Poisson brackets and magnetic field I'm a maths student trying to teach myself some physics so sorry if I'm missing something simple here. I think the main problem is lack of experience with the Levi-Cevita symbol.
We have a particle in a magnetic field $\mathbf{B}=\nabla\times\mathbf{A}$, described by the Hamiltonian (taking $c=1$):
\begin{equation*}
 H=\frac{1}{2m}(p-eA(\mathbf{r}))^2=\frac{m}{2}\mathbf{\dot{r}}^2
 \end{equation*}
I am told that the Poisson bracket structure reads $\{m\dot{r}_a,m\dot{r}_b\}=e\epsilon_{abc}B_c$, and have tried to prove this as follows:
\begin{equation*}
\left\lbrace m \, \dot{r}_a, m \, \dot{r}_b \right\rbrace = \left\lbrace p_a - e \, A_a(\mathbf{r}), p_b - e \, A_b(\mathbf{r}) \right\rbrace \\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= \left\lbrace p_a, p_b \right\rbrace - e \, \left\lbrace p_a, A_b(\mathbf{r})\right\rbrace + e \, \left\lbrace p_b, A_a(\mathbf{r}) \right\rbrace + e^2 \, \left\lbrace A_a(\mathbf{r}), A_b(\mathbf{r}) \right\rbrace
\end{equation*}
Now the first and last terms in here are zero, and so this simplifies to:
\begin{equation*}
\left\lbrace m \, \dot{r}_a, m \, \dot{r}_b \right\rbrace = e\frac{\partial{A_b(\mathbf{r})}}{\partial{r_a}} - e\frac{\partial{A_a(\mathbf{r})}}{\partial{r_b}}
\end{equation*}
using the fact that $\left\lbrace p_a, f(\mathbf{r}) \right\rbrace = -\frac{\partial f(\mathbf{r})}{\partial r_a}$. I'm not too sure where to go from here, I'm guessing there is some sort of manipulation with the Levi-Civita symbol needed? Thanks in advance.
 A: I'm going to use Einstein summation notation throughout.
You're almost there. You just need to use
$$
{\bf B} = \nabla \times {\bf A}
$$
or, equivalently,
$$
\begin{eqnarray}
B_k &=& \frac{\partial}{\partial r_i} A_j \epsilon_{i j k} \\
&=& \frac{1}{2}\left(\frac{\partial}{\partial r_i} A_j \epsilon_{i j k} + \frac{\partial}{\partial r_j} A_i \epsilon_{j i k}\right) \\
&=& \frac{1}{2}\left(\frac{\partial}{\partial r_i} A_j - \frac{\partial}{\partial r_j} A_i \right) \epsilon_{i j k}\\
\end{eqnarray}
$$
There I used the antisymmetry of $\epsilon_{i j k}$ under exchange of any two indices.
Now using the first equation here,
$$
\begin{eqnarray}
B_k \epsilon_{k r s} &=& \frac{1}{2}\left(\frac{\partial}{\partial r_i} A_j - \frac{\partial}{\partial r_j} A_i \right) \epsilon_{k i j} \epsilon_{k r s} \\
&=& \frac{1}{2}\left(\frac{\partial}{\partial r_i} A_j - \frac{\partial}{\partial r_j} A_i \right) \left(\delta_{i r} \delta_{j s} - \delta_{i s} \delta_{j r}\right) \\
&=& \frac{1}{2}\left(\frac{\partial}{\partial r_r} A_s - \frac{\partial}{\partial r_s} A_r \right) - \frac{1}{2}\left(\frac{\partial}{\partial r_s} A_r - \frac{\partial}{\partial r_r} A_s \right) \\
&=& \frac{\partial}{\partial r_r} A_s - \frac{\partial}{\partial r_s} A_r
\end{eqnarray}
$$
