# Classifying canonical transformation and scaling transformation

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.

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

• 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. Jul 18, 2022 at 18:36
• 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 $\begin{pmatrix} 0& -I_n \\ I_n&0\end{pmatrix}$ Jul 18, 2022 at 19:05
• Thank you for the clarification. Jul 18, 2022 at 19:55