Poincare Generators in terms of Position and Momentum The $10$ generators of the Poincare group $P(1;3)$ are $M^{\mu\nu}$ and $P^\mu$. These generators can be determined explicitly in the matrix form. However, I have found that $M^{\mu\nu}$ and $P^\mu$ are often written in terms of position $x^\mu$ and momentum $p^\mu$ as
$$ M^{\mu\nu} = x^\mu p^\nu - x^\nu p^\mu $$
and
$$ P^\mu = p^\mu$$
How do we get these relations?
 A: One obtains those expressions by considering a particular action of the Poincare group on fields.
Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$.  Let $\mathcal F$ denote the space of such fields.  Define an action $\rho_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows
\begin{align}
  \rho_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a))
\end{align}
Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity.  Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra).  We can use this generator to define a corresponding infinitesimal generator for group action $\rho_\mathcal F$ as follows:
\begin{align}
  G_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|_{\epsilon = 0}
\end{align}
Example - translations. Consider the translation generators $P^\mu$ which have the property
\begin{align}
  e^{-ia_\mu P^\mu}x = x+a
\end{align}
The generator of $\rho_\mathcal F$ corresponding to $P^0$, for instance, is
\begin{align}
  (P^0)_\mathcal F(\phi)(x) 
&= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\
&= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\
&= i\partial^0\phi(x)
\end{align}
where $e_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives
\begin{align}
  (P^\mu)_\mathcal F = i\partial^\mu.
\end{align}
Example - Lorentz boosts.
If you use this same procedure for Lorentz boost generators, you will find that
\begin{align}
  (M^{\mu\nu})_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu
\end{align}
Disclaimer about signs etc. There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.
A: The generators of isometry, also generators of the Poincare group, are the Killing vectors, hence we need the Killing vectors of Minkowski spacetime, $ds^2 = -dt^2 + dx^2 + dy^2+ dz^2$. The defining equation of the Killing vectors in terms of their components ($\xi = \xi^\mu \partial_\mu$ is a Killing vector with components $\xi^\mu$) is $$\partial_{(\mu} \xi_{\nu)}=0,$$ where the parentheses denote symmetrization over the enclosed indices. We differentiate once, then cyclically permute the indices, $$\begin{split}\partial_c\partial_a \xi_b + \partial_b\partial_c \xi_a &=0,\\
\partial_a\partial_b\xi_c + \partial_c\partial_a\xi_b&= 0,\\
\partial_b\partial_c\xi_a + \partial_a\partial_b\xi_c &=0,\end{split}$$ which is a linear system with unknowns $\partial_a\partial_b\xi_c$ and its permutations. The only solution of the system is the trivial $\partial_a\partial_b\xi_c=0$, from which we obtain $\xi_a = a_{ab} x^b + b_a$ with $a_{ab}$ and $b_a$ constants. From the defining equation we obtain $a_{ab} = -a_{ba}$ and, since the Killing vectors are unique up to multiplication with a constant and translation, we obtain $$    \begin{aligned}
      \partial_{t} && \partial_{x} && \partial_{y} && \partial_{z},\\
      -x\partial_{t} - t\partial_{x}, && -y\partial_{t} -
      t \partial_{y} && - z\partial_{y} - t\partial_{z},\\
      x\partial_{y} - y\partial_{x} && y\partial_{z} - z\partial_{x} &&
      z\partial_{x} - x \partial_{z}.&&
    \end{aligned}$$
On the first line are the translation generators, on the second the boost generators and on the third the rotation generators.
