# 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?

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.

• Can you please tell me what the commutation relation is: $[x^\mu, P^\nu] = ?$ Jul 15, 2014 at 20:52
• @Ome Since that's a distinct question and would be useful for other users, you should ask it in another post! Jul 15, 2014 at 20:54

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.