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Lets assume we have a very simple transformation in 1 Dimension from $(x, p_x)\rightarrow (y,p_y)$ given as $$\begin{aligned} y &= cx \\ p_y &= c^{-1} p_x \end{aligned}$$

Is this a strictly canonical transformation, an extended canonical transformation or a scaling transformation?

Let the Hamiltonian in $x,p_x$ be $$ H(x, p_x) = \frac{p_x^2}{2m} + V(x) $$

A quick test with Poisson brackets $$\{f,H \}_{(x,p_x)}\equiv \frac{\partial f}{\partial x}\frac{\partial H}{\partial p_x} - \frac{\partial f}{\partial p_x}\frac{\partial H}{\partial x} $$

yields $$\begin{aligned} \dot y &= \{y,H \}_{(x,p_x)} = \frac{c}{m}p_x = c \dot x \\ \dot p_y &= \{p_y,H \}_{(x,p_x)} = -c^{-1}V'(x) = c^{-1}\dot p_x \\ \end{aligned}$$

which looks like it satisfies the strict canonical transformation conditions.

Is this a special case of a scaling transformation that is at the same time a canonical transformation? And if it is a canonical transformation, what is a generating function for this transformation? I fail at matching it with the four common types and have a hard time applying the formalism for canonical transformations to it.

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First of all, let my clarify my notation. In this answer I'll use

  • $(p,q)$ as coordinates before the transformation;
  • $(P,Q)$ as coordinates after the transformation.

In this notation, your transformation reads $$Q=cq \qquad P=c^{-1}p$$ this is indeed a scale transformation and it preserves the canonical structure of Hamilton's equations$^1$. In other words, there exists a Hamiltonian in the new set of coordinates that satisfies Hamilton's equations. According to some authors this is the definition of canonical transformation.

Others, the book I quoted below for example, define as canonical a transformation whose jacobian matrix is symplectic.$^2$ You can verify that the jacobian matrix is

$$\begin{pmatrix} c^{-1} & 0 \\ 0 & c \end{pmatrix}$$ is symplectic i.e.

$$\begin{pmatrix} c^{-1} & 0 \\ 0 &c\end{pmatrix}^T\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} c^{-1} & 0 \\ 0 & c \end{pmatrix}=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $$

This is not just luck. More generally, a scale transformation is defined as follows$^3$ $$Q_i=c_i q \qquad P_i=d_i p_i \qquad\forall i$$ with $c_i,d_i\in\mathbb{R}$ such that $c_id_i=\lambda\in\mathbb{R}$.

If $\lambda=1$, it turns out it is canonical also in this sense$^4$ (symplectic jacobian matrix). Of course, in our case we have just one coordinate and $d=c^{-1}$, so our condition is satisfied.

Finally, regarding the generating function, consider a generating function of the second type $F_2(q,P)=cqP$ that is the generating function of identical transformation multiplied by $c$. Using the definition of generating functions $$Q=\frac{\partial F_2}{\partial P}=cq \iff Q=cq\\ p=\frac{\partial F_2}{\partial q}=cP\iff P=c^{-1}q$$ that is your transformation.


$^1$ See Analytical Mechanics. Fasano&Marmi. Chapter 10, section 10.2. Example 2.3.

$^2$ See Analytical Mechanics. Fasano&Marmi. Chapter 10, Definitions 1.2 and 2.2.

$^3$ No summation on repeated indexes is understood.

$^4$ See Analytical Mechanics. Fasano&Marmi. Chapter 10, section 10.2. Example 2.7

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  • $\begingroup$ The reference and answer is very helpful. There is one point that is still murky for me. Which matrix is the symplectic one? The jacobian $$\begin{pmatrix} c^{-1} & 0 \\ 0 & c \end{pmatrix}$$ is clearly not skew-symmetric and thus not symplectic. I think I am misunderstanding something obvious here. $\endgroup$
    – Hans Wurst
    Commented Jul 18, 2022 at 18:36
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    $\begingroup$ I expanded that part. Symplectic does not mean skew symmetric. Identity matrix is symplectic, for example. As for the definition of symplectic, you can check footnote 2, definition 1.2. A $2n\times 2n$matrix $A$ is symplectic iff $A^T\mathcal{J}A=\mathcal{J}$ where $\mathcal{J}$ is $\begin{pmatrix} 0& -I_n \\ I_n&0\end{pmatrix} $ $\endgroup$ Commented Jul 18, 2022 at 19:05
  • $\begingroup$ Thank you for the clarification. $\endgroup$
    – Hans Wurst
    Commented Jul 18, 2022 at 19:55

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