Solving the Vertical Plane Pendulum: Energy Levels & Corrections

eman2009
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Homework Statement


a mass m is attached by a massless rod of length l to a pivot , which allows 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?

Homework Equations





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?
 
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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
 
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
 
like what? try expanding cos(theta) for small theta, the potential should look familiar
 
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
 
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...
 
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?
 
try the taylor series expansion and show some working
 
for what i use tylar expansion can you clear it please
 
  • #10
taylor expansions

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