1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Quantum Mechanics: Even/Odd Stationary States

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m in the infinite square well of width a at time t = 0 is in a linear superposition of the ground- and the first excited- eigenstates, specifically it has the wave function

    $$| \Psi(x,t) > = A[ | \psi_1 > + e^{i \phi} | \psi_2 >$$

    Find the expectation value <x(t)> of the position of the particle at time t.

    2. Relevant equations

    [tex]< f(x,t) | g(x,t) > = \int_{-a/2}^{a/2} f(x,t)^{*}g(x,t)\ dx[/tex]
    $$ <x(t)> = < \Psi(x,t) | x \Psi(x,t) > $$

    3. The attempt at a solution

    Obtaining a solution for this problem isn't extraordinarily difficult-- given the stationary states [itex] \psi_1[/itex] and [itex] \psi_2 [/itex], we can simply plug them into the above equations and-- after having normalized for 'A'-- solve for <x(t)>. My question is a more methodological one, regarding the even-ness and odd-ness of the integrands. After going through a few steps to solve for <x>, we would obtain the following solution:

    $$ <x> = A < \psi_1 | x \psi_1 > + B | \psi_1 | x \psi_2 > + C < \psi_2 | x \psi_1 > + D < \psi_2 | x \psi_2 > $$

    where A, B, C, and D, are imaginary constants.
    In the interest of time, I decided to consider the even-ness and odd-ness of the above integrands: ## \psi_{0,2,4,...} ## correspond to odd functions, and ## \psi_{1,3,5,...} ## correspond to even functions. 'x' should also be considered an odd function, since I have chosen the integral to run from -a/2 to a/2. Hence, we can determine whether each integrand is ultimately even or odd (even times even is even; odd times even is odd; etc.), and immediately set the odd integrands to zero (integrating an odd function centered at the origin yields a value of zero). This would mean: A = 0 and D = 0. However, this turns out to give me the wrong answer, as it seems I'm supposed to consider the value of *all* the integrals.

    I tried to keep the mathematical calculations in this question to a minimum; I would just like to know whether the above methodology is mathematically intact (I almost hope it isn't...), and-- if so-- why it doesn't work out in this case.

    Thanks for any help.
  2. jcsd
  3. Mar 23, 2013 #2
    The methodology seems pretty much fine. Check the endpoints of the well in the problem statement to make sure there is symmetry across x=0. If so, it would not be A and D which were zero, but the inner products [itex] <\psi_1|x|\psi_1 >[/itex] and [itex] <\psi_2|x|\psi_2 >[/itex]. Perhaps this is for the general case where the endpoints are not symmetrically located across x = 0. Also A in your last equation is not the same A as in your first equation, but perhaps you know this. B and C would depend on phi, which depends on time.

    $$ <x(t)> = < \Psi(x,t) | x \Psi(x,t) > = A^*A( <\psi_1| + e^{-i\phi} <\psi_2|)x( | \psi_1 > + e^{i \phi} | \psi_2 >) $$

    $$ A^*A( <\psi_1|x|\psi_1 > + e^{i \phi} <\psi_1|x| \psi_2> + e^{-i\phi} <\psi_2|x|\psi_1 > + <\psi_2 |x| \psi_2 >) $$
  4. Mar 23, 2013 #3
    The nature of the infinite square well is such that there is always symmetry about the center (inside, the stationary states represent standing waves). In my particular case, I have chosen the center to lie at x=0 so that I may employ the odd-ness of the function f(x) = x and pursue the methods outlined above, which naturally elicit the following: [itex] < \psi_1 | x \psi_1 > = < \psi_2 | x \psi_2 > = 0 [/itex].
    However, anylitically solving each integral will yield a value of zero for all of them.

    To be sure, the infinite square well of width 'a' has the following stationary states:

    [tex] | \psi_n > = \sqrt{ \frac{2}{a} } sin( n \pi x / a) [/tex]


    [tex] < \psi_1 | x \psi_1 > = \frac{2}{a} \int_{-a/2}^{a/2} xsin^{2}( \pi x / a)\ dx = 0 [/tex]

    [tex] < \psi_2 | x \psi_2 > = \frac{2}{a} \int_{-a/2}^{a/2} xsin^{2}( 2 \pi x /a)\ dx = 0 [/tex]

    [tex] < \psi_1 | x \psi_2 > = < \psi_2 | x \psi_1 > = \frac{2}{a} \int_{-a/2}^{a/2} xsin( \pi x /a)sin(2 \pi x /a)\ dx = 0 [/tex]

    (Results obtained from WolframAlpha)
    The former two results are expected in light of the integrands' oddness, but the lattermost result is confounding, and constitutes the basis of my question. I'm sure there's something wrong with my 'even/odd-ness' approach (I wouldn't doubt the validity of explicit integration), I'm just not sure what.
    (Honestly, I'd just stick to explicit integration, but this might not turn out so well on a 50-minute midterm :\)

    Edit: Nvm.. Even if 'zero' was the true result, this would void the possibility of any time-oscillatory terms (zero truncates everything )which are otherwise present in a system running from 0 to a...
    Last edited: Mar 23, 2013
  5. Mar 23, 2013 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Check to see if these wavefunctions satisfy the boundary conditions at x = ±a/2.
  6. Mar 23, 2013 #5
    Wow-- completely overlooked that.

    Thanks a lot!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted