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Canonical transformation problem

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data

    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]


    2. Relevant equations



    3. 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 in to 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].
     
    Last edited: Dec 10, 2013
  2. jcsd
  3. Dec 10, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    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 [tex]P_1=\frac{p_1-p_2}{2q^1}[/tex]
     
  4. Dec 10, 2013 #3
    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]
     
    Last edited: Dec 10, 2013
  5. Dec 10, 2013 #4

    TSny

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    Homework Helper
    Gold Member

    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]
     
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