Canonical transformation problem

[mjordan2nd]

Homework Statement

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

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

to

[tex] K = P_1^2 + P_2.[/tex]

Homework Equations

The Attempt at a Solution

I have shown that

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

[tex]P_2 = p_2 + \frac{\partial F}{\partial q^2}[/tex]

is the most general canonical transformation for the momenta, where [itex]F=F(q^1, q^2)[/itex]. 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

[tex] p_2 = P_2 - \frac{\partial F}{\partial q^2},[/tex]
[tex] p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.[/tex]

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

[tex] K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.[/tex]

Since we want K to be

[tex] K = P_1^2 + P_2,[/tex]

this means

[tex]\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.[/tex]
[tex]F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.[/tex]

Plugging this into the general transformation I derived I find that

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

My equation for [itex]P_2[/itex] is consistent with the solutions manual, but my equation for [itex]P_1[/itex] is not. According to the solutions manual

[tex]P_1=\frac{p_1+p_2}{2q^1}.[/tex]

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

 

[TSny]

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

The inverse transformation is

[tex] p_2 = P_2 - \frac{\partial F}{\partial q^2},[/tex]
[tex] p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.[/tex]

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

[tex] K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.[/tex]

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

[tex] K = P_1^2 + P_2,[/tex]

this means

[tex]\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.[/tex]
[tex]F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.[/tex]

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

According to the solutions manual

[tex]P_1=\frac{p_1+p_2}{2q^1}.[/tex]

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

 

[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

[tex]P_2=p_2-(q^1+q^2)^2[/tex]

[tex]P_1=\frac{1}{2q^1}(p_1-p_2)[/tex]

 

[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

[tex]P_2=p_2-(q^1+q^2)^2[/tex]

[tex]P_1=\frac{1}{2q^1}(p_1-p_2)[/tex]

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

 

[paranormal]

I am unsure what the correct answer is.

As a scientist, it is important to carefully check all calculations and assumptions made in the solution of a problem. It is possible that there may have been a mistake in the mathematical steps, or there may be different approaches to solving the problem that could lead to different answers. It may also be helpful to consult with other experts in the field or to refer to additional resources for guidance. Ultimately, it is important to understand the underlying principles and concepts involved in the problem, rather than just focusing on the final answer. This will help to ensure a deeper understanding and ability to apply the knowledge in future problems.