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?
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.
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.
Consider the following transform, given in infinitesimal form, on coordinates $(𝐫,t,u)$, where $𝐫 = (x,y,z)$, by $$Δ𝐫 = 𝞈×𝐫 - 𝞄t + 𝝴,\quad Δt = -α𝞄·𝐫 + τ,\quad Δu = 𝞄·𝐫 + ψ,$$ where $(𝞈,𝞄)$ are identified, respectively, as an infinitesimal rotation and boost and $(𝝴,τ,ψ)$ as infinitesimal translations on the coordinates. This is actually a one-parameter family of transforms, parametrized by $α$.
We'll see what that means in a moment. It is connected to the following. For coordinate differentials, we have the transforms: $$Δd𝐫 = 𝞈×d𝐫 - 𝞄dt,\quad Δdt = -α𝞄·d𝐫,\quad Δdu = 𝞄·d𝐫.$$ Out of this arise two invariants: $$|d𝐫|^2 + 2 dt du + α(du)^2,\quad ds = dt + αdu,\quad (s ≡ t + αu).$$
Let the coordinate transforms be written as Poisson brackets $$Δ(\_) = \left\{\_,Λ\right\},\quad Λ = 𝞈·𝐉 + 𝞄·𝐊 + 𝝴·𝐏 - τH + ψμ.$$ Then we have the following associations: $$\begin{align} Λ &⇔ -(𝞈×𝐫 - 𝞄t + 𝝴)·∇ - (-α𝞄·𝐫 + τ)\frac{∂}{∂t} - (𝞄·𝐫 + ψ)\frac{∂}{∂u}\\ &= 𝞈·(-𝐫×∇) + 𝞄·\left(t∇ + 𝐫\left(α\frac{∂}{∂t} - \frac{∂}{∂u}\right)\right) + 𝝴·(-∇) - τ\left(\frac{∂}{∂t}\right) + ψ\left(-\frac{∂}{∂u}\right). \end{align}$$ Component-by-component, this leads to the following associations: $$ 𝐉 ⇔ -𝐫×∇,\quad 𝐊 ⇔ t∇ + 𝐫\left(α\frac{∂}{∂t} - \frac{∂}{∂u}\right),\quad 𝐏 ⇔ -∇,\quad H ⇔ \frac{∂}{∂t},\quad μ ⇔ -\frac{∂}{∂u}. $$
Defining the relations $$\left\{x^a,x^b\right\} = 0,\quad \left\{x^a,\frac{∂}{∂x^b}\right\} = -δ^a_b,\quad \left\{\frac{∂}{∂x^a},x^b\right\} = δ^b_a,\quad \left\{\frac{∂}{∂x^a},\frac{∂}{∂x^b}\right\} = 0,$$ we can treat the associations as identities $$ 𝐉 = -𝐫×∇,\quad 𝐊 = t∇ + 𝐫\left(α\frac{∂}{∂t} - \frac{∂}{∂u}\right),\quad 𝐏 = -∇,\quad H = \frac{∂}{∂t},\quad μ = -\frac{∂}{∂u}, $$ and write down the following brackets: $$ \left\{J_i,J_j\right\} = ε^k_{ij}J_k,\quad \left\{J_i,K_j\right\} = ε^k_{ij}K_k,\quad \left\{J_i,P_j\right\} = ε^k_{ij}P_k,\\ \left\{K_i,K_j\right\} = -αε^k_{ij}J_k,\quad \left\{K_i,P_j\right\} = δ_{ij}M,\quad \left\{P_i,P_j\right\} = 0,\\ \left\{J_i,H\right\} = 0,\quad \left\{K_i,H\right\} = P_i,\quad \left\{P_i,H\right\} = 0,\\ \left\{J_i,μ\right\} = 0,\quad \left\{K_i,μ\right\} = 0,\quad \left\{P_i,μ\right\} = 0,\quad \left\{H,μ\right\} = 0,\quad M ≡ μ + αH. $$ This is a one-parameter family of Lie algebras, with the only dependency on the parameter $α$ being with the $\left\{K,K\right\}$ and $\left\{K,P\right\}$ brackets.
From these, we can also write down the following identity for $M$: $$M = α\frac{∂}{∂t} - \frac{∂}{∂u},$$ and the following brackets for $M$: $$ \left\{J_i,M\right\} = 0,\quad \left\{K_i,M\right\} = αP_i,\quad \left\{P_i,M\right\} = 0,\quad \left\{H,M\right\} = 0,\quad \left\{μ,M\right\} = 0. $$
In the case $α = 0$, it is the Lie algebra for the Galilei group, lifted to the Bargmann group by central extension with the central charge $μ$. Both $M$ and $μ$ coincide and correspond to the mass. The geometric invariants reduce to: $$dx^2 + dy^2 + dz^2 + 2 dt du,\quad ds = dt,$$ the roles played by $s$ and $t$ coincide and correspond to time, as well as to "proper time".
The generators correspond to angular momentum $𝐉$, moment $𝐊$, linear momentum $𝐏$, kinetic(+internal) energy $H$ and mass $M = μ$.
In the case $α > 0$, the quadratic invariant may be rewritten by substituting $u$ by $s$, as: $$dx^2 + dy^2 + dz^2 - \frac{dt^2}{α} + \frac{ds^2}{α}.$$ If we continue to identify the invariant $ds$ as the differential for proper time, and set the quadratic invariant to 0, then the result will be a reduction to the geometry given by: $$ds^2 = dt^2 - α\left(dx^2 + dy^2 + dz^2\right).$$ This is the Minkowski metric, written as an absolute time metric, where $α = 1/c^2$, and $c$ is the in-vacuo speed of light.
The Lie algebra is a deformation of the Bargmann group at $α = 0$, to a 1-generator extension of the Poincaré group at $α = 1/c^2 > 0$.
Perform the corresponding substitutions on the differential operators: $$\left(\frac{∂}{∂t}\right)_u = \left(\frac{∂}{∂s}\right)_t + \left(\frac{∂}{∂t}\right)_s,\quad \left(\frac{∂}{∂u}\right)_t = α\left(\frac{∂}{∂s}\right)_t\quad⇒\quad α\left(\frac{∂}{∂t}\right)_u - \left(\frac{∂}{∂u}\right)_t = α\left(\frac{∂}{∂t}\right)_s. $$ This leads to the following revised identities: $$ 𝐉 = -𝐫×∇,\quad 𝐊 = t∇ + α𝐫\frac{∂}{∂t},\quad 𝐏 = -∇,\quad H = \frac{∂}{∂s} + \frac{∂}{∂t},\quad μ = -α\frac{∂}{∂s},\quad M = α\frac{∂}{∂t}. $$ The generator $M$ now plays the role of "moving mass" and is usually written as "total energy" $E = Mc^2$, i.e. $M = αE$.
The Lie algebra splits into two subalgebras generated respectively by $(𝐉,𝐊,𝐏,E)$ - the Lie algebra for the Poincaré group - and $(μ)$, where $(H,M)$ are replaced by $(E,μ)$. The generator for Poincaré group Lie algebra reduce to: $$ 𝐉 = -𝐫×∇,\quad 𝐊 = t∇ + α𝐫\frac{∂}{∂t},\quad 𝐏 = -∇,\quad E = \frac{∂}{∂t}. $$ Adopting the coordinates $\left(x^0,x^1,x^2,x^3\right) = (ct,x,y,z)$, now with $c = 1/\sqrt{α}$, and writing $∂_a = ∂/∂x^a$, the generators can be written as: $$ 𝐉 = \left(x^3 ∂_2 - x^2 ∂_3, x^1 ∂_3 - x^3 ∂_1, x^2 ∂_1 - x^1 ∂_2\right),\\ 𝐊c = \left(x^0 ∂_1 + x^1 ∂_0, x^0 ∂_2 + x^2 ∂_0, x^0 ∂_3 + x^3 ∂_0\right),\\ \frac{E}{c} = ∂_0,\quad 𝐏 = \left(-∂_1, -∂_2, -∂_3\right). $$
With respect to the metric, rewritten as $$d(cs)^2 = d(ct)^2 - dx^2 - dy^2 - dz^2,$$ the indices raise as $$\left(∂^0,∂^1,∂^2,∂^3\right) = \left(∂_0,-∂_1,-∂_2,-∂_3\right),$$ so that we can write: $$ 𝐉 = \left(x^2 ∂^3 - x^3 ∂^2, x^3 ∂^1 - x^1 ∂^3, x^1 ∂^2 - x^2 ∂^1\right),\\ 𝐊c = \left(x^1 ∂^0 - x^0 ∂^1, x^2 ∂^0 - x^0 ∂^2, x^3 ∂^0 - x^0 ∂^3\right),\\ \frac{E}{c} = ∂^0,\quad 𝐏 = \left(∂^1, ∂^2, ∂^3\right). $$
Unlike the previous expressions, though, this depends on which convention is adopted for the metric. If the opposite signs are used $$dx^2 + dy^2 + dz^2 - d(ct)^2 = -d(cs)^2,$$ then all the signs for the index-raised forms of $∂^a$ in $(𝐉,𝐊c,𝐏,E/c)$ will flip.
