Help with Canonical Poisson Brackets & EM Field

[Matthew_]

We were introduced the lagrangian for a particle moving in an eletromagnetic field (for context, this was a brief introduction before dealing with Zeeman effect) as $$\mathcal{L}=\dfrac{m}{2}(\dot{x}^2_1+\dot{x}^2_2+\dot{x}^2_3)-q\varphi+\dfrac{q}{c}\vec{A}\cdot\dot{\vec{x}}.$$ A "velocity-dependent generalized potential" appears. For a constant magnetic field oriented along the ##x_3## axis:$$\vec{A}=\frac{B}{2}(x_1\hat{u}_2-x_2\hat{u}_1).$$
Now, the conjugated momenta of the generalized coordinates are: $$p_1=m\dot{x}_1-\dfrac{qB}{2c}x_2,$$ $$p_2=m\dot{x}_2+\dfrac{qB}{2c}x_1,$$ $$p_3=m\dot{x}_3.$$
It was claimed that canonical poisson brackets hold, namely ##\left\{p_i,p_j\right\}=0##. I have no idea why this is the case tho, since the derivation of the lagrangian with respect of the generalized velocities gives a clear dependence between the coordinates and the adjoint momenta. Evaluating ##\left\{p_1,p_2\right\}## I think I should get something like (summation over j is omitted): $$\left\{p_1,p_2\right\}=\frac{\partial p_1}{\partial x_j}\frac{\partial p_2}{\partial p_j}-\frac{\partial p_1}{\partial p_j}\frac{\partial p_2}{\partial x_j}=\frac{\partial p_1}{\partial x_2}-\frac{\partial p_2}{\partial x_1}=-\frac{qB}{c}\neq 0.$$ Is there some reason why this does not work?

 

[vanhees71]

The canonical Poisson brackets hold by definition, because for phase-space functions ##A## and ##B## by definition
$$\{A,B \}=\frac{\partial A}{\partial q_j} \frac{\partial B}{\partial p_j} - \frac{\partial B}{\partial q_j} \frac{\partial A}{\partial p_j}.$$
To calculate Poisson brackets you have to express all quantities with the ##q_j## and ##p_j## as independent variables, i.e., you have to eliminate the ##\dot{q}_j## using the ##p_j## and ##q_j## first. So it's the other way, i.e., now the Poisson brackets between the velocities are non-zero now. You have
$$\dot{\vec{x}}=\frac{1}{m}(\vec{p}-q \vec{A}/c),$$
i.e.,
$$[\dot{x}_j,\dot{x}_k]=\frac{1}{m^2} \left [-\frac{q}{c} \frac{\partial A_j}{\partial x_i} \delta_{ki} +\delta_{ij} \frac{q}{c} \frac{\partial A_k}{\partial x_i} \right] = \frac{q}{mc} \left (\frac{\partial A_k}{\partial x_j}-\frac{\partial A_j}{\partial x_i} \right) = \frac{q}{mc} \epsilon_{jki} B_i,$$
where I've used that ##\vec{B}=\vec{\nabla} \times \vec{A}##.

The Hamiltonian formulation is derived by first calculating the Hamiltonian as a function of the coordinates and canonical momenta,
$$H=\vec{x} \cdot \vec{p}-L=\frac{m}{2} \dot{\vec{x}}^2+q \varphi=\frac{1}{2m} (\vec{p}-q/c \vec{A})^2+q \varphi.$$
Then the equations of motion are the Hamilton canonical equations,
$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}} = \{\vec{x},H \}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}}=\{\vec{p},H \}.$$
Note that the canonical momenta ##\vec{p}## are not the mechanical momenta, ##\vec{\pi}=m \dot{\vec{x}}=\vec{p}-q \vec{A}/c##.

 

[Matthew_]

The canonical Poisson brackets hold by definition, because for phase-space functions ##A## and ##B## by definition
$$\{A,B \}=\frac{\partial A}{\partial q_j} \frac{\partial B}{\partial p_j} - \frac{\partial B}{\partial q_j} \frac{\partial A}{\partial p_j}.$$
To calculate Poisson brackets you have to express all quantities with the ##q_j## and ##p_j## as independent variables, i.e., you have to eliminate the ##\dot{q}_j## using the ##p_j## and ##q_j## first. So it's the other way, i.e., now the Poisson brackets between the velocities are non-zero now. You have
$$\dot{\vec{x}}=\frac{1}{m}(\vec{p}-q \vec{A}/c),$$
i.e.,
$$[\dot{x}_j,\dot{x}_k]=\frac{1}{m^2} \left [-\frac{q}{c} \frac{\partial A_j}{\partial x_i} \delta_{ki} +\delta_{ij} \frac{q}{c} \frac{\partial A_k}{\partial x_i} \right] = \frac{q}{mc} \left (\frac{\partial A_k}{\partial x_j}-\frac{\partial A_j}{\partial x_i} \right) = \frac{q}{mc} \epsilon_{jki} B_i,$$
where I've used that ##\vec{B}=\vec{\nabla} \times \vec{A}##.

The Hamiltonian formulation is derived by first calculating the Hamiltonian as a function of the coordinates and canonical momenta,
$$H=\vec{x} \cdot \vec{p}-L=\frac{m}{2} \dot{\vec{x}}^2+q \varphi=\frac{1}{2m} (\vec{p}-q/c \vec{A})^2+q \varphi.$$
Then the equations of motion are the Hamilton canonical equations,
$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}} = \{\vec{x},H \}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}}=\{\vec{p},H \}.$$
Note that the canonical momenta ##\vec{p}## are not the mechanical momenta, ##\vec{\pi}=m \dot{\vec{x}}=\vec{p}-q \vec{A}/c##.

Thank you, this was illuminating