Question 9.2: Solve the equation of motion for the one-dimensional harmoni......

Now H = p²/2m + mω²q²/2 and the Hamilton-Jacobi equation is written

\frac{1}{2m}\left(\frac{\partial S}{\partial q}\right)^{2} + \frac{m \omega^{2} }{2} q^{2} +\frac{\partial S}{\partial t}=0.  (9.26)

As in the free particle case, Eq. (9.26) is separable in the form

S = W(q) − αt (9.27)

where W obeys

\frac{1}{2m}\left(\frac{dW}{d q}\right)^{2} + \frac{m \omega^{2} }{2} q^{2}=\alpha.   (9.28)

The positive constant α coincides with the constant value of the Hamiltonian (equal to the total energy, in the present case) inasmuch as dW/dq = ∂S/∂q = p. A solution to (9.28) is

W =\int{\sqrt{2m\alpha- m^{2}\omega^{2}q^{2}} dq} .   (9.29)

As a consequence,

S(q, α, t) =\int{\sqrt{2m\alpha- m^{2}\omega^{2}q^{2}} dq- \alpha t},   (9.30)

and the solution of the equation of motion for q is obtained from

\beta= \frac{\partial S}{\partial \alpha}= m \int{\frac{dq}{\sqrt{2m\alpha- m^{2}\omega^{2}q^{2}}}-t} = \frac{1}{\omega} \sin^{-1} \left(\sqrt{\frac{m\omega^{2}}{2\alpha}}q\right)-t.   (9.31)

Solving this equation for q we find, with δ = ωβ,

q(t) =\sqrt{\frac{2\alpha}{m\omega^{2}}} \sin \left(\omega t +\delta\right),   (9.32)

which is the well-known general solution to the oscillator equation of motion.