Necc. and suff. conditions for a canonical transformation.

Question

Let $\mathbf{P} = C^{−1}\mathbf{p} + B\mathbf{q}, \mathbf{Q} = C\mathbf{q}$, where $C$ is a symmetric nonsingular matrix. Determine necessary and sufficient conditions on $C$ for the transformation $(\mathbf{q}, \mathbf{p}) \mapsto (\mathbf{Q}, \mathbf{P})$ to be canonical.

Where we are expected to assume that everything is of the appropriate dimensions for the transformation to make sense. Noting that symplecticity (word?) of the Jacobian of the transformation is enough, and this boils down to showing that the poisson structure must be preserved for our choice of $C$. i.e.:

$\{Q_i,Q_J\}_{q,p}=0 =\{P_i,P_j \}_{q,p}$

$\{Q_i,P_j\}_{q,p}=\delta_{ij}$

Where $\{.,.\}_{q,p} = \dfrac{\partial (.)}{\partial q_k}\dfrac{\partial (.)}{\partial p_k} - \dfrac{\partial (.)}{\partial p_k}\dfrac{\partial (.)}{\partial q_k}$

But I'm struggling to derive conditions on $C$ that would satisfy an examiner (this is a past examination question). Any hints on the general direction one should go in as far as the index fiddling is concerned? FWIW, I've shown that the first of the above conditions holds identically (I hope!).

Thanks in advance!

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EDIT - To elaborate further, given that Qmechanic has recommended that I continue trying what I've already tried:

$\{P_i,P_j \}_{q,p} = 0$

$\iff \partial_{q_k}[A_{ia}p_a + B_{ia}q_a] \partial_{p_k}[A_{jb}p_b + B_{jb}q_b] -\partial_{p_k}[A_{ia}p_a + B_{ia}q_a] \partial_{q_k}[A_{jb}p_b + B_{jb}q_b] =0 $

$\iff \left(A_{ia}\dfrac{\partial p_a}{\partial q_k}A_{jk} - A_{ik}A_{ja}\dfrac{\partial p_a}{\partial q_k}\right) + \left(B_{ik}A_{jk} - A_{ik}B_{jk}\right) + \left(A_{ia} \dfrac{\partial p_a}{\partial q_k}B_{jb}\dfrac{\partial q_b}{\partial p_k} - B_{ia} \dfrac{\partial q_a}{\partial p_k}A_{jb}\dfrac{\partial p_b}{\partial q_k}\right) + \left(B_{ik}B_{ja}\dfrac{\partial q_a}{\partial p_k} - B_{ia}\dfrac{\partial q_a}{\partial p_k}B_{jk}\right) = 0$

Where $A=C^{-1}$, and summation convention applies. I'm struggling to see how all but the second term should vanish? And also hoping there is a better way to do this problem without excessive fiddling with the indices!

(Also, my apologies if I'm misusing up/down indices here - I have yet to cover how they work.)

Thanks.

EDIT: Ah, I failed to note that $q_i,p_j$ are independent of one another… Which then yields the desired conditions.

Answer 1

Hints:

1. Consider a coordinate transformation of the form $$P_i ~:=~ p_j A^j{}_i + B_{ij}q^j , \qquad Q^i := C^i{}_j q^j. \tag{1}$$

2. A canonical transformation is in this context a symplectomorphism. It should take Darboux coordinates $(q^i,p_j)$ into Darboux coordinates $(Q^i,P_j)$, so that $$ \{Q^i,Q^j\}_{q,p}~=~0, \qquad \{Q^i,P_j\}_{q,p}~=~\delta^i_j, \qquad \{P_i,P_j\}_{q,p}~=~0. \tag{2}$$

3. Inserting ansatz (1) into eqs. (2) leads to $$ \text{Automatically satisfied}, \qquad CA ~=~{\bf 1}_{n \times n} ,\qquad BA~=~(BA)^t,\tag{3}$$ respectively.