# Are Poisson brackets preserved during a canonical transformation?

Fix a Hamiltonian $$H(q, p, t)$$.

Definition: A transformation $$(q, p, t)\mapsto (Q(q, p, t), P(q, p, t), t)$$ is said to be canonical iff for the Kamiltonian $$K$$ defined as $$H(q, p, t)=K(Q(q, p, t), P(q, p, t), t)$$, $$(q(t), p(t))$$ satisfies Hamilton's equations implies that $$(Q(q(t), p(t), t), P(q(t), p(t), t))$$ satisfies Kamiltonian's equations.

Then I've been able to show that $$\frac{dQ_i}{dt} = \frac{\partial K}{\partial Q_k}\{Q_i, Q_k\} + \frac{\partial K}{\partial P_k}\{Q_i, P_k\} + \frac{\partial Q_i}{\partial t}.\tag{1}$$ Now if the transformation is to be canonical, then we must have the above equal to $$\frac{\partial K}{\partial P_i}$$

Question: Can I conclude that $$\frac{\partial Q_i}{\partial t} = 0$$, $$\{Q_i, Q_k\} = 0$$, and $$\{Q_i, P_k\}=\delta_{ik}$$?

• Feb 20, 2021 at 8:43

The definition is not correct.

A canonical transformation is defined by requiring that the Poisson bracket are preserved.

Then it follows that the Hamiltonian form of the equation of motion is preserved as well with a new Hamiltonian. The converse is generally false.

Consider, for a pair of reals such that $$a\cdot b\neq 0$$,

$$P:=ap$$, $$Q:=bq$$ and $$K(Q,P):=abH(q,p)$$.

In this case the Hamiltonian form of the equation of motion is preserved, but

$$\{Q,P\} = ab \{q,p\}= ab$$ instead of $$1$$.

Let us consider the spacetime of phases $$\mathbb{R} \times F$$, where $$F$$ is the space of phases and $$\mathbb{R}$$ the temporal axis (a better description would use a fiber bundle).

Definition. A bijective and bi-differentiable coordinate transform between two local charts $$T= t+c\:,\quad Q=Q(t,q,p)\:, \quad P=P(t,q,p)$$ on $$\mathbb{R}\times F$$ (where from now on $$t$$ and $$T$$ are always the natural coordinate on the temporal axis $$\mathbb{R}$$ up to an additive constant) is said to be canonical, if it preserves the Poisson brackets: $$\{f,g\}_{t, q,p} = \{f,g\}_{T,Q,P}$$ for every choice of smooth functions $$f$$ and $$g$$.

A canonical transformation is completely canonical if it has the form $$T= t+c\:,\quad Q=Q(q,p)\:, \quad P=P(q,p)\:.$$ $$\diamondsuit$$

N.B: The above preservation of Poisson bracket is equivalent to requiring that $$\{Q^k,Q^h\}_{t,q,p}= \{P_k,P_h\}_{t,q,p}=0\:,\quad \{Q^k,P_h\}_{t,q,p}= \delta^k_h\:.$$

We have the following result.

Proposition 1 [Preservation of Hamilton equations]. Let a bijective and bi-differentiable coordinate transform $$T= t+c\:,\quad Q=Q(t,q,p)\:, \quad P=P(t,q,p)$$ between two local charts on $$\mathbb{R}\times F$$ be canonical and suppose that the domain of the former coordinate system has the form $$I\times G$$, where $$I$$ is an open interval and $$G\subset F$$ is an open connected and simply connected set.

If $$H=H(t,q,p)$$ is a (smooth) Hamilton function, then there is a second Hamilton function $$K=K(T,Q,P)$$ such that the solutions of the Hamilton equations referred to $$H$$, translated into the new coordinates, are solutions of the Hamilton equations referred to $$K$$ and vice versa.

$$K$$ is determined by $$H$$ up to an additive arbitrary function of $$T$$.

If the coordinate transformation is completely canonical, then it is always possible to choose $$K(T,Q,P)=H(t,q,p)$$. $$\blacksquare$$

Remark. The preservation of Hamiltonian form of the equation of motion does not imply that the transformation of coordinate is canonical as I illustrated with the example above.

There are equivalent definitions of canonical transformations.

Proposition 2 [Equivalent conditions to canonicity]. Consider a bijective and bi-differentiable coordinate transform $$T= t+c\:,\quad Q=Q(t,q,p)\:, \quad P=P(t,q,p)$$ between two local charts on $$\mathbb{R}\times F$$

Suppose that the domain of the former chart has the form $$I\times G$$, where $$I$$ is an open interval and $$G\subset F$$ is an open connected and simply connected set.

The following facts are equivalent

1. The transformation of coordinates is canonical.

2. $$\sum_{k=1}^n dp_k \wedge dq^k = \sum_{k=1}^n dP_k \wedge dQ^k$$.

3. For every function $$H=H(t,q,p)$$ there is a function $$K=K(T,Q,P)$$ such that $$\sum_{k=1}^n p_kdq^k - H dt = \sum_{k=1}^n P_kdQ^k - K dT + df$$ for some smooth function $$f$$.

4. The Jacobian matrix $$\frac{\partial (P,Q)}{\partial (p,q)}$$ belongs to $$Sp(n,\mathbb{R})$$ everywhere. $$\blacksquare$$

Remarks

1. $$\sum_{k=1}^n dp_k \wedge dq^k$$ is said symplectic form on $$F$$.

2. $$\sum_{k=1}^n p_kdq^k - H dt$$ is called Poincaré-Cartan 1-form, and the confition 3 above is namebd Lie's condition.

• The correct definition of canonical transformation I wrote above is equivalent to the requirement that for every $H$ there is a $K$ such that the solutions of H equations are solution of the K equations and viceversa. Feb 20, 2021 at 8:35
• If the transformations explicitly depend on time, then $K\neq H$. Otherwise they coincide (when passing to the new variables). Feb 20, 2021 at 8:37
• Sorry I was too sloppy: the equivalent condition is that for every Hamiltonian $H$ there is another Hamiltonian K such that the two Poincaré Cartan forms coincide up to a total differential. This is a stronger requirement than the one I wrote above. $pdq -Hdt = PdQ -Kdt + df$. Feb 20, 2021 at 8:55
• Well, the definition is just the preservation of Poisson brackets. This definition is equivalent to the requirement that for every H there is K such that the identity you wrote is true for some df. Feb 20, 2021 at 9:13
• Yes, that condition is both the starting point for the construction of generating functions and for the development of Hamilton-Jacobi theory. Feb 20, 2021 at 19:59