# Canonical Transformation

1. Jun 10, 2012

### M. next

I have posted before this, an example in which I struggled through.
Now am gnna ask something more general, for me and for the students who suffer from studying a material alone.

If you were asked to prove that the time-independent transformation P=.. and Q=.. is canonical. And finding the generating function.
There are two methods as I know so far.
1) By applying pdq-PdQ=dF
2) By using $\partial$F/$\partial$q=p - $\partial$F/$\partial$Q=-P

(in accordance to what we are asked for: F(q,Q) F(q,P) ...)

My questions are:
In 1) What should we be aware of, Can we face a problem in concluding the F at the end?
In 2) What are the steps! One by one? Why do I see in some problems that after partial differentiation at the beginning they try to manipulate coordinates, instead of q, Q - instead of p, q or so on.. (am not being specific). Why? On what basis?

Do me this favor, please - Is canonical transformation this hard? Or is it steps that should be followed?

Best Regards,

2. Jun 10, 2012

### vanhees71

If you have the transformation given in explicit form

$$q=q(Q,P), \quad p=p(Q,P),$$

the most simple way to prove that this is a canonical transformation (local symplectomorphism) is to show that the Poisson brackets are the canonical ones, i.e.,

$$[q,p]=1,$$

where the Poisson bracket is defined by the partial derivatives wrt. $Q$ and $P$.

If you want to find the generating function in its original form, i.e., as a function

$$F=F(q,Q)$$

you just solve for

$$p=\frac{\partial F}{\partial q}, \quad P=-\frac{\partial F}{\partial Q}.$$

3. Jun 10, 2012

### M. next

It works everytime? And what concerning the methods I mentioned? You mentioned a new method I suppose [q,p]=1, right? Shed lights on my methods please.

Yes and what if I want in your general example, F(q,P)?

4. Jun 10, 2012

### vanhees71

Your methods 1) and 2) are fine, but as I said, if you have the transformation given explicitly, to check whether it's canonical you should check the integrability conditions in terms of the Poisson brackets.

Of course, you can write the generating function with any pair of old and new phase-space coordinates you like. The original form is that where you use $q$ and $Q$. If you want, e.g., $q$ and $P$, you make the appropriate Legendre transformation, i.e., you set

$$F(q,Q)=g(q,P)-Q P,$$

because then you get

$$\mathrm{d} q \partial_q F+\mathrm{d} Q \partial_Q F=\mathrm{d} q \partial_q g+(\partial_P g-Q)\mathrm{d P}-P \mathrm{d} Q.$$

Comparison of the left- and right-hand side of this equation yields\

$$p=\partial_q F=\partial_q g, \quad P=-\partial_Q F, \quad Q=\partial_P g.$$

Canonical transformations are not so difficult, but one has to get used to the concepts about them. A good source is Landau/Lifschitz Vol. 1.

5. Jun 10, 2012

### M. next

Thanks, I have the book, it is a very good book, but kind of condensed. Thanks again.