The following representation describes how a field $\varphi$ transforms under the Poincaré group $\mathcal{P}$.

$$\mathsf{S} : \left\lbrace \begin{aligned} \mathcal{P} \times C^{\infty}(\mathcal{M}) &\longrightarrow C^{\infty}(\mathcal{M})\\ \big((\mathbf{a},\mathbf{\Lambda}), \varphi \big)\ \ &\longmapsto \mathsf{S}_{(\mathbf{a},\mathbf{\Lambda})}\cdot \varphi (x^{\mu}) := \varphi\big((\Lambda^{-1})^{\mu}{}_{\nu}\, (x- a)^{\nu}\big) \end{aligned} \right.$$


  1. This is equivalent to giving a map $\mathcal{P} \longrightarrow \mathrm{End}\big( C^{\infty}(\mathcal{M}) \big)$. ("Module"="Representation"). The reason one has $(\Lambda^{-1})^{\mu}{}_{\nu}$ and $- a^{\mu}$ is: -1) geometrically in analogy with the simple case of translations, if the graph of a function is to be moved to the right by $a$, the argument is $(x-a)$. - 2) algebraically, to have a left action as opposed to right action.
  2. There are other ways a field may transform, mathematically: there are other representations of $\mathcal{P}$ on $C^{\infty}(\mathcal{M})$, the present one is what physicists call scalar, as in "scalar field".

For a Lorentz transformation $\mathbf{\Lambda} \in SO(1,3)\subset\mathcal{P}$ close to the identity one usually writes the following expansion

$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + \omega^{\mu}{}_{\nu} + o(\omega)\qquad \qquad (Eq. 1)$$ (from which one obtains $\mathbf{\Lambda} \in SO(1,3)\ \Leftrightarrow \omega^{\mu\nu}= - \omega^{\nu\mu}$. $\eta^{\rho\sigma}$ involved) but for the representation above one writes $$\mathsf{S}_{\mathbf{\Lambda}} = \Big( \mathsf{Id} - \frac{i}{2\hbar} \omega^{\mu\nu} \boldsymbol{\mathsf{J}}_{\mu\nu} \Big) + o(\omega) \qquad\qquad (Eq. 2)$$

The action of the generators $\boldsymbol{\mathsf{J}}_{\mu\nu}$ of $\mathcal{P}$ on $C^{\infty}(\mathcal{M})$ are then differential operators: $$\mathsf{S}_{\mathbf{\Lambda}} \cdot \varphi (x^{\mu}) =\varphi\big((\Lambda^{-1})^{\mu}{}_{\nu}\, x^{\nu} \big) $$ Using $(\Lambda^{-1})^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} - \omega^{\mu}{}_{\nu} + o(\omega)$ on the r.h.s. one obtains

$$ \Big(\mathsf{Id} - \frac{i}{2\hbar} \omega^{\mu\nu} \boldsymbol{\mathsf{J}}_{\mu\nu}\Big)\cdot \varphi (x^{\mu}) = \varphi \Big( (\delta^{\mu}{}_{\nu} - \omega^{\mu}{}_{\nu} + o(\omega))\, x^{\nu} \Big) \\ = \varphi (x^{\mu})\, -\, \omega^{\mu}{}_{\nu}\, x^{\nu}\, \frac{\partial \varphi}{\partial x^{\mu} } = \varphi (x^{\mu})\, -\, \omega^{\mu\nu}\, x_{\nu}\, \partial_{\mu} \varphi $$ and thus $$ \frac{1}{2i \hbar} \omega^{\mu\nu} \boldsymbol{\mathsf{J}}_{\mu\nu}\cdot \varphi = -\, \omega^{\mu\nu}\, x_{\nu}\, \partial_{\mu} \varphi$$ Of course there is a sum over all $\mu \nu$. As presented here $\boldsymbol{\mathsf{J}}_{\mu\nu}$ cannot be totally determined but one can choose it antisymmetric (by exchange of $\mu \leftrightarrow \nu$. For a fixed choice $\mu_0\nu_0,\ \boldsymbol{\mathsf{J}}_{\mu_0\nu_0}$ is a differential operator, antisymmetry has no sense) because a symmetric part would not contribute by antisymmetry of $\omega^{\mu\nu}$. Hence (antisymmetrizing the r.h.s.) $$ \frac{1}{2i \hbar} \boldsymbol{\mathsf{J}}_{\mu\nu} = \frac{1}{2} \Big( - x_{\nu}\, \partial_{\mu} + x_{\mu}\, \partial_{\nu}\Big)\quad \Longleftrightarrow\quad \boldsymbol{\mathsf{J}}_{\mu\nu} = i\hbar \Big(x_{\mu}\, \partial_{\nu} - x_{\nu}\, \partial_{\mu}\Big) \quad (Eq. 3)$$ Similarly the generator of translations is found to be $\ \boldsymbol{\mathsf{P}}_{\mu}= - i\hbar \partial_{\mu}$

Question: I was double checking that the commutation relations of this representation of the Poincaré algebra did coincide with those of the defining representation (up to the $i\hbar$ factor) but THEY DIFFER BY A MINUS SIGN!! Where does it come from?

Relation in the defining rep.: $$\begin{gather} [\mathbf{P}_{\mu}, \mathbf{P}_{\nu}] = 0\\ \left[\mathbf{J}_{\mu\nu}, \mathbf{P}_{\rho} \right] = \eta_{\nu\rho}\, \mathbf{P}_{\mu} - \eta_{\mu\rho}\, \mathbf{P}_{\nu}\\ [\mathbf{J}_{\mu\nu}, \mathbf{J}_{\rho\sigma}] = \eta_{\mu\sigma}\, \mathbf{J}_{\nu\rho} - \eta_{\nu\sigma}\, \mathbf{J}_{\mu\rho} + \eta_{\nu\rho}\, \mathbf{J}_{\mu\sigma} - \eta_{\mu\rho}\, \mathbf{J}_{\nu\sigma} \end{gather}$$ ($(Eq .1)\ \Leftrightarrow\ \mathbf{\Lambda}= \mathbf{Id}_4 + \omega^{\mu}{}_{\nu}\, \mathbf{E}_{\mu}{}^{\nu} = \mathbf{Id}_4 + \omega^{\mu\nu}\, \eta_{\nu\rho} \mathbf{E}_{\mu}{}^{\rho} = \mathbf{Id}_4 + \frac{1}{2}\omega^{\mu\nu}\, \mathbf{J}_{\mu\nu}$ $\ \Rightarrow\ \mathbf{J}_{\mu\nu} = \eta_{\nu\rho} \mathbf{E}_{\mu}{}^{\rho} - \eta_{\mu\rho} \mathbf{E}_{\nu}{}^{\rho} $ with $\mathbf{E}_{\alpha}{}^{\beta}$ the $4\times 4$ matrix with $1$ on line $\alpha$, column $\beta$ and $0$ elsewhere.) I also double checked these relations because they depend on conventions (Calculate using the inclusion $\mathcal{P}\subset GL_5(\mathbb{R})$)! These are associated to the following parametrization $$\mathbf{R}_{z}(\epsilon) = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \cos \epsilon & - \sin \epsilon & 0\\ 0 & \sin \epsilon & \cos \epsilon & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$ $$ \mathbf{\Lambda}_x \Big(\epsilon =\beta = \frac{v}{c}\Big) = \begin{pmatrix} \gamma=(1-\epsilon^2)^{-1/2} & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$ of rotations and boosts whose generators $\mathbf{J}^z, \mathbf{K}^x$ are related to the $\mathbf{J}_{\mu\nu}$ by $$\mathbf{J}^i := - \frac{1}{2}\, \epsilon^{ijk}\, \mathbf{J}_{jk}\ ,\ \mathbf{K}^i := \mathbf{J}_{0i} \quad \Longleftrightarrow\quad \mathbf{J}_{ij} = \epsilon_{ijk}\, \mathbf{J}^k\ , \ \mathbf{J}_{0i} = \mathbf{k}^i$$ with the convention that $\epsilon^{ijk}= - \epsilon_{ijk}$ (because I use $\eta =\mathrm{diag}(1,-1,-1,-1)$)!!! Formally, if one pretends the generators are numbers then $$\boldsymbol{\mathsf{J}}_{\mu\nu} = \begin{pmatrix} 0 & \boldsymbol{\mathsf{K}}^1 & \boldsymbol{\mathsf{K}}^2 & \boldsymbol{\mathsf{K}}^3 \\ -\boldsymbol{\mathsf{K}}^1 & 0 & \boldsymbol{\mathsf{J}}^3 & -\boldsymbol{\mathsf{J}}^2 \\ -\boldsymbol{\mathsf{K}}^2 & -\boldsymbol{\mathsf{J}}^3 & 0 & \boldsymbol{\mathsf{J}}^1 \\ -\boldsymbol{\mathsf{K}}^2 & \boldsymbol{\mathsf{J}}^2 & -\boldsymbol{\mathsf{J}}^1 & 0 \end{pmatrix} $$ (Remark: I again "triple" checked, not the same as between $F_{\mu\nu}$ and $\mathbf{E}, \mathbf{B}$. Rather $-\mathbf{B}$...)

Cancelling the $i\hbar$, one finds for example $$\left[\frac{1}{i\hbar}\mathbf{J}^1, \frac{1}{i\hbar}\mathbf{P}_2\right] = \left[\big(x_2\, \partial_3 - x_3\, \partial_2\big) , -\partial_2\right] = \partial_3 = -\frac{1}{i\hbar}\mathbf{P}_3$$ THERE IS AN EXTRA MINUS SIGN

  • $\begingroup$ One could actually have the same relation between $\boldsymbol{\mathsf{J}}_{\mu\nu}$ and $\boldsymbol{\mathsf{J}}^i, \boldsymbol{\mathsf{K}}^j$ as the one between $F_{\mu\nu}$ and the electromagnetic field by saying that rotations are rather parametrized by $\begin{pmatrix} \cos \epsilon & \sin \epsilon \\ -\sin\epsilon & \cos \epsilon\end{pmatrix}$. This looks ad hoc but finally it is coherent with the fact that the boost correspond here to "passive" transformation (i.e. coordinate in a new referential fct of the coord. in the old one). Above I chose "active" rotation, in this comment "passiv $\endgroup$
    – Noix07
    Jun 4, 2017 at 21:34

1 Answer 1


I finally got it!! Here is the detail I missed $$x_i := \eta_{i\mu}\, x^{\mu} = - x^i\quad \text{and}\quad \partial_{\mu}:= \frac{\partial}{\partial x^{\mu}} $$

(again, convention $\eta= \mathrm{diag}(1,-1,-1,-1)$) so that $$ \partial_0\, x_0=1 \quad \text{but}\quad \partial_i\, x_i = - \partial_i\, x^i = -1$$.

Anyway I checked it on a simple case, $\varphi\in C^{\infty}(\mathbb{R}^2)$: $$\mathbf{R}_{\epsilon}^{-1}:=\begin{pmatrix} \cos \epsilon & -\sin\epsilon \\ \sin \epsilon & \cos\epsilon \end{pmatrix}^{-1}= \begin{pmatrix} \cos \epsilon & \sin\epsilon \\ -\sin \epsilon & \cos\epsilon \end{pmatrix} =\mathbf{Id} - \epsilon\, \mathbf{J} + o(\epsilon)$$

with $\ \mathbf{J} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Now for the representation on $C^{\infty}(\mathbb{R}^2)$: $$ \boldsymbol{\mathsf{S}}_{\epsilon}\cdot \varphi \begin{pmatrix} x\\ y \end{pmatrix} = \varphi \left(\mathbf{R}_{\epsilon}^{-1}\cdot \begin{pmatrix} x\\ y \end{pmatrix} \right) =\varphi \begin{pmatrix} x + \epsilon\, y + o(\epsilon)\\ -\epsilon\, x +y + o(\epsilon)\end{pmatrix} $$

$$\Longleftrightarrow\quad\left(\boldsymbol{\mathsf{Id}} + \epsilon\, \boldsymbol{\mathsf{J}} \right)\cdot \varphi \begin{pmatrix} x\\ y \end{pmatrix} = \varphi \begin{pmatrix} x\\ y \end{pmatrix} + \epsilon\, y\, \partial_x \varphi \begin{pmatrix} x\\ y \end{pmatrix} - \epsilon\, x\, \partial_y \varphi \begin{pmatrix} x\\ y \end{pmatrix}$$

$$\Longleftrightarrow\quad \boldsymbol{\mathsf{J}} = y\, \partial_x\ -\, x\, \partial_y $$ which is suppose to correspond to $\boldsymbol{\mathsf{J}}^3 = \boldsymbol{\mathsf{J}}_{12}= x_1\, \partial_2\ -\, x_2\, \partial_1 $. It does because of the first eq. of the answer!


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