Uniqueness of canonical transformations

[neelakash]

The following question seems to be simple enough...Anyway, I hope if someone could confirm what I am thinking.

Is canonical transformation in mechanics unique? We know that given \ (q, p)\rightarrow\ (Q, P), \ [q,p] = [Q,P] = constant and Hamilton's equations of motion stay the same in the new co-ordinates.

My question is: given \ q\rightarrow Q in a canonical transoformation, is the map \ p\rightarrow P uniquely determined? Seems yes to me, but I do not find an off-hand argument in favour.

Can anyone tell how to derive this map?

-Regards,
Neel

 

[vanhees71]

This is a simplified special case, i.e.,
Q=Q(q).
Now we have
\{Q,P \}=Q'(q) \frac{\partial P}{\partial p} \stackrel{!}{=} 1.
This means that
\frac{\partial P}{\partial p}=-\frac{1}{Q'(q)}.
This means that
P(q,p)=-\frac{p}{Q'(q)}+\tilde{P}(q).
with an arbitrary function \tilde{P}(q) alone. Thus, the canonical transformation is determined up to this arbitrary function only.

 

[neelakash]

Thanks for the reply...I think I see your point; in such a case, apparently p\rightarrow P is a linear map...By the way, if it was intended, I could not understand the appearance of '!' and '-' sign.

 

[vanhees71]

Argh! The minus sign is simply wrong.

The exclamation mark over the equality sign indicates that this is a constraint to make the transformation a canonical one (i.e., a symplectomorphism on phase space). One can show that a transformation is canonical if and only if the canonical Poisson-bracket relations hold for the new variables. So the correct answer to your question is

Q=Q(q), \quad P(q,p)=+\frac{p}{Q'(q)}+\tilde{P}(q).

By the way, this is the special case of a "point transformation", which is the same as changing from a generalized coordinate q to an arbitrary new one, Q within the Lagrangian formulation of analytical mechanics.

The important point to realize is that the Hamilton formulation admits a larger group of transformations, namely the canonical transformations!

 

[neelakash]

I agree fully...The scope of canonical transformations is a larger than the so-called symmetry transformations...Thank you very much for the explanations...