# Canonical Transformation

1. Sep 18, 2006

### Emanuel84

Hi, I tried to solve this problem, but I was unsuccessful :grumpy:

Here is the problem:

Given the transformation:

$\left \{ \begin{array}{l} Q = p^\gamma \cos(\beta q) \\ P = p^\alpha \sin(\beta q) \end{array} \right.$

a) Determine the values of the constants $\alpha$, $\beta$ and $\gamma$ for which this transformation is canonical.

b) In correspondence of these values, find a generating function of the transformation.

How can I solve this problem? Firstly, I used the Poisson bracket condition for canonicity:

$[Q,P]_{q,p} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}$.

Afterwards I supposed:

$pdq-PdQ$

to be an exact differential.

Still, I didn't manage to find $\alpha$, $\beta$ and $\gamma$, as if I missed a condition...

2. Sep 18, 2006

### Epicurus

current conservation. This transformation should leave momentum and energy invariant , which can be found via noethers theorem. You have constants of motion for which the poisson bracket will be zero. Put in these constraints and you should have the required equations.

3. Sep 19, 2006

### Emanuel84

err...how can I deduce momentum and energy?

I didn't study Noether's theorem yet, so I don't think this problem should use it...

4. Sep 19, 2006

### xman

Remember in order for a transformation to be canonical requires the Poisson bracket to be unity, i.e. [A,B] = 1. Use this to determine the constants, you'll find it is not that bad. For the second part use the definitions of the relationships between canonical variables to determine the generating function.

5. Sep 19, 2006

### ptabor

Correction:
{q_k, p_j} = kronecker delta (j, k) = 0 or 1 depending on j, k
[ 0 if j != k, 1 if j = k ]

From Shankar, Chapter 2, page 95

6. Sep 19, 2006

### Emanuel84

I know that:

$[Q_j, P_k] = \delta_{jk}$

so, in this case, since $Q = Q_1$ and $P = P_1$ , we have:

$[Q_1, P_1] = 1 = [Q, P]$

These are the calculations I made:

$[Q,P]_{q,p} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} = 1$

$\left \{ \begin{array}{l} Q = p^\gamma \cos(\beta q) \\ P = p^\alpha \sin(\beta q) \end{array} \right.$

$\frac{\partial Q}{\partial q} = -\beta p^\gamma \sin(\beta q)$

$\frac{\partial P}{\partial p} = \alpha p^{\alpha - 1} \sin(\beta q)$

$\frac{\partial Q}{\partial p} = \gamma p^{\gamma -1} \cos(\beta q)$

$\frac{\partial P}{\partial q} = \beta p^\alpha \cos(\beta q)$

Rearranging terms, Poisson bracket condition leads to:

$-p^{\alpha + \gamma - 1} \beta \left[ \alpha \sin^2(\beta q) - \gamma \cos^2(\beta q) \right] = -p^{\alpha + \gamma - 1} \beta \left[ -\gamma + \left( \alpha + \gamma \right) \sin^2(\beta q) \right] = 1$

I don't think this is a sufficient condition in order to determine the constants.

Last edited: Sep 19, 2006
7. Sep 19, 2006

### Emanuel84

On the other hand, one can think to use the definition of canonical trasformation:

A time-independent transformation $Q = Q(q,p)$ , and $P = P(q,p)$ is called canonical if and only if there exists a function $F(q,p)$ such that:

$dF(q,p) = p_i dq_i - P_i(q,p) dQ_i(q,p)$

In other words, since I have only $Q_1 = Q$ and $P_1 = P$ the former condition becomes:

$pdq - PdQ = pdq - P\left( \frac{\partial Q}{\partial q}dq + \frac{\partial Q}{\partial p}dp \right) = \left(p - P\frac{\partial Q}{\partial q}\right) dq + \left(-P \frac{\partial Q}{\partial p}\right) dp = dF$

Substituting the values of $P$ , $\frac{\partial Q}{\partial q}$ and $\frac{\partial Q}{\partial p}$:

$\left[ p + p^{\alpha + \gamma} \beta \sin^2(\beta q) \right] dq + \left[ -\frac{1}{2} p^{\alpha + \gamma - 1} \gamma \sin(2 \beta q) \right] dp$

has to be an exact differential.

Remembering that a differential form:

$dF = A(q,p)dq + B(q,p)dp$

is exact when:

$\frac{\partial A}{\partial p} = \frac{\partial B}{\partial q}$

I obtain another condition:

$1 + p^{\alpha + \gamma - 1} \beta (\alpha + \gamma) \sin^2(\beta q) = -p^{\alpha + \gamma - 1} \beta \gamma \cos(2\beta q) = -p^{\alpha + \gamma - 1} \beta \gamma \left[1 - 2\sin^2(\beta q)\right]$

After some calculations I have:

$- p^{\alpha + \gamma - 1} \beta \left[ \gamma + \left( \alpha - \gamma \right) \sin^2(\beta q) \right] = 1$

Comparing this equation with the last one of the previous post:

$-p^{\alpha + \gamma - 1} \beta \left[ -\gamma + \left( \alpha + \gamma \right) \sin^2(\beta q) \right] = - p^{\alpha + \gamma - 1} \beta \left[ \gamma + \left( \alpha - \gamma \right) \sin^2(\beta q) \right]$

So:

$-\gamma + \left( \alpha + \gamma \right) \sin^2(\beta q) = \gamma + \left( \alpha - \gamma \right) \sin^2(\beta q)$

Finally:

$\gamma \left[ \sin^2(\beta q) - 1 \right] = 0 \Longleftrightarrow \gamma = 0$

Is this correct?

At this point, I simply don't know what to do...

Last edited: Sep 19, 2006
8. Sep 19, 2006

### Physics Monkey

Hi Emanuel,

Don't give up yet, you've already done the hard part. The easiest way to attack this problem is to impose the condition that $$[Q,P]_{PB} = 1 .$$ You have already calculated this quantity in post 6, but you have an error in your final equation. It should be a plus, not a minus, in front of the $$\gamma \cos^2{\beta q}$$ term. The corrected equation is more than enough to determine all three parameters because it must be true for every q and p! In other words, the LHS is a function of q and p, but the RHS is totally independent of q and p, so clearly this is only going to happen if you choose very special values of$$\alpha$$, $$\beta$$, and $$\gamma$$.

Hope this helps.

Last edited: Sep 19, 2006
9. Sep 20, 2006

### Emanuel84

Thank you, I solved the problem...it was a banal error sign, as you said. :tongue: