Question 8.3: Complete the phase space change of variables P =1/b²( p² + a......

We need to find Q = Q(q, p) so as to define a canonical transformation together with (8.32). Writing

p = \left(b^{2}P − a^{2}q^{2}\right)^{{1}/{2}}= b\left(P −c^{2}q^{2}\right)^{{1}/{2}}, c = \frac{a}{b},   (8.33)

and comparing this with (8.16) we find that it is appropriate to search for a generating function of the type F_{2}(q, P). Therefore, we must have

p_{i} =\frac{\partial F_{2}}{\partial q_{i}}  , Q_{i} =\frac{\partial F_{2}}{\partial P_{i}} ,  i = 1, . . . , n , (8.16)

p = \frac{\partial F_{2}}{\partial q}= b\left(P −c^{2}q^{2}\right)^{{1}/{2}} \Longrightarrow F_{2}= b \int{\sqrt{P −c^{2}q^{2}} } dq . (8.34)

This integral is elementary, but it is more convenient not to perform the integration at this stage because we are actually interested in the other half of the canonical transformation given by

Q = \frac{\partial F_{2}}{\partial P}=\frac{b}{2}\int{\frac{dq }{\left(P −c^{2}q^{2}\right)^{{1}/{2}}}}=\frac{b}{2c} \sin^{-1} \left(\frac{cq}{\sqrt{P}}\right)= \frac{b^{2}}{2a} \sin^{-1} \left(\frac{aq}{b\sqrt{P}}\right),   (8.35)

where we have discarded a “constant” of integration C(P) in order to deal with the simplest canonical transformation possible. Thus,

Q =\frac{b^{2}}{2a} \sin^{-1}\left\{\frac{aq}{\left(p^{2} +a^{2}q^{2}\right)^{{1}/{2}}} \right\} . (8.36)

The inverse transformation has the form

q =\frac{b}{a} \sqrt{P}\sin \left(\frac{2aQ}{b^{2}} \right), p = b\sqrt{P}\cos\left(\frac{2aQ}{b^{2}} \right).   (8.37)

The Hamiltonian for the oscillator can be written in the form

H(q, p) = \frac{1}{2m} \left(p^{2} +m^{2}\omega ^{2}q^{2}\right),   (8.38)

which suggests that we choose a = mω in (8.32) and, in order to simplify equations (8.37), b =\sqrt{2a}=\sqrt{2m\omega}. Proceeding like this and taking into account that F_{2} does not explicitly depend on time, we get

K(Q, P) = H(q, p) = \frac{p^{2}}{2m}+ \frac{m\omega ^{2}}{2}q^{2}= \frac{\omega}{2m\omega}\left(p^{2} +m^{2}\omega ^{2}q^{2}\right)=\omega P,   (8.39)

from which

\dot{Q}=\frac{\partial K}{\partial P}=\omega \Longrightarrow Q = ωt + δ , (8.40a)

\dot{P}=-\frac{\partial K}{\partial Q}=0 \Longrightarrow P = constant =\frac{ E}{ \omega}, (8.40b)

where E denotes the total energy of the oscillator, the constant value of H. Returning to the original variable q with the help of (8.37), we finally obtain

q =\sqrt{\frac{2E}{m\omega^{2}} }\sin \left(\omega t+δ\right),  (8.41)

which is the usual solution to the oscillator equation of motion.