Canonical Transformation / Poisson Brackets

[Zag]

Question:

(A) Show that the following transformation is a canonical transformation:
[itex]Q = p + aq[/itex]
[itex]P = (p - aq)/(2a)[/itex]

(B) Find a generating functions for this transformation.


Attempt of Solution:
Alright, so this seems to be a very straight forward problem. Transformations are canonical if their Poisson Brackets satisfy:

[itex][P_{\alpha},Q_{\beta}]_{PB} = -\delta_{\alpha\beta}[/itex]
[itex][Q_{\alpha},Q_{\beta}]_{PB} = 0[/itex] and
[itex][P_{\alpha},P_{\beta}]_{PB} = 0[/itex]

For the given problem the system is one-dimensional, so the Poisson Brackets' identities reduce to a single relevant expression:

[itex][P,Q]_{PB} = -1[/itex]

We can now calculate this Poisson Bracket explicitly since we were given the transformation connecting the variables [itex]Q[/itex], [itex]P[/itex], [itex]q[/itex] and [itex]p[/itex]. The calculation is shown below:

[itex][P,Q]_{PB} = \frac{\partial P}{\partial q}\frac{\partial Q}{\partial p} - \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} = -\frac{1}{2}a - a\frac{1}{2a}[/itex]

[itex]\therefore[/itex] [itex][P,Q]_{PB} = \frac{a}{2} - \frac{1}{2}[/itex]

Comments and Issues:
It is clear from this calculation that the transformation is not canonical for a general [itex]a[/itex], since the result of the Poisson Bracket is not [itex]-1[/itex]. In fact, the only value of [itex]a[/itex] for which the transformation would be canonical is [itex]a = 1[/itex].

Now, the problem is that the question doesn't mention anything about finding the value of [itex]a[/itex] for which the transformation is canonical, it simply asks the student to show that the transformation is canonical. So I am not sure whether there is something wrong with my solution or the problem statement is formulated in an ambiguous.

The second part of the problem could also be easily solved if one assumes [itex]a = 1[/itex], otherwise - according to my solution - it would be impossible to find a generating function for a general [itex]a[/itex] since in that case the transformation would not even canonical to begin with.

Any thoughts or comments on this issue would be greatly appreciated!

Thank you very much guys!
Zag

 

[Orodruin]

##\frac{\partial Q}{\partial p}= 1 \neq a##

The transformation is canonical.

 

[Zag]

Nice! Thanks Orodruin, I indeed mixed my partial derivative results there. It works now! :)

 

[Juan Carlos]

it's simple derivation's mistake

 

[paranormal]

ros.

Hi Zagros, thank you for your attempt at solving this problem. Your solution is correct in terms of showing that the transformation is not canonical for a general a. However, as you mentioned, the question does not specify a value for a, so it is possible that there may be a typo or ambiguity in the problem statement. It would be best to clarify this with your instructor or the person who assigned you this problem.

Regarding the second part, you are correct that finding a generating function for a general a would be impossible since the transformation is not canonical in that case. So it is possible that the question meant to ask for a generating function for a = 1.

In any case, it is important to clarify the problem statement to ensure that you are solving the correct problem. Good job on your solution!