# Canonical transformation problem

1. Dec 10, 2013

### mjordan2nd

1. The problem statement, all variables and given/known data

Let $Q^1 = (q^1)^2$, $Q^2 = q^1+q^2$, $P_{\alpha} = P_{\alpha}\left(q,p \right)$, $\alpha = 1,2$ be a CT in two freedoms. (a) Complete the transformation by finding the most general expression for the $P_{\alpha}$. (b) Find a particular choice for the $P_{\alpha}$ that will reduce the Hamiltonian

$$H = \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2$$

to

$$K = P_1^2 + P_2.$$

2. Relevant equations

3. The attempt at a solution

I have shown that

$$P_1 = \frac{1}{2q^1} \left( p_1 + \frac{\partial F}{\partial q^1} - p_2 - \frac{\partial F}{\partial q^2} \right),$$

$$P_2 = p_2 + \frac{\partial F}{\partial q^2}$$

is the most general canonical transformation for the momenta, where $F=F(q^1, q^2)$. This is consistent with the solution manual. For part b, however, the answer I get for an intermediate step is inconsistent with the solutions manual, and I don't understand why. Given that the transformation is canonical, all I need to do to find the transformed Hamiltonian K is find the inverse transformation and plug it in to the Hamiltonian H. The inverse transformation is

$$p_2 = P_2 - \frac{\partial F}{\partial q^2},$$
$$p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.$$

Plugging this into H, and renaming H to K since it's in terms of the transformed coordinates we have

$$K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.$$

Since we want K to be

$$K = P_1^2 + P_2,$$

this means

$$\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.$$
$$F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.$$

Plugging this into the general transformation I derived I find that

$$P_1 = \frac{1}{2q^1} \left(p_1-p_2-(q^1)^2 \right),$$
$$P_2 = (q^1+q^2)^2+p_2.$$

My equation for $P_2$ is consistent with the solutions manual, but my equation for $P_1$ is not. According to the solutions manual

$$P_1=\frac{p_1+p_2}{2q^1}.$$

So my question is, where did I go wrong. I have worked out the problem twice, and get the same answer for $P_1$.

Last edited: Dec 10, 2013
2. Dec 10, 2013

### TSny

I think you dropped some terms when you substituted for $\left( \frac{p_1 - p_2}{2q^1} \right)^2$ in H.

Note that the C here is possibly a function of $q^1$.

I might have made a mistake, but I get $$P_1=\frac{p_1-p_2}{2q^1}$$

3. Dec 10, 2013

### mjordan2nd

Wow, don't know how I managed to do that twice. I think I see my mistake now. Thank you.

Edit: I also get flipped minus signs from the book's answers

$$P_2=p_2-(q^1+q^2)^2$$

$$P_1=\frac{1}{2q^1}(p_1-p_2)$$

Last edited: Dec 10, 2013
4. Dec 10, 2013

### TSny

I get $$P_2=p_2+(q^1+q^2)^2$$ and $$P_1=\frac{1}{2q^1}(p_1-p_2)$$

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