Wednesday 26 August 2015

homework and exercises - Question about canonical transformation


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


$$H=\frac{1}{2}p^2,$$



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.




Answer



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.



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