# How to check if a generating function produces an identity transformation without substituting the CT equations in the Hamiltonian?

In chapter 9, Goldstein ($$3^{rd}$$ ed.) includes a discussion and a few "trivial special cases" of Canonical Transformation which keeps the form of the Hamiltonian unchanged and named it Identity Transformation. Recently I've encountered another generating function- $$F_1(q,Q)= {1\over2}m\omega \cot\theta (Q^2+q^2)-{m \omega qQ \over \sin \theta} ,$$ which also generates an identity transformation for the following Hamiltonian- $$H={p^2 \over 2m}+{1 \over 2}m \omega^2 q^2.$$ I must add that $$F_1$$ merely generates a "rotation+some scaling" in phase space- $$Q=q\cos\theta-{p \over m\omega}\sin\theta,$$ $$P=m\omega q\sin\theta+p\cos\theta.$$I should also mention that $$\theta$$ is not a function of time. Anyway, one has to go through a lengthy calculation (such as deriving the CT $$eq^n$$ from GF+ expressing the old variable in terms of the new + substituting it in the old Hamiltonian+ deriving the new H) before discovering that it is just an identity transformation. And this motivated me to ask the following questions-

1. Is there a way to know beforehand whether a generating function produces an identity transformation or not?
2. Identity transformation adds no benefits to the problem as it leaves the form of the Hamiltonian unchanged and produces the same $$eq^n$$s of motion. So why do we study them?

First, the scaling you mention amounts to failure to clean up your variables, by defining $$\tilde p\equiv \frac{p}{m\omega},$$ likewise for $$\tilde P$$, and $$\tilde H \equiv \frac{H}{m\omega^2}= \frac{1}{2} (\tilde p^2+ q^2).$$

It is then apparent that the transformation $$Q=q \cos \theta -\tilde p \sin\theta\\ \tilde P= \tilde p \cos\theta + q \sin\theta, \tag{0}$$ the rotation you observed, is a continuous symmetry of your $$\tilde H$$, a rotational scalar.

It is also manifestly a canonical transformation without even consideration of generating functions, wont to confuse you, since it preserves Poisson Brackets, essentially by inspection, $$\{ Q,\tilde P \}=1,$$ (of course, $$\{ Q,Q \}=0, \quad \{ \tilde P,\tilde P \}=0$$.)

But, in your particular case, the generator which produces your canonical transformation by Lie Poisson commutation, $$Q= e^{\{ G, \bullet \}} ~~ q = q +\{ G, q \} + \frac{1}{2} \{ G , \{ G, q \} \} +\frac{1}{3!}\{G,\{ G, \{ G, q\}\}\}+ ... \\ \tilde P = e^{\{ G, \bullet \}} ~~ \tilde p = \tilde p +\{ G, \tilde p \} + \frac{1}{2} \{ G , \{ G, \tilde p \} \} +... \tag{1}$$ happens to be $$G= \theta \tilde H,$$ so, of course, it Poisson-commutes with itself, $$\{ G, \tilde H \}=0$$. (Do check that, since $$\{G,q\}=-\theta \tilde p$$, etc, the above series sum to the exact finite rotation (0).) Consequently, the transformation it generates leaves the Hamiltonian invariant. Your criterion is Poisson-commutation with the Hamiltonian.

• But this is exactly how this Hamiltonian, through Lie iteration of Hamilton's equations, acts to generate motion, here a mere phase-space rotation. (Coincidentally, here, you are replicating the general fact that motion is a canonical transformation, crucially important in QM.)

To sum up, in this language, given a generator $$G(q,\tilde p)$$ Poisson-commuting with the Hamiltonian, it will be a symmetry thereof, so the (provably canonical, below) transformations it generates will leave the Hamiltonian invariant.

I am not sure I understand your second point, since I cannot fathom your "no benefits to the problem". Canonical changes of variables allow you to simplify problems to the point where their solutions are virtually self-evident. Have you gotten to the action-angle variables part of your course? The fact that motion is a canonical transformation and preserves the local dynamics of the theory in question is also central.

• Proof of the generic finite transformation (1) being canonical, as per comment. (Geeky)

For generic $$G=\theta g(q,p)$$, not just the above, is the following quantity =1? $$Y(\theta)= \{Q,P \} \equiv \{ e^{\{\theta g , \bullet \}} ~q ~,~ e^{\{ \theta g, \bullet \}} ~p \} \\ \equiv 1+\theta y_1 +\theta^2 y_2+\theta^3 y_3+ ...$$

Consider, for any $$\theta$$-independent function $$f(q,p)$$, $$\frac{\partial (e^{\{ \theta g, \bullet \}}~ f)} {\partial \theta}= \{ g ~,~ e^{\{ \theta g, \bullet \}} ~f \},$$ implying, by the Jacobi identity, $$\frac{\partial Y(\theta) } {\partial \theta}= \{\{ g,Q\},P\}+ \{Q,\{g,P\}\}=\{g,Y(\theta)\}.$$

So, then, $$y_{n+1}=\{ g,y_n\}$$, constraining all higher coefficients to zero, since $$y_1=\{g,1\}=0$$, hence $$Y(\theta)=1$$, independent of $$\theta$$.

In the same breath, you may prove the all-orders converse Noether's theorem, $$\{g,H\}=0 \Longrightarrow H(Q,P)=H(q,p)$$, paradoxically easier to prove in QM (!), but available in Arnold's classic text. Here, I will just all but remind you the lowest-order result is evident by inspection, $$H(Q,P)=H(q,p)+ \theta ~ \left (-\frac{\partial H}{\partial q} \frac{\partial g}{\partial p}+\frac{\partial H}{\partial p}\frac{\partial g}{\partial q} \right )+ O(\theta^2)\\ = H(q,p)+ \theta ~ \{ g,H\} + O(\theta^2)=H(q,p)+ O(\theta^2).$$ The $$O(\theta^2)$$ vanishes as well, since, by $$\{Q,P\}=1$$, it does not matter which basis our PBs are in; and hence, likewise and by the above proof, $$\partial H(Q,P)/\partial \theta= ... = \{g,H\}=0$$, to all orders in $$\theta$$.

• I believe the OP meant by "no benefits to the problem" that as the form the Hamiltonian is same by Identity CT, solving Hamiltonian EOM involves the same level of difficulty. So, *no added benefit to the problem" – Mockingbird Aug 30 at 6:19
• "To sum up, in this language, given a generator G(q,p~) Poisson-commuting with the Hamiltonian, it will be a symmetry thereof, and the (easily provably canonical) transformations it generates will leave the Hamiltonian invariant." Can you clarify this part? And, if possible, suggest me study material on this topic of symmetry and invariance of the Hamiltonian. TIA – Mockingbird Aug 30 at 6:51
• @Mockingbird Well, in this system, the generator is the Energy, so it does not constrict the phase-space hypersurface of motion more than the Hamiltonian: a mere circle. It reminds you the phase is arbitrary and can serve as the origin of time, or the orientation of phase-space axes. In less trivial problems, the additional constraints provided by such "constants of the motion" simplify the problem by excluding inaccessible regions of phase space. – Cosmas Zachos Aug 30 at 11:32
• @Mockingbird Long story. Marsden & Ratiu are quite helpful, plus the dozens of questions here and here on the converse Noether thm. To see the finite tfmation is canonical, see, e.g., here, app A. – Cosmas Zachos Aug 30 at 11:40
• Can you tell me the particular part of the book where the symmetry arguement is covered? – Mockingbird Aug 31 at 1:50