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?


3 Answers 3


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.

  • $\begingroup$ Can you please tell me what the commutation relation is: $[x^\mu, P^\nu] = ?$ $\endgroup$
    – rainman
    Jul 15, 2014 at 20:52
  • $\begingroup$ @Ome Since that's a distinct question and would be useful for other users, you should ask it in another post! $\endgroup$ 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 $$Δ𝐫 = πžˆΓ—π« - πž„t + 𝝴,\quad Ξ”t = -Ξ±πž„Β·π« + Ο„,\quad Ξ”u = πž„Β·π« + ψ,$$ where $(𝞈,πž„)$ are identified, respectively, as an infinitesimal rotation and boost and $(𝝴,Ο„,ψ)$ as infinitesimal translations on the coordinates. This is actually a one-parameter family of transforms, parametrized by $Ξ±$.

We'll see what that means in a moment. It is connected to the following. For coordinate differentials, we have the transforms: $$Ξ”d𝐫 = πžˆΓ—d𝐫 - πž„dt,\quad Ξ”dt = -Ξ±πž„Β·d𝐫,\quad Ξ”du = πž„Β·d𝐫.$$ Out of this arise two invariants: $$|d𝐫|^2 + 2 dt du + Ξ±(du)^2,\quad ds = dt + Ξ±du,\quad (s ≑ t + Ξ±u).$$

Let the coordinate transforms be written as Poisson brackets $$Ξ”(\_) = \left\{\_,Ξ›\right\},\quad Ξ› = πžˆΒ·π‰ + πž„Β·πŠ + 𝝴·𝐏 - Ο„H + ψμ.$$ Then we have the following associations: $$\begin{align} Ξ› &⇔ -(πžˆΓ—π« - πž„t + 𝝴)Β·βˆ‡ - (-Ξ±πž„Β·π« + Ο„)\frac{βˆ‚}{βˆ‚t} - (πž„Β·π« + ψ)\frac{βˆ‚}{βˆ‚u}\\ &= 𝞈·(-π«Γ—βˆ‡) + πž„Β·\left(tβˆ‡ + 𝐫\left(Ξ±\frac{βˆ‚}{βˆ‚t} - \frac{βˆ‚}{βˆ‚u}\right)\right) + 𝝴·(-βˆ‡) - Ο„\left(\frac{βˆ‚}{βˆ‚t}\right) + ψ\left(-\frac{βˆ‚}{βˆ‚u}\right). \end{align}$$ Component-by-component, this leads to the following associations: $$ 𝐉 ⇔ -π«Γ—βˆ‡,\quad 𝐊 ⇔ tβˆ‡ + 𝐫\left(Ξ±\frac{βˆ‚}{βˆ‚t} - \frac{βˆ‚}{βˆ‚u}\right),\quad 𝐏 ⇔ -βˆ‡,\quad H ⇔ \frac{βˆ‚}{βˆ‚t},\quad ΞΌ ⇔ -\frac{βˆ‚}{βˆ‚u}. $$

Defining the relations $$\left\{x^a,x^b\right\} = 0,\quad \left\{x^a,\frac{βˆ‚}{βˆ‚x^b}\right\} = -Ξ΄^a_b,\quad \left\{\frac{βˆ‚}{βˆ‚x^a},x^b\right\} = Ξ΄^b_a,\quad \left\{\frac{βˆ‚}{βˆ‚x^a},\frac{βˆ‚}{βˆ‚x^b}\right\} = 0,$$ we can treat the associations as identities $$ 𝐉 = -π«Γ—βˆ‡,\quad 𝐊 = tβˆ‡ + 𝐫\left(Ξ±\frac{βˆ‚}{βˆ‚t} - \frac{βˆ‚}{βˆ‚u}\right),\quad 𝐏 = -βˆ‡,\quad H = \frac{βˆ‚}{βˆ‚t},\quad ΞΌ = -\frac{βˆ‚}{βˆ‚u}, $$ and write down the following brackets: $$ \left\{J_i,J_j\right\} = Ξ΅^k_{ij}J_k,\quad \left\{J_i,K_j\right\} = Ξ΅^k_{ij}K_k,\quad \left\{J_i,P_j\right\} = Ξ΅^k_{ij}P_k,\\ \left\{K_i,K_j\right\} = -Ξ±Ξ΅^k_{ij}J_k,\quad \left\{K_i,P_j\right\} = Ξ΄_{ij}M,\quad \left\{P_i,P_j\right\} = 0,\\ \left\{J_i,H\right\} = 0,\quad \left\{K_i,H\right\} = P_i,\quad \left\{P_i,H\right\} = 0,\\ \left\{J_i,ΞΌ\right\} = 0,\quad \left\{K_i,ΞΌ\right\} = 0,\quad \left\{P_i,ΞΌ\right\} = 0,\quad \left\{H,ΞΌ\right\} = 0,\quad M ≑ ΞΌ + Ξ±H. $$ This is a one-parameter family of Lie algebras, with the only dependency on the parameter $Ξ±$ being with the $\left\{K,K\right\}$ and $\left\{K,P\right\}$ brackets.

From these, we can also write down the following identity for $M$: $$M = Ξ±\frac{βˆ‚}{βˆ‚t} - \frac{βˆ‚}{βˆ‚u},$$ and the following brackets for $M$: $$ \left\{J_i,M\right\} = 0,\quad \left\{K_i,M\right\} = Ξ±P_i,\quad \left\{P_i,M\right\} = 0,\quad \left\{H,M\right\} = 0,\quad \left\{ΞΌ,M\right\} = 0. $$

In the case $Ξ± = 0$, it is the Lie algebra for the Galilei group, lifted to the Bargmann group by central extension with the central charge $ΞΌ$. Both $M$ and $ΞΌ$ 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 $M = ΞΌ$.

In the case $Ξ± > 0$, the quadratic invariant may be rewritten by substituting $u$ by $s$, as: $$dx^2 + dy^2 + dz^2 - \frac{dt^2}{Ξ±} + \frac{ds^2}{Ξ±}.$$ 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: $$ds^2 = dt^2 - Ξ±\left(dx^2 + dy^2 + dz^2\right).$$ This is the Minkowski metric, written as an absolute time metric, where $Ξ± = 1/c^2$, and $c$ is the in-vacuo speed of light.

The Lie algebra is a deformation of the Bargmann group at $Ξ± = 0$, to a 1-generator extension of the PoincarΓ© group at $Ξ± = 1/c^2 > 0$.

Perform the corresponding substitutions on the differential operators: $$\left(\frac{βˆ‚}{βˆ‚t}\right)_u = \left(\frac{βˆ‚}{βˆ‚s}\right)_t + \left(\frac{βˆ‚}{βˆ‚t}\right)_s,\quad \left(\frac{βˆ‚}{βˆ‚u}\right)_t = Ξ±\left(\frac{βˆ‚}{βˆ‚s}\right)_t\quadβ‡’\quad Ξ±\left(\frac{βˆ‚}{βˆ‚t}\right)_u - \left(\frac{βˆ‚}{βˆ‚u}\right)_t = Ξ±\left(\frac{βˆ‚}{βˆ‚t}\right)_s. $$ This leads to the following revised identities: $$ 𝐉 = -π«Γ—βˆ‡,\quad 𝐊 = tβˆ‡ + α𝐫\frac{βˆ‚}{βˆ‚t},\quad 𝐏 = -βˆ‡,\quad H = \frac{βˆ‚}{βˆ‚s} + \frac{βˆ‚}{βˆ‚t},\quad ΞΌ = -Ξ±\frac{βˆ‚}{βˆ‚s},\quad M = Ξ±\frac{βˆ‚}{βˆ‚t}. $$ The generator $M$ now plays the role of "moving mass" and is usually written as "total energy" $E = Mc^2$, i.e. $M = Ξ±E$.

The Lie algebra splits into two subalgebras generated respectively by $(𝐉,𝐊,𝐏,E)$ - the Lie algebra for the PoincarΓ© group - and $(ΞΌ)$, where $(H,M)$ are replaced by $(E,ΞΌ)$. The generator for PoincarΓ© group Lie algebra reduce to: $$ 𝐉 = -π«Γ—βˆ‡,\quad 𝐊 = tβˆ‡ + α𝐫\frac{βˆ‚}{βˆ‚t},\quad 𝐏 = -βˆ‡,\quad E = \frac{βˆ‚}{βˆ‚t}. $$ Adopting the coordinates $\left(x^0,x^1,x^2,x^3\right) = (ct,x,y,z)$, now with $c = 1/\sqrt{Ξ±}$, and writing $βˆ‚_a = βˆ‚/βˆ‚x^a$, the generators can be written as: $$ 𝐉 = \left(x^3 βˆ‚_2 - x^2 βˆ‚_3, x^1 βˆ‚_3 - x^3 βˆ‚_1, x^2 βˆ‚_1 - x^1 βˆ‚_2\right),\\ 𝐊c = \left(x^0 βˆ‚_1 + x^1 βˆ‚_0, x^0 βˆ‚_2 + x^2 βˆ‚_0, x^0 βˆ‚_3 + x^3 βˆ‚_0\right),\\ \frac{E}{c} = βˆ‚_0,\quad 𝐏 = \left(-βˆ‚_1, -βˆ‚_2, -βˆ‚_3\right). $$

With respect to the metric, rewritten as $$d(cs)^2 = d(ct)^2 - dx^2 - dy^2 - dz^2,$$ the indices raise as $$\left(βˆ‚^0,βˆ‚^1,βˆ‚^2,βˆ‚^3\right) = \left(βˆ‚_0,-βˆ‚_1,-βˆ‚_2,-βˆ‚_3\right),$$ so that we can write: $$ 𝐉 = \left(x^2 βˆ‚^3 - x^3 βˆ‚^2, x^3 βˆ‚^1 - x^1 βˆ‚^3, x^1 βˆ‚^2 - x^2 βˆ‚^1\right),\\ 𝐊c = \left(x^1 βˆ‚^0 - x^0 βˆ‚^1, x^2 βˆ‚^0 - x^0 βˆ‚^2, x^3 βˆ‚^0 - x^0 βˆ‚^3\right),\\ \frac{E}{c} = βˆ‚^0,\quad 𝐏 = \left(βˆ‚^1, βˆ‚^2, βˆ‚^3\right). $$

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 $βˆ‚^a$ in $(𝐉,𝐊c,𝐏,E/c)$ will flip.


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