# Casimir operator and particle worldline

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?

• $[C,x]$ is $[\partial_t^2-\partial_x^2,x]$. You should know how to compute that. – G. Smith Sep 12 '19 at 17:04

## 1 Answer

The answer to question (2) is really trivial and I don't know how I did not realize it before. This should be the answer:

$$[p_0,x]f(t,x)=(\partial_tx-x\partial_t)f=(\partial_tx)f+x\partial_tf-x\partial_tf={\partial x \over \partial t}f=0 \Rightarrow [p_0,x]=0$$
$$[p_1,x]f(t,x)= f \Rightarrow [p_1,x]=1$$

And vice versa for $$[p_0,t]$$,$$[p_1,t]$$. Therefore:

$$[C,x]=[p_0^2,x]-[p_1^2,x]=p_0[p_0,x]+[p_0,x]p_0-p_1[p_1,x]-[p_1,x]p_1=-2p_1$$
$$[C,t]=2p_0$$.

I still can't figure out how to answer to question (1). Any tip would be appreciate.

EDIT:

I tried to answer by myself and this is what came out.

(1.2) I have found another definition of the Poisson bracket:

$$\{f,g\} := \sum_{i=1}^{n} ({\partial f \over \partial p_i}{\partial g \over \partial q_i}-{\partial f \over \partial q_i}{\partial g \over \partial p_i})$$

Then Hamilton's equations can be written in this short form:

$$\dot x_i= \{H,x_i\}$$

where $$i=1,...,2n$$ and $$x_i$$ denotes any of the coordinate functions $$(q_i,p_i)$$. With this definition I have the result I was looking for, because you can prove that Poisson bracket and Lie bracket are in this relation:

$$X_{\{f,g\}}(h):=\{\{f,g\},h\} =[X_f,X_g](h)$$

which basically means that we can use Lie brackets instead of Poisson brackets.

The previous definition I was using was:

$$\{f,g\} := \sum_{i=1}^{n} ({\partial f \over \partial q_i}{\partial g \over \partial p_i}-{\partial f \over \partial p_i}{\partial g \over \partial q_i})$$

and in this case we have this relation:

$$[X_f,X_g]=-X_{\{f,g\}}$$.

If you want to use the second definition and you want to use Lie bracket instead of Poisson bracket, you have to consider the minus sign.

(1.1) Probably the original statement was wrong, because it continues saying that, once you have found $${\partial x \over \partial s}= -2p_1$$ and $${\partial t \over \partial s}= 2p_0$$, you can find the velocity with $${\partial x \over \partial t}=- {p_1 \over p_0}$$, but of course velocity definition has the total derivative. Also, the use of the total derivative would be in accordance with what you get from the geodesic equation.