the expectation value of the particles momentum is zero since this is an average of the momentum as it moves back and forth.
    Am I reading that correctly? If not, please explain.

    Thanks.

    edited twice for dumb errors.
     
    Last edited: Dec 12, 2004
  5. Dec 13, 2004 #4

    Doc Al

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    No. "Infinite" refers to depth of the box (walls are infinitely high) not the width. (Often this is called an infinite square well potential.) As opposed to a finite box, where the sides are some finite magnitude; the finite square well potential problem is harder to solve.
     
  6. Dec 13, 2004 #5

    There is an error between these two lines. You are raising the momentum operator to the nth power, not the wave function.
     
  7. Dec 13, 2004 #6
    Thanks a lot for the tip.

    In that case I think I need to remove the exponent of n on the cos function.
    Is that right?
    If so, that probably makes it much easier to deal with.
    I will try to work through that scenario later and post my result.
    If that is the wrong approach, please let me know before I kill myself on it. :rofl:

    Thanks.
     
  8. Dec 13, 2004 #7

    Janitor

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    Hypnotoad is correct. The rightmost part of your integrand is going to be a cosine to the first power or a sine to the first power, depending on whether n is odd or even.
     
  9. Dec 13, 2004 #8

    dextercioby

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    I believe it can be put in a form containing only sine,irrelevant of the even/odd form of the "n".BTW,the "n" in the argument is not the same with the exponent of the operators.So le't call it "m" the one from the argumrnt of the wave function:
    He has to compute this derivetive:
    [tex] \frac{d^{n}}{dx^{n}} \cos({\frac{m\pi x}{L}}) [/tex]
    Where "m" and "n" are natural numbers,nonconnected.
    [tex] \frac{d}{dx} \cos x =(-)^{1} cos (x+\frac{\pi}{2}) [/tex]
    [tex] \frac{d^{2}}{dx^{2}} \cos x =(-)^{2} cos (x+\frac{2\pi}{2}) [/tex]
    So i could prove by induction that
    [tex] \frac{d^{n}}{dx^{n}} \cos x =(-)^{n} cos (x+\frac{n\pi}{2}) [/tex]
    Now,i change the argument [itex] x\rightarrow \frac{m\pi x}{L} [/itex] and get
    [tex] \frac{d^{n}}{dx^{n}} \cos({\frac{m\pi x}{L}}) =(-)^{n}(\frac{m\pi}{L})^{n}
    cos ({\frac{m\pi x}{L}+\frac{n\pi}{2}}) [/tex]
    which can be put in a form involving [itex] \sin [/itex]
    and from there,the integral can be made easily.

    Daniel.
     
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