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(Apologies if HW questions are not allowed -- I couldn't really find a definite answer on this)


Let $Q^1 = (q^1)^2, Q^2 = q^1+q^2, P_{\alpha} = P_{\alpha}\left(q,p \right), \alpha = 1,2$ be a CT in two freedoms.

(a) Complete the transformation by finding the most general expression for the $P_{\alpha}$.

(b) Find a particular choice for the $P_{\alpha}$ that will reduce the Hamiltonian

$$H = \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2$$


$$K = P_1^2 + P_2.$$


I have shown that

$$P_1 = \frac{1}{2q^1} \left( p_1 + \frac{\partial F}{\partial q^1} - p_2 - \frac{\partial F}{\partial q^2} \right), $$

$$P_2 = p_2 + \frac{\partial F}{\partial q^2}$$

is the most general canonical transformation for the momenta, where $F=F(q^1, q^2)$. This is consistent with the solution manual. For part b, however, the answer I get for an intermediate step is inconsistent with the solutions manual, and I don't understand why. Given that the transformation is canonical, all I need to do to find the transformed Hamiltonian K is find the inverse transformation and plug it in to the Hamiltonian H. The inverse transformation is

$$p_2 = P_2 - \frac{\partial F}{\partial q^2},$$ $$p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.$$

Plugging this into H, and renaming H to K since it's in terms of the transformed coordinates we have

$$K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.$$

Since we want K to be

$$K = P_1^2 + P_2,$$

this means

$$\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.$$ $$F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.$$

Plugging this into the general transformation I derived I find that

$$P_1 = \frac{1}{2q^1} \left(p_1-p_2-(q^1)^2 \right),$$ $$P_2 = (q^1+q^2)^2+p_2.$$

My equation for $P_2$ is consistent with the solutions manual, but my equation for $P_1$ is not. According to the solutions manual


My question is, is my methodology essentially correct, and if so did I go wrong in the algebra or did I make some sort of mistake in how I solved the problem.

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Homework questions are frowned upon because they are generally of interest only to one person (the one doing the homework) and we prefer questions that will be interesting to a large audience. –  John Rennie Dec 10 '13 at 10:12
I think this is conceptual enough, actually. –  Dimensio1n0 Dec 10 '13 at 15:18

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