$$ [x_l,[\dot x_j,\dot x_k]]+[\dot x_l,[\dot x_j,x_k]]+[\dot x_l,[x_j,\dot x_k]]= \\ = -\frac{1}{m}[x_l,[x_j,F_k]]+\underbrace{[\dot x_j,\delta_{kl}]}_\text{0}+\underbrace{[\dot x_k,\delta_{lj}]}_\text{0}=\\ =[x_l,[x_j,F_k]]=0 $$ 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. For more on the invariance paradox check the comments below on the NOTES AND DISCUSSIONS section. Deriving identity (3) in $t$ $$m\frac{d[x_j,\dot x_{k}]}{dt}=m\frac{d(x_j\dot x_{k}-\dot x_{k}x_j)}{dt}= \\ = m( \dot x_j\dot x_{k}+x_j\ddot x_{k}-\ddot x_{k}x_j-\dot x_{k} \dot x_j)= \\ = m( x_j\ddot x_{k}-\ddot x_{k}x_j+\dot x_j\dot x_{k}-\dot x_{k} \dot x_j)=\\ =[x_j,F_k]+m[\dot x_{j}, \dot x_k] = 0 \\ $$ 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. $$ [x_j,F_k]= -\frac{i\hbar}{m}\epsilon_{jkl}H_l \equiv\\ -m[\dot x_j,\dot x_k]=-\frac{i\hbar}{m}\epsilon_{jkl}H_l $$ Now if we multiply both sides by $\epsilon_{jki}$ $$ -m\epsilon_{jki}[\dot x_j,\dot x_k]=-\frac{i\hbar}{m}\underbrace{\epsilon_{jkl}\epsilon_{jki}}_{2\delta_{li}}H_l\equiv \\ H_i=-\frac{im^2}{2\hbar}\epsilon_{jki}[\dot x_j,\dot x_k] $$ $$ -[x_k,F_j]=m[\dot x_k, \dot x_j]=-m[\dot x_j, \dot x_k]=[x_j,F_k] $$ 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. 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 $x_j=x_k$. $$ -\frac{im}{\hbar}\epsilon_{jkl}[E_j+\epsilon_{jmn}\dot x_m H_n , \dot x_k]=\\ =-\frac{im}{\hbar}(\epsilon_{jkl}[E_j,\dot x_k]+\underbrace{\epsilon_{jkl}\epsilon_{jmn}}_{\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm}}[\dot x_m H_n , \dot x_k])=\\ =-\frac{im}{\hbar}(\epsilon_{jkl}[E_j,\dot x_k]+[\dot x_k H_l , \dot x_k]-[\dot x_l H_k , \dot x_k]) $$ 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. $\operatorname{div} H = 0$. 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). 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.