Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was going through my professor's notes about Canonical transformations. He states that a canonical transformation from $(q, p)$ to $(Q, P)$ is one that if which the original coordinates obey Hamilton's canonical equations then so do the transformed coordinates, albeit for a different Hamiltonian. He then considers, as an example the Hamiltonian


with a transformation:

$$Q = q,$$ $$P = \sqrt{p} - \sqrt{q}.$$

The notes state that this transformation is locally canonical with respect to $H$, and that in the transformed coordinates the new Hamiltonian is:

$$K = \frac{1}{3} \left( P + \sqrt{Q} \right)^3.$$

I don't understand how we know that this is locally canonical, or what it really even means to be locally canonical. Also, where do we get K from? Considering that the inverse transformation would be:

$$q=Q,$$ $$p=\left( P + \sqrt{Q} \right)^2,$$

Why isn't the new Hamiltonian this:

$$K= \frac{1}{2} \left(P + \sqrt{Q} \right)^4,$$

where all I've done is plug the inverted transformation into the original Hamiltonian?

I'm a bit confused by all this. Would appreciate any help.

share|cite|improve this question
up vote 1 down vote accepted

The original coordinates satisfy the equations of motion when the integral of $p\, \dot{q} - H(p,q)$ is minimized, and the new coordinates satisfy the equations of motion when the integral of $P\, \dot{Q} - K(P,Q)$ is minimized. There is no requirement that $H$ and $K$ be numerically equal.

The transformation is canonical if the Poisson bracket remains invariant.

The EOMs are

$\dot{p} = 0$

$\dot{q} = p$

and from the new Hamiltonian, we get

$\dot{P} = -(P+\sqrt{Q})^2 \frac{1}{2\sqrt{Q}} = - \frac{p}{2\sqrt{q}} = \frac{d}{dt} \left(\sqrt{p} - \sqrt{q} \right)$

$\dot{Q} = (P+\sqrt{Q})^2 = \dot{q}$

thus the equations of motion are numerically equal.

share|cite|improve this answer
As joshphysics pointed out, the Poisson bracket in this case isn't invariant. – lionelbrits Dec 5 '13 at 23:45

Cool question!

Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors.

If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined as follows (I paraphrase)

Goldstein Definition: A transformation $f:\mathcal P\to\mathcal P$ on phase space $\mathcal P$ is canonical provided there exists a phase space function $K$ such that if $(q(t), p(t))$ is a solution to Hamilton's equations generated by $H$, then $(Q(t), P(t)) = f(q(t), p(t))$ is a solution to Hamilton's equations generated by $K$.

This is essentially the definition given by lionelbrits in his answer.

On the other hand, if you look, for example, in Spivak's mechanics text, then you'll find the following definition:

Spivak's Definition: A transformation $f:\mathcal P \to \mathcal P$ on phase space is canonical provided it preserves the symplectic form.

In more concrete terms (namely in canonical coordinates), Spivak's definition can be stated as follows:

The transformation $f(q,p) = (f^q(q,p), f^p(q,p))$ is canonical if and only if its Jacobian (derivative) matrix preserves the symplectic matrix $J$, namely \begin{align} f'(p,q)\,J\,f'(p,q)^t = J \end{align} where \begin{align} J=\begin{pmatrix} 0 & I_n \\ -I_n & 0 \\ \end{pmatrix},\qquad f' = \begin{pmatrix} \frac{\partial f^q}{\partial q} & \frac{\partial f^q}{\partial p} \\ \frac{\partial f^p}{\partial q} & \frac{\partial f^p}{\partial p} \\ \end{pmatrix} \end{align} where $2n$ is the dimension of phase space and $I_n$ is the $n\times n$ identity matrix.

It also turns out that

If a transformation is canonical in the sense defined by Spivak, then it is canonical is the sense of Goldstein with $K = H\circ f^{-1}$

but the converse is not true. In fact, this example you brink up is a counterexample to the converse! What lionelbrit showed in his answer is that the example you have written is a canonical transformation in the sense of Goldstein, but, as you should try to convince yourself (I did), the function $K = H\circ f^{-1}$ that you wrote down by inverting the transformation and plugging back into $H$ leads to Hamilton's equations that are not satisfied by $(Q(t), P(t)) = f(q(t), p(t))$. You can show this directly by writing down the equations of motion. You can also show this by computing the Jacobian of the transformation and showing that it does not preserve the symplectic matrix. In fact, you should find that the Jacobian is given by \begin{align} f'(q,p)=\begin{pmatrix} 1 & 0 \\ -\frac{1}{2\sqrt{q}} & \frac{1}{2\sqrt{p}} \\ \end{pmatrix} \end{align} and that \begin{align} f'(q,p) J f'(q,p)^t = \frac{1}{2\sqrt{p}} J \end{align} In other words, the Jacobian of the transformation preserves the symplectic matrix up to a multiplicative factor.

Speculation. I'm going to go out on a limb and guess that your professor calls Goldstein's definition a "local canonical transformation" and Spivak's definition a "canonical transformation." If we adopt this terminology, then it's clear from our remarks that the $K$ he gives shows that your example is a local canonical transformation, but that the transformation is not canonical.

share|cite|improve this answer
Are there good reasons to understand and use transformations that are locally canonical but not canonical? – Jess Riedel Jun 4 '15 at 20:38
@JessRiedel Not sure be be honest. I don't personally know of a physical situation for which the more general definition of canonical transformation is useful/necessary. – joshphysics Jun 5 '15 at 4:18
@joshphysics. Thank you for the illustrative discussion. We may further note that in this example, the Kamiltonian $K=p^{3/2}/3$ can be expressed as a function of the Hamiltonian $H=p^2/2$, and $$f'(q,p) J f'(q,p)^t = (dK/dH) J.$$ This condition might be used to generalize the symplectic condition. – hbp Oct 15 '15 at 8:11

I) The restricted$^1$ transformation (RT)

$$\tag{1} (q,p)~\longrightarrow~ (Q,P) ~:=~(q, \sqrt{p} - \sqrt{q})$$

of OP's professor with inverse RT

$$\tag{2} (Q,P)~\longrightarrow~ (q,p) ~:=~(Q, (P+ \sqrt{Q})^2) ,$$

and with Hamiltonian $H=\frac{p^2}{2}$ and Kamiltonian $K=\frac{p^3}{3}$ is indeed interesting. Apparently we should assume that $p,q,Q\geq 0$ and $P+\sqrt{Q}\geq 0$.

II) As joshphysics essentially writes in his answer, the RT (1) is not a symplectomorphism, because the Poisson bracket is not preserved if $p\neq \frac{1}{4}$:

$$\tag{3} \{Q,P\} ~=~\frac{\{q,p\}}{2\sqrt{p}}~\neq~\{q,p\}~=~1. $$

III) As lionelbrits shows in his answer, the RT (1) do transform the Hamilton's eqs. into Kamilton's eqs, which according to Wikipedia (December 2013) is the defining property of a canonical transformation (CT). Goldstein, Landau and Lifshitz (Ref. 1 and 2) disagree with such a definition of CT. Ref. 1 and 2 state that form invariance is only a necessary but not a sufficient condition for being a canonical transformation (CT).

IV) Both Refs. 1 and 2 define a CT as satisfying

$$\tag{4} (p\dot{q}-H)-(P\dot{Q}-K)~=~\frac{dF}{dt},$$

or equivalently

$$ \tag{5} (p\mathrm{d}q-H\mathrm{d}t)-(P\mathrm{d}Q -K\mathrm{d}t) ~=~\mathrm{d}F,$$

for some function $F$. Or equivalently (ignoring possible topological obstructions),

$$ \tag{6} \mathrm{d}\left(p\mathrm{d}q-P\mathrm{d}Q +(K-H)\mathrm{d}t\right)~=~0. $$

For OP's example the condition (6) does not hold

$$ \tag{7} \mathrm{d}\left((p-\sqrt{p}+\sqrt{q})\mathrm{d}q +(\frac{p^3}{3}-\frac{p^2}{2})\mathrm{d}t\right) ~\neq~0. $$

So OP's example is not a CT according to Refs. 1 and 2.


  1. H. Goldstein, Classical Mechanics, Chapter 9. See text under eq. (9.11).

  2. L.D. Landau and E.M. Lifshitz, Mechanics, $\S45$. See text between eqs. (45.5-6).


The word restricted means that the transformation $(q,p)\longrightarrow (Q,P)$ has no explicit time dependence.

share|cite|improve this answer
+1: For further illuminating the issue and especially for using the term "Kamilton;" did you come up with that? – joshphysics Dec 7 '13 at 0:02
@joshphysics: Ha-Ha. No, Ref. 1 mentions Kamiltonian in a footnote on the page before eq. (9.11). – Qmechanic Dec 7 '13 at 0:08
Ha! I never noticed that! Very funny. – joshphysics Dec 7 '13 at 0:16
Shall we take refs 1. and 2. to be the canonical definition of a canonical transformation? – lionelbrits Dec 7 '13 at 0:33

See V.I.Arnold Mathematical Methods of Classical Mechanics chapter 44 E for definitions and proofs. Also check the footnotes up to p.241. In particular, Landau & Lifshitz conflates the two definitions of a canonical transformation and has several errors on the subject.

If we define a canonical transformation as the diffeomorphism on a symplectic manifold that preserves the symplectic structure, then the other definition follows, but the two are not equivalent.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.