Where Did I Go Wrong in My Canonical Transformation Problem?

[mjordan2nd]

Homework Statement

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.

Homework Equations

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 into 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.

 

[TSny]

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

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.

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

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.

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

According to the solutions manual

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

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

 

[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)

 

[TSny]

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)

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