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Could you justify at the end of solution II and solution III the chain rule? Why are the derivative of q_i with respect to Q_k zero?

Hallo, why is the derivate of P_s with respect to Q_j equal zero? you do this in the last step of solution III. If this is true, then your equation (3), derivate of P_i w.r.t. p_k, would be equal zero too. maybe it could be useful in solution II and III, that eq. (3) is zero...

Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):

Claim: For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).

Proof:

We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).

So we get:

Equation 1:

$$ 1 = \Phi_{p Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{p Q}} $$

Equation 2:

Holding \( p \) fixed and derive by \( q \):

$$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$

Equation 3:

Holding \( q \) fixed and derive by \( p \):

$$ P_p |_q = - \Phi_{QQ}Q_q |_p \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$

where we used equation 1 again.

Equation 4:

Holding \( p \) fixed and derive by \( q \):

$$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} - \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$

using equation 2.

We are now prepared to calculate:

$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{p Q}} \left( \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)

By the script:

$$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$

with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and

$$ J = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix} $$

and this results in showing all the things we wanted to show.

Cheerio, A.