$$ [x_l,[\dot x_j,\dot x_k]]+[\dot x_l,[\dot x_j,x_k]]+[\dot x_l,[x_j,\dot x_k]]= \\ = -\frac{1}{m}[x_l,[x_j,F_k]]+\underbrace{[\dot x_j,\delta_{kl}]}_\text{0}+\underbrace{[\dot x_k,\delta_{lj}]}_\text{0}=\\ =[x_l,[x_j,F_k]]=0 $$ And by using the geodesic equations with a force, can one get Lorentz Invariant equations similar to Maxwell's ones ? A quick comment : one can also recover Maxwell/Proca's equation by using PauliLubanwski pseudo-vector of the Poincaré Group (see Ryder's book for instance). Question: isn't the idea *not* to assume equation (4)? It seems circular to me to prove (14) using (4). It should be enough to have (13) as the definition of H. This expression means that $H_l$ is not a function of $x_l$, otherwise they would not commute because $[x_l,\dot x_l]= \frac{i \hbar}{m}$. Now if $H_l$ is independent of $x_l$ the divergence of $H$ is zero because $\frac{\partial H_l}{\partial x_l}=0$. $$\text{div} H = \sum_l \frac{\partial H_l}{\partial x_l}=0$$ Which is one of Maxwell's equations. For more on the invariance paradox check the comments below on the NOTES AND DISCUSSIONS section. Deriving identity (3) in $t$ $$m\frac{d[x_j,\dot x_{k}]}{dt}=m\frac{d(x_j\dot x_{k}-\dot x_{k}x_j)}{dt}= \\ = m( \dot x_j\dot x_{k}+x_j\ddot x_{k}-\ddot x_{k}x_j-\dot x_{k} \dot x_j)= \\ = m( x_j\ddot x_{k}-\ddot x_{k}x_j+\dot x_j\dot x_{k}-\dot x_{k} \dot x_j)=\\ =[x_j,F_k]+m[\dot x_{j}, \dot x_k] = 0 \\ $$ $$ -[x_k,F_j]=m[\dot x_k, \dot x_j]=-m[\dot x_j, \dot x_k]=[x_j,F_k] $$ Freeman Dyson is an English-born American theoretical physicist and mathematician. He is professor emeritus at the Institute for Advanced Study and worked on several areas including quantum electrodynamics, solid-state physics, astronomy and nuclear engineering.  He met Richard Feynman at Cornell University in 1947 while he was working with Hans Bethe. You can see [here](https://www.youtube.com/watch?v=rlaPLvETBug&feature=youtu.be&t=77) Freeman Dyson talking about the first time he interacted with Richard Feynman. To prove that $[x_j,F_k]=-\frac{i \hbar}{m}\epsilon_{jkl} H_l$ we first start by using expression (4) $$ [x_j,F_k]= [x_j,E_k]+\epsilon_{kim}[x_j,\dot x_i H_m] $$ Now, since $[x_j,F_k]=-[x_k,F_j]$ it means that $[x_j,F_k]$ is antisymmetric and so it has to be proportional to the Levi-Civita symbol (which is also antisymmetric). This means $[x_j,E_k]=0$ (E is a function of t and x only) and $$ [x_j,F_k]= \epsilon_{kim}[x_j,\dot x_i H_m] $$ To calculate $[x_j,\dot x_i H_m]$ we recall that for any 3 operators A,B and C: $[A,BC]=[A,B]C+B[A,C]$ and so $$ [x_j,\dot x_i H_m]=[x_j,\dot x_i] H_m + \dot x_i[x_j, H_m]=\\ =\frac{i \hbar}{m}\delta_{ji}H_m+\dot x_i[x_j, H_m] $$ $$ [x_j,F_k]=\frac{i \hbar}{m}\epsilon_{kjm}H_m+\epsilon_{kim}\dot x_i[x_j, H_m]=\\ =-\frac{i \hbar}{m}\epsilon_{jkm}H_m+\epsilon_{kim}\dot x_i[x_j, H_m] $$ Using property (11) we see that $$ [x_l,[x_j,F_k]]= -\frac{i \hbar}{m}\epsilon_{jkm}[x_l,H_m]+\epsilon_{kim}[x_l,\dot x_i[x_j, H_m]] $$ $$ [x_l,\dot x_i[x_j, H_m]]=\frac{i\hbar}{m}\delta_{li}[x_j,H_m]+\dot x_i(x_l[x_j,H_m]-[x_j,H_m]x_l) $$ The only way for $[x_l,[x_j,F_k]]$ to be $0$ is if $[x_j, H_m]=0$ which means that all terms will vanish. Again as with $E$, this means that $H$ is a function of time and x only. $$ -\frac{im}{\hbar}\epsilon_{jkl}[E_j+\epsilon_{jmn}\dot x_m H_n , \dot x_k]=\\ =-\frac{im}{\hbar}(\epsilon_{jkl}[E_j,\dot x_k]+\underbrace{\epsilon_{jkl}\epsilon_{jmn}}_{\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm}}[\dot x_m H_n , \dot x_k])=\\ =-\frac{im}{\hbar}(\epsilon_{jkl}[E_j,\dot x_k]+[\dot x_k H_l , \dot x_k]-[\dot x_l H_k , \dot x_k]) $$ Cristian is correct, Luis' derivation is circular. Instead, one should simply *define* $H_l : = \frac{mi}{2\hbar}\epsilon_{pql}[x_p,F_q]$. You can then recover (13) by multiplying both sides by $\epsilon_{jkl}$, and using the standard identity $\epsilon_{jkl}\epsilon_{pql} = \delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp}$ and equation (12). This is incorrect - $H_l$ is ultimately going to be the magnetic field, which definitely does depend on $x_i$ in general. Instead, what is being used here is that from (3) it follows by induction that $[\dot{x}_i, (x_k)^n] = -\frac{i\hbar}{m}n (x_k)^{n-1}\delta_{ik}$ for all $n\ge 1$, which implies $[\dot{x}_i, f] = -\frac{i\hbar}{m}\frac{\partial f}{\partial x_i}$ for any (real analytic) function $f$ of the coordinates $x$ (and possibly also $t$). Hence $[\dot{x}_l, H_l] = 0$ is equivalent to $-\frac{i\hbar}{m}\frac{\partial H_l}{\partial x_l} = 0$, i.e. $\operatorname{div} H = 0$. $$ [x_j,F_k]= -\frac{i\hbar}{m}\epsilon_{jkl}H_l \equiv\\ -m[\dot x_j,\dot x_k]=-\frac{i\hbar}{m}\epsilon_{jkl}H_l $$ Now if we multiply both sides by $\epsilon_{jki}$ $$ -m\epsilon_{jki}[\dot x_j,\dot x_k]=-\frac{i\hbar}{m}\underbrace{\epsilon_{jkl}\epsilon_{jki}}_{2\delta_{li}}H_l\equiv \\ H_i=-\frac{im^2}{2\hbar}\epsilon_{jki}[\dot x_j,\dot x_k] $$ There's also an article about Feynman's derivation of the Schrödinger equation that has been annotated on Fermat's Library. To check it click [here](http://fermatslibrary.com/s/feynmans-derivation-of-the-schrodinger-equation). Since the space coordinate operators $x_1,x_2,x_3$ commute with each other $[x_j,x_k]=0$. To prove (22) we need to recall that the momentum operator in x-space is represented by $\dot x = p_x = \frac{\hbar}{i}\frac{\partial}{\partial x}$ and so: $$ [x_j,p_k]\Psi(x,t) =x_jp_k\Psi(x,t)-p_kx_j\Psi(x,t)= \\ =x_j\frac{\hbar}{i}\frac{\partial}{\partial x_k}\Psi(x,t)-\frac{\hbar}{i}\frac{\partial}{\partial x_k}x_j\Psi(x,t)=\\ =x_j\frac{\hbar}{i}\frac{\partial \Psi(x,t)}{\partial x_k}-\frac{\hbar}{i}\frac{\partial x_j}{\partial x_k}\Psi(x,t)-x_j\frac{\hbar}{i}\frac{\partial \Psi(x,t)}{\partial x_k}=\\ =-\frac{\hbar}{i}\frac{\partial x_j}{\partial x_k}\Psi(x,t)=- \frac{\hbar}{i}\delta_{jk}\Psi(x,t)=i \hbar \delta_{jk}\Psi(x,t) $$ Where $\delta_{jk}$ is the Kronecker delta: $$\delta_{jk} = \begin{cases} 0 &\text{if } j \neq k, \\ 1 &\text{if } j=k. \end{cases} $$ Note that since the coordinates are independent $\frac{\partial x_j}{\partial x_k}$ is not zero only when $x_j=x_k$.