# Hamiltonian conjugate dynamic variables

Tags:
1. Feb 13, 2017

### thecourtholio

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. Feb 13, 2017

### jambaugh

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.

3. Feb 13, 2017

### thecourtholio

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?

4. Feb 13, 2017

### jambaugh

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted