Are Poisson brackets preserved during a canonical transformation?

Question

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 $\{Q_i, P_k\}=\delta_{ik}$?

Answer 1

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 $1$.

ADDENDUM.

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.