1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uncertainty of a wave packet

  1. Aug 6, 2015 #1
    The problem stated below is from Liboff "Introductory Quantum Mechanics" (2nd Edition), exercise 5.4.
    1. The problem statement, all variables and given/known data
    A pulse ## 1m ## long contains ##1000 \alpha ## particles. At ## t=0## each ##\alpha## particle is in the state:
    [tex]
    \psi (x,0)=\frac{1}{10}\exp (ik_ox)
    [/tex]
    for [itex]|x|\leq 50cm[/itex] and zero elsewhere. It is given that [itex]k_o=\pi/50[/itex]
    (a) At [itex]t=0[/itex], how many [itex]\alpha[/itex] particles have momentum in the interval [itex]0\leq\hbar k\leq \hbar k_o[/itex] ?
    (b) At which values of momentum will ##\alpha## particles not be found at ##t=0##?
    (c) Describe an experiment to "prepare" such a state.
    (d) Construct ##\Delta x## and ##\Delta p## for this state, formally. What is ##\Delta x\Delta p##? (Hint: To calculate ##\Delta p##, use ##|b(k)|^2##)

    2. Relevant equations


    3. The attempt at a solution
    As I get it, we are in a Hilbert Space spanned by the eigenfunctions of the momentum operator, which constitute a continuous set. Thus, the state wavefunction should be written as:
    [tex]
    \psi (x,0)=\int_{-\infty}^{+\infty} b(k)\phi_k dx
    [/tex]
    where
    [tex]
    \phi_k =\frac{1}{\sqrt{2\pi}}e^{ikx}
    [/tex]
    Thus, we have (let L=100cm): ##\displaystyle b(k)=\int_{-\infty}^{+\infty}\psi (x,0)\phi^{\star}_k dx=...=\sqrt{\frac{2}{\pi L}}\frac{\sin[(k_o-k)L/2]}{k_o-k}##
    (a) That being said, the number of particles in the given interval is just the integral:
    [tex]
    N_{interval}=N_{total}\int_0^{k_o}|b(k)|^2dk
    [/tex]
    (b) We will not find particles in the states whom ##b(k)## equals to zero, that is: ##\sin[(k_o-k)L/2]=0\rightarrow k=\frac{2n\pi}{L}-k_o\rightarrow p_k=\hbar k=\cdots## with ##n\in\mathbb Z##
    (c) I suppose that we could get a laser beam of definite wavelenght, which means of definitive momentum, and just send a pulse of ##100cm## in lenght. The state of such a laser beam (thought in the whole x axis) should be an eigenstate of the momentum (the one with ##k_o## should be chosen obviously). When we send the pulse (turn on and off the laser), we should get the given wavefunction. Is that any close to being right?
    (d) I have encountered many problems here. I used the definition for the uncertainty, that is:
    [tex]
    (\Delta p)^2=\left<\hat{p}^2\right>-\left<\hat{p}\right>^2=\left<\psi \right|\hat{p}^2\left|\psi\right>-\left<\psi\right|\hat{p}\left|\psi\right>^2
    [/tex]
    Computing the above, I get:
    ##\displaystyle \left<\psi \right|\hat{p}^2\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}^2\psi dx=(-i\hbar)^2(ik_o)^2\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o^2\hbar^2##
    And the other one: ##\displaystyle \left<\psi \right|\hat{p}\left|\psi\right>=\int_{-\infty}^{+\infty}\psi^{\star}\hat{p}\psi dx=(-i\hbar)(ik_o)\int_{-L/2}^{L/2}\psi^{\star}\psi dx=k_o\hbar##
    But this gives ##\Delta p =0## ! This could be true if the uncertainty in ##x## was infinite. However, it is easily seen that this is not the case because: ##\displaystyle \left<x\right>=\int_{-L/2}^{L/2} x|\psi|^2dx=0## and ##\displaystyle \left<x^2\right>=\int_{-L/2}^{L/2} x^2|\psi|^2dx=\frac{L^2}{12}##
    My first issue is that Heisenberg's uncertainty principle is violated, which means that somewhere lies a mistake. I see that the uncertainty in the momentum cannot be zero because I prooved in (a) that there is a continuous spectrum of the momentum eigenstates which constitute our wavefunction. In addition to that, I tried to compute the integrals in the case where the exponential is all over the x axis, namely when we have as state function an eigenstate of the momentum (where I should get zero uncertainty in the momentum). Both integrals (obviously) go to infinity and cannot be substracted. Last but not least, we are supposed to have ##C^{\infty}## functions as wavefunctions (at least ##C^2##). This postulate gives us the boundary conditions in problems like the infinite well. However ##e^z## does not equal zero for any ##z\in\mathbb C##. How can we have, then, such a wavefunction?
    I want to know, first of all, which of my ideas are right and which are false. I study quantum mechanics on my own and there is no one else to ask for help. I am supposed to take the course at the University this Fall and I am trying to prepare myself a bit. I have a lot of other questions as well, but this one has got on my nerves for 2 days now and I cannot sleep, because I think I am deprived of basic knowledge.
    Thank you for your time and forgive my english. It is not my mother tongue.
     
  2. jcsd
  3. Aug 6, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    In (a), you calculated that the beam is not monochromatic. It cannot be, otherwise it would extend to infinity. A true monochromatic source cannot send pulses.
    There is no realistic way to get extremely sharp edges like here, but cutting a monochromatic beam from both sides from is probably the best approach.

    (d): Your ψ is not differentiable at the ends of the beam. You cannot ignore that region, it leads to the deviations from your calculations. While it should be possible to use delta distributions here, it I guess it is easier to use the momentum spectrum that you found already.
     
  4. Aug 8, 2015 #3
    Thank you very much for your reply! Sometimes I cannot see the obvious!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Uncertainty of a wave packet
  1. Wave packet (Replies: 1)

  2. Wave packet (Replies: 2)

  3. Gaussian Wave packet (Replies: 8)

Loading...