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Infinite square well expectation value problem

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


    2. The attempt at a solution

    Honestly, I don't even know where to begin.
    I assumed V<0, V>L is V=∞ and 0<V<L is V=0

    I tried setting up the expectation value formula

    1/2∫x|ψ*Pψ|2+(x|ψ2)*P x|ψ|2

    but what is ψ? is it √(2/L) Sin ((n2πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?

    Thanks for viewing my question!
     
    Last edited: Dec 13, 2012
  2. jcsd
  3. Dec 14, 2012 #2
    [itex]\psi[/itex] is the wave function for the given problem, that you find by solving Schrodinger equation for a given potential
     
  4. Dec 14, 2012 #3

    vela

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    You mean V=∞ when x<0 or x>L.

    How did you get this?

    You do use a solution for the infinite square well, but the function you wrote isn't quite correct.
     
  5. Dec 15, 2012 #4
    Thanks for responding guys! I very much appreciate it!

    Ahh yes, sorry for the crappy notation


    Uhhh Now that I've done some more studying I can see that its wrong but I was, at the time, hoping that the expectation values would be ∫<xp>+<px> = ∫<xp+px> and the expectation values of xp, px will just add together like magic.

    Which is wrong! So i've redone the problem in an attempt to simply find the operator first

    <xp> = -ihx(dψ/dx) and
    <px> = -ih (d(xψ)/dx)
    1/2<xp+px> = (-ih)/2 (x(dψ/dx)+(d(xψ)/dx)

    Now the expectation value should be <1/2(xp+px)>= ∫ψ*(1/2)<xp+px>ψ dx?

    do I just plug in 1/2<xp+px> (assuming its correct) and plug in ψ (the solution to the infinite square well) and take the integral from 0 to L?

    Sorry! I was very wrong haha
    √(2/L) Sin ((nπx)/L)
     
  6. Dec 15, 2012 #5
    One thing that might help you:

    Using [x,p] = ihbar, or xp - px = ihbar, or px = -ihbar + xp, you can write

    xp + px = xp + xp - ihbar = 2xp - ihbar, which has an easier to calculate expectation value. (you should check my work, I may have made a mistake)

    In general an expectation value is:

    <Q> = ∫ψ*Qψdx, where Q on the right is the operator for the observable you want to calculate. ψ is the wavefunction of the particle, which should be supplied by your textbook (it looks like you have the right one though).

    Removed by moderator

    Where I used the normalization condition ∫ψ*ψdx = 1.

    I'll leave it to you do actually do the integral, and to check my work.
     
    Last edited by a moderator: Dec 16, 2012
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