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Energy level

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    a mass m is attached by a massless rod of length l to a pivot , which allowes swing freely in a vertical plane under the influence of gravity .let the anglethita betweenthe rod the vertical .
    find the energy levels of the system?
    find the lowest -order correction to the ground state energy with small angle?

    2. Relevant equations



    3. The attempt at a solution
    if i can solve it classically first and then find the energy from schrodenger eqn , but i'm not sure if the hamiltonain work in this case????
     
  2. jcsd
  3. Feb 25, 2009 #2

    lanedance

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    hmmm... have a look what the simplified potential would look like (ie for small angles) should hopefully give you a hint
     
  4. Feb 26, 2009 #3
    v=-mg cos thita
    T= p^2/2m
    H=T+V
    put in schrodenger eqn solve for E then find the energy level......??????
    and for small angle thita =0 is it like this
     
  5. Feb 26, 2009 #4

    lanedance

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    like what? try expanding cos(theta) for small theta, the potential should look familiar
     
  6. Feb 27, 2009 #5
    V=mgl
    cos thita=1 for small thita
    howabout the boundry condition ...... if the angle is small is it
    FI =FI(thita+2bi)
    or
    FI(0)=0
    FI the wave function
     
  7. Feb 27, 2009 #6

    lanedance

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    i'm not sure what you last post means

    try writing what the force - F = -dV/dx,

    do you know taylor series? if so you could expand both the potential (cos term) & the force (sin) for small theta

    i think it will look soemthing like
    F ~ -k.x

    similarly
    V ~ k.x^2

    look familiar? looking simply harmonic to me...
     
  8. Feb 28, 2009 #7
    i think i got the answer is it totally differnet
    look...v=mgl(1-costhita)
    H=1/2 ml^2thita'^2+1/2 mglthita^2
    and E=(n-1/2)hw but i don't know from where get the E?
    the lowest correction is H'=v-1/2 ml thita^2=1/24mglthita^4 , how ?
    E'0=3/4 alpha^4(-1/24mgl/l^3)......how?
     
  9. Feb 28, 2009 #8

    lanedance

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    try the taylor series expansion and show some working
     
  10. Mar 1, 2009 #9
    for what i use tylar expansion can you clear it please
     
  11. Mar 2, 2009 #10

    lanedance

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    taylor expansions

    cos(t) ~1 + t^2/2 +o(t^4)
    sin(t) ~ t + o(t^3)
     
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