1
$\begingroup$

I want to prove that the Poisson bracket from Hamiltonian mechanics satisfies the Jacobi identity and I want to do so using the matrix $$(J^{ij})=\begin{pmatrix}0 & -I_2 \\ I_2 & 0\end{pmatrix},$$ where $I_2$ is the 2$\times$2 identity matrix. It is obvious to see that $J^{ij}=-J^{ji}$. The matrix $J$ can be used to write Hamilton's equations for position and conjugate momentum compactly as $$\dot{x}=J\cdot\nabla H;\quad x:=(p_1,\cdots,p_n,q_1,\cdots,q_n).$$

For any two observables $A,B$, the Poisson bracket may be written as $$\{A,B\}=J^{ij}\partial_iA\partial_jB.$$

Now I want to use this to prove the Jacobi identity $$\{A,\{B,C\}\}=-\{B,\{C,A\}\}-\{C,\{A,B\}\}.$$

Using the formula above for writing the Poisson bracket in terms of $J$, we get: $$\{A,\{B,C\}\}=J^{ij}J^{kl}\partial_iA\partial_j(\partial_kB\partial_lC)=J^{ij}J^{kl}(\partial_iA\partial^2_{jk}B\partial_lC+\partial_iA\partial_kB\partial^2_{jl}C).$$ Now we use: $$\partial_iA\partial^2_{jk}B=\partial_k(\partial_iA\partial_{j}B)-\partial^2_{ik}A\partial_jB,\\ \partial_iA\partial^2_{jl}C=\partial_l(\partial_iA\partial_{j}C)-\partial^2_{il}A\partial_jC.$$ Substituting this into our result above leaves us with: $$J^{ij}J^{kl}\left(\partial_k(\partial_iA\partial_{j}B)\partial_lC +\partial_l(\partial_iA\partial_{j}C)\partial_kB\right) -J^{ij}J^{kl}\left(\partial^2_{ik}A\partial_jB\partial_lC +\partial^2_{il}A\partial_kB\partial_jC\right).$$ The first term is easily transformed into: $$J^{kl}\partial_k(J^{ij}\partial_iA\partial_{j}B)\partial_lC +J^{kl}\partial_kB\partial_l(J^{ij}\partial_iA\partial_{j}C)=\{\{A,B\},C\}+\{B,\{A,C\}\}\\ =-\{B,\{C,A\}\}-\{C,\{A,B\}\}.$$

But from here on, I have a bit of a problem to show, that the other term vanishes: $$J^{ij}J^{kl}\left(\partial^2_{ik}A\partial_jB\partial_lC +\partial^2_{il}A\partial_kB\partial_jC\right)=0.$$ First, I pull $\partial^2_{ik}A$ out of the bracket by switching the indices $l,k$ in the second term. Using $J^{kl}=-J^{lk}$, we obtain: $$J^{ij}\partial^2_{ik}A\left(J^{kl}\partial_jB\partial_lC -J^{lk}\partial_kB\partial_lC\right).$$ Notice that $\partial_kB$ becomes $\partial_lB$. Now we switch $j,k$, so we can pull $\partial_jB\partial_lC$ out of the bracket as well. We are left with: $$\partial^2_{ik}A\partial_jB\partial_lC\left(J^{ij}J^{kl} -J^{ik}J^{lj}\right).$$

How do I now show that this vanishes? Any hints or ideas?

$\endgroup$
1
$\begingroup$

When you pull $\partial^2_{ik}A$ out of the brackets you actually get $$J^{ij}\partial^2_{ik}A\big(J^{kl}\partial_j B\partial_l C + J^{lk} \partial_l B \partial_j C\big).$$ Then you use $J^{lk} = -J^{kl}$ and switch $l \rightleftarrows j$ in the second term to get $$\partial^2_{ik}A \partial_j B\partial_l C\big(J^{ij}J^{kl}-J^{il}J^{kj}\big).$$ Finally use that $\partial_i$ and $\partial_k$ commute and do $i \rightleftarrows k$ to one of the terms to conclude the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.