Question

Now let's treat the simple pendulum quantum mechanically (a) In the small x limit, construct a formula for the energy levels. (b) If x isn't so small, we should include a second term (proportional to x) in the expansion of the potential energy function. Treating this second term as a perturbation, set up an expression for the correction it provides to the zero-point energy and determine the sign of this correction (i.e., + or -). (c) Explain how your result in (b) could also follow from a heuristic (i.e., "stylish hand- waving") uncertainty principle argument.

Answer 1

If we treat simple pendulum as a quantum mechanically In the small x limit potential energy v = 1/2 k x^2 so Hamiltonian H = P^^2_x/2m + 1/2 m w^2 x^^2 where k = m w^2 we Introduce creation and annihilation operator since H^^/hwithstroke w = P^^2_x/2m hwithstroke w + 1/2 m w/hwithstroke x^^2 So annihilation operator a^^= squareroot m w/2 hwithstroke x^^+ i P^^_x/squareroot 2 m hwithstroke w and creation operator^+ = squareroot m w/2 hwithstroke x^^- i P^^_x/squareroot 2m hwithstroke Now consider a^^+ a^^= (squareroot m w/2 hwithstroke x^^- i P^^_x/squareroot 2m hwithstroke w) (squareroot m w/2 hwithstroke x^^+ i P^^_x/squareroot 2m hwithstroke w) \r\n a^^+ a^^= m w/2 hwithstroke x^^2 + P^^2_x/2m hwithstroke w - i/2 hwithstroke (P^^_x x^^) + i/2 hwithstroke x^^P^^_x a^^+ a^^= 1/hwithstroke w (m^2 w^2/2 x^^2 + P^^2_x/2m) + i/2 hwithstroke (x^^P^^_x - P^^_x x^^) because [x^^, P^^_x] = x^^P^^_x