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I'm self studying through a book that has the following question. The book gives the answer, but I'm trying to understand why:

Under what condition is the following transformation NOT canonical? $$Q = q + ip, P = q - ip.$$

The book says this transformation is not canonical when $H = K$, where $H$ is the hamiltonian under the original coordinates, and $K$ is the hamiltonian under the transformed coordinates. But I'm having trouble seeing why this is true. It didn't seem to help to work through the Poisson brackets with the different coordinates. Can anyone shed any light on this, or on the concept of complex canonical transformations in general?

EDIT: Also, the problem is 5.3 from Hamill's Student's Guide to Lagranigans and Hamiltonians.

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    $\begingroup$ Don't we have $\{Q,P\} = -2i$, which would mean that the transformation is not canonical? Also, I don't see how the canonical-ness of a transformation depends on the Hamiltonian. It either is, or isn't. $\endgroup$ – Javier Aug 14 at 22:04
  • $\begingroup$ Well, maybe someone who knows more than me will come along and correct me, but it seems like the transformation is simply not canonical. The book is pretty cryptic, though. $\endgroup$ – Javier Aug 16 at 15:12
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Patrick Hamill, actually, borrowed that problem from Goldstein (see- $3^{rd}$ ed. Chapter 9, derivations 1), and unfortunately, misphrased it. So I'm just quoting the problem from Goldstein, and my interpretation of it, and its solution for your part of that problem with a personal advice.

One of the attempts at combining the two sets of Hamilton’s equations into one tries to take q and p as forming a complex quantity. Show directly from Hamilton’s equations of motion that for a system of one degree of freedom the transformation $$Q=q+ip,$$ $$P=Q^*$$ is not canonical if the Hamiltonian is left unaltered. Can you find another set of coordinates $Q'$, $P'$ that are related to $Q$,$P$ by a change of scale only, are they canonical?

My interpretation of the unaltered Hamiltonian: Given a Hamiltonian, and if you perform this transformation the form of the Hamiltonian will remain the same that is $H(Q,P)=H(q,p)$. As the transformation has no time dependent term, so no PD of generating function wrt time is taken into account.
Now applying inverse transformation $$q={1 \over 2}(Q+P),$$ $$p={1 \over 2i}(Q-P)$$ And we expect $$\dot Q= \dot q +i\dot p.$$ Now from Hamilton's equation of motion $$\dot Q= {\partial H \over \partial P}={\partial H \over \partial q}{\partial q \over \partial P}+{\partial H \over \partial p}{\partial p \over \partial P}={- \dot p . {1 \over 2}}+ {{-1 \over 2i} \dot q}={{-1 \over 2} (\dot p-i \dot q)}$$ which is obviously not what we expected and this proves the noncanonicity. The condition part of your question plays its role when we consider the second part of Goldstein's problem whose solution is given in Homer Reed.

Please don't use Hamill for self study. It's full of error. If there's anything useful in that book, that's its bibliography. Choose a mainstream textbook mentioned there.

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  • $\begingroup$ Thank you so much. That was incredibly helpful. $\endgroup$ – TKT Oct 11 at 5:01

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