I'm studying Killing vectors in 2d Minkowski space-time, with signature $(+,-)$, the usual metric given by
$ds^2=dt^2-dx^2$.
I have found these Killing vectors:
$\xi^{(1)}=(1,0)=\partial_t\equiv p_0$
$\xi^{(2)}=(0,1)=\partial_x\equiv p_1$
$\xi^{(3)}=(x,t)=x\partial_t+t\partial_x\equiv N$
and this Casimir operator:
$C\equiv g^{\mu\nu}p_{\mu}p_{\nu}=p_0^2-p_1^2$.
Then I have the following statement: "Let's take a free particle, with world-line $(x(s),t(s))$, where $s$ is a generic parameter. We can use our Casimir operator as an Hamiltonian operator $H=C$, so:
${\partial x \over\partial s }=[C,x]=-2p_1$
${\partial t \over\partial s }=2p_0$
${\partial p_1 \over\partial s }=0={\partial p_0 \over\partial s }$."
My questions are: (1) How do I prove that ${\partial x \over\partial s }=[C,x]$? (2) How $[C,x]$ works? Do I have to look at $x$ as an operator or as a function of $s$? I don't know how to prove that $[C,x]=-2p_1$.
Thank you in advance.
EDIT:
(1) I found out that you can write Hamilton's eom in terms of the Poisson bracket (which is, for our purposes, Lie bracket):
$\dot q_i = {\partial H \over \partial p_i}= [q_i,H]$
where, in our case, $q_0=t$ and $q_1=x$.
Question (1.1): Can I read $\dot q_i$ as the partial derivative with respect to $s$?
Therefore:
$\dot q_1 =\dot x = {\partial H \over \partial p_1}= [q_1,H]$
But:
${\partial H \over \partial p_1}={\partial C \over \partial p_1}=-2p_1$
$[q_1,H]=[x,C]=-[C,x]=2p_1$
And I have the same problem with $t$ (so this is not signature related).
Question (1.2): How do I solve this last discrepancy?