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Hamiltonian conjugate dynamic variables

  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a charge ##q##, with mass ##m##, moving in the ##x-y## plane under the influence of a uniform magnetic field ##\vec{B}=B\hat{z}##. Show that the Hamiltonian $$ H = \frac{(\vec{p}-q\vec{A})^2}{2m}$$ with $$\vec{A} = \frac{1}{2}(\vec{B}\times\vec{r})$$ reduces to $$ H(x,y,p_x,p_y) = \frac{(p_x+\frac{1}{2}qBy)^2}{2m} + \frac{(p_y-\frac{1}{2}qBx)^2}{2m}$$
    Demonstrate
    $$ Q = \frac{(p_x+\frac{1}{2}qBy)}{qB} \qquad \qquad P = (p_y-\frac{1}{2}qBx) $$
    are conjugate dynamic variables, given ##x, p_x, y, p_y## are, then express $$H(Q,P)$$ in terms of ##m## and the cyclotron frequency, ##\omega \frac{qB}{m}##

    Show next that
    $$ P' = \frac{(p_x-\frac{1}{2}qBy)}{qB} \qquad \qquad Q' = (p_y+\frac{1}{2}qBx) $$
    Are yet another, linearly-independent, conjugate pair whose brackets with ##Q,P## necessarily vanish, i.e.
    $$ [Q,Q'] = [Q,P'] = [P,Q'] = [P,P'] = 0 $$
    Argue from the foregoing that ##Q',P'## must be constants of the motion
    2. Relevant equations
    Most are listed in problem statement. Definition of poisson bracket (PB): $$ [Q,P] = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} $$
    Fundamental PBs: ## [q_i,q_k] = [p_i,p_k] = 0, \ \ [q_i,p_k] = \delta_{ik}##

    3. The attempt at a solution
    My main question is, how exactly do I show that ##Q,P## are conjugate dynamical variables? Is this just by evaluating the PBs ##[Q_i, Q_k], [P_i, P_k], \text{ and } [Q_i,P_k]## and proving they preserve the fundamental PBs? So far, I have found
    $$ [Q_i, Q_k]_{q,p} = 0 $$
    And
    $$ [P_i,P_k]_{q,p} = 0$$
    But for ##[Q_i,P_k]_{q,p}## I find
    $$ [Q_i,P_k]_{q,p} = \frac{1}{2}+\frac{1}{2}qB \neq 1 (or \neq \delta_{ik})$$
    So is this not what I need to be doing or am I just evaluating it wrong?

    And as for expressing the Hamiltonian as functions of ##Q## and ##P##, what do I need to do? My prof kind of worked out finding ##H(q(Q,P), p(Q,P))## for a harmonic oscillator by seeking a transformation of the form ##H(q(Q,P),p(Q,P)) = \frac{f^2(P)}{2m}##, but I'm having trouble figureing out how to do it in the reverse, like ##H(Q(q,p), P(q,p))## for this problem.
     
  2. jcsd
  3. Feb 13, 2017 #2

    jambaugh

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    I'm not clear why, given the problem statement, you are using subscripts on the proposed conjugate variables of P and Q. There are just the two as I read the problem.
    I get the correct Poisson bracket relations on [Q,P] so check your work.
     
  4. Feb 13, 2017 #3
    I was just using the indecies as good practice so i remember to use all indecies of Q or P later if they need it. But for [Q,P] I find
    $$[Q,P] = \bigg[\frac{\partial Q}{\partial x}\frac{\partial P}{\partial p_x} - \frac{\partial Q}{\partial p_x}\frac{\partial P}{\partial x}\bigg] + \bigg[ \frac{\partial Q}{\partial y}\frac{\partial P}{\partial p_y} - \frac{\partial Q}{\partial p_y}\frac{\partial P}{\partial y}\bigg]$$
    Where ##\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial p_x} = \frac{\partial Q}{\partial p_y} = 0## which make the PB
    $$[Q,P] = -\frac{\partial Q}{\partial p_x}\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\frac{\partial P}{\partial p_y} = -(\frac{1}{qB})(-\frac{1}{2}qB) + (\frac{1}{2}qB)(1) = \frac{1}{2}+ \frac{1}{2}qB$$ so where am I going wrong?
     
  5. Feb 13, 2017 #4

    jambaugh

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    [itex] \partial Q / \partial y = \tfrac{1}{2}[/itex] , [itex] \partial P/\partial p_y = 1[/itex] So total bracket is [itex]\tfrac{1}{2} + \tfrac{1}{2} = 1[/itex] . Or did I mess up?
     
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