Question 9.1: Solve the equation of motion for the free particle in one di......

Since H = p²/2m, the Hamilton-Jacobi equation takes the form

\frac{1}{2m}\left(\frac{\partial S}{\partial q}\right)^{2} + \frac{\partial S}{\partial t}=0.  (9.18)

A complete integral of this equation can be obtained by separation of variables in the form of a sum:

S = W(q) + T(t) . (9.19)

Inserting S of this form into (9.18) yields

\frac{1}{2m}\left(\frac{dW}{dq}\right)^{2}=-\frac{dT}{dt}.   (9.20)

If we fix t and vary q the left-hand side of (9.20) should vary. However, it cannot vary because it is equal to the right-hand side, which remains fixed. Therefore, both sides of (9.20) are equal to the same positive constant we denote by α. Thus, we are left with the two ordinary differential equations

\frac{dW}{dq}= \sqrt{2m \alpha } , -\frac{dT}{dt}= \alpha ,   (9.21)

whence

S(q, α, t) =\sqrt{2m \alpha }q-\alpha t,  (9.22)

where we have abandoned merely additive constants of integration. The solution of the equation of motion for q is obtained from

\beta= \frac{\partial S}{\partial \alpha}= \sqrt{\frac{m}{2 \alpha}}q-t,   (9.23)

and from

p =\frac{\partial S}{\partial q}=\sqrt{2m \alpha }= constant =p_{0}. (9.24)

Therefore,

q = \sqrt{\frac{2\alpha}{m}}\beta+ \sqrt{\frac{2\alpha}{m}}t \equiv q_{0}+\frac{p_{0}}{m}t ,   (9.25)

which is the elementary solution to the free particle problem.