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Homework Help: Derivation Problem

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data

    The analytical expression of [tex]\Delta[/tex]x[tex]\Delta[/tex]p for a particle in a box is:

    [tex]\Delta[/tex]x[tex]\Delta[/tex]p = h/2pi[tex]\sqrt{(n\pi)^{2} - 6}[/tex] / [tex]\sqrt{12}[/tex]
    for any quantum number, n

    2. Relevant equations

    ([tex]\Delta[/tex]x)[tex]^{2}[/tex] = <x[tex]^{2}[/tex]> - <x>[tex]^{2}[/tex]

    and ([tex]\Delta[/tex]p)[tex]^{2}[/tex] = <p[tex]^{2}[/tex]> - <p>[tex]^{2}[/tex]

    [tex]\Psi[/tex] = [tex]\sqrt{2/L}[/tex] sin(nxpi/L)

    3. The attempt at a solution

    so i tried to find (delta x)^2 and multiplied it with (delta p)^2 and rooted it, but i still have "L" left over in my derivation which doesnt work out...
     
  2. jcsd
  3. Mar 4, 2008 #2
    Can you show the calculations for the uncertainties in x and p? What you're doing is correct from what you've described.
     
  4. Mar 4, 2008 #3
    <x^2> = integral of (x^2)(psi^2) = 2/ L integral (x^2)sin^2(nxpi/L) dx

    integrate by parts from zero to L andd
    <x^2> = L^2 - L + 1

    <p^2> = integral (conjugate psi)(momentum operator)^2(psi)dx
    <p^2> = -h/Lpi integral (conjugate psi)(d^2 psi/ dx^2)dx
    after taking derivative and the integral
    <p^2> = [ (h bar)(n)(pi)/(L) ] ^2

    <p>^2 i found to be zero
    and <x>^2 i found to be L^2 / 4

    so ( delta x)^2 = L^2 - L + 1 - L^2 / 4 = 3L^2 /4 - L + 1
    and (delta p)^2 = [(h bar)(n)(pi)/(L)] ^2

    (delta x)(delta p) = sqrt [(3L^2/4 - L + 1)[(hbar)(n)(pi)/(L)]^2]

    you can see that L is un removable...it doesnt really work out for me..
     
  5. Mar 4, 2008 #4

    pam

    User Avatar

    psi(x) must be zero at the ends, so your -L+1 is wrong.
    Check the dimensions. <x*2> has to ~L^2.
     
  6. Mar 4, 2008 #5
    <x^2> = 2/L integral (x^2)sin^2 (nxpi/L) dx
    = 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - integral x sin^2 (nxpi/L)] from 0 to L
    = 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - integral of sin^2 (nxpi/L)] from 0 to L
    =2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - (x/2 - (L/4npi)sin (2nxpi/L] from 0 to L
    = 2/L [ L^3 /2 - L^2/2 + L /2]
    = L^2 - L + 1

    i dont think i did anythign wrong in my integration...well clearly something is wrong

    but i found this
    6a1042615e1f7e4a192736255848f2f1.png

    its from -a/2 to a/2..so how do i make that form 0 to L?
    its the same right?
     
    Last edited: Mar 4, 2008
  7. Mar 4, 2008 #6
    so this pic basically solves my problem
    its jsut that...
    i wouldnt know how to integrate to get that in the first place
    and
    <x>^2 = L^2 / 4...and i cant incorporate that..
    because well sqrt( <x^2> * (delta p)^2) = the answer that they want
    but i need (delta x)^2(delta p)^2

    (delta x)^2 = <x^2> - <x> ^2...and if i do that its no longer correct...
     
    Last edited: Mar 4, 2008
  8. Mar 5, 2008 #7
    Things to note: [tex]\langle x \rangle = \langle p \rangle = 0[/tex] -- can you show this?

    Further, you should be able to integrate [tex]\int x^n sin(x)\,dx[/tex] as an indefinite integral. Can you do so for n=1 or n=2?
     
  9. Mar 5, 2008 #8
    <x> doesnt equal <p>
    in my textbook they show that,
    <x> = L/2 and <p> = 0..
    and that integral doesnt really relate because x^n sin x...when i have (x^n)(sin^2 x)
     
  10. Mar 5, 2008 #9
    True -- my bad when posting too quickly. But you should still be able to do those integrals by parts...
     
  11. Mar 5, 2008 #10
    6a1042615e1f7e4a192736255848f2f1.png

    from my integration by parts for <x^2> i get (L^2)/3 when accoring to that equation
    i should be gettin L^2[(n^2)(pi^2) - 6] / [12(n^2)(pi^2)]

    a = L, but you hafta multiply it by 2/L, since the wavefunction is sqrt(2/L) sin(nxpi/L)
     
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