The Pauli-Lubanski operator is defined as $${W^\alpha } = \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta}{M_{\mu \nu }},\qquad ({\varepsilon ^{0123}} = + 1,\;{\varepsilon _{0123}} = - 1)$$ where $M_{\mu\nu}$ is the generators of Lorentz group.

The commutation relation between generators of Poincare group is know as $$i[{M^{\mu \nu }},{M^{\rho \sigma }}] = {\eta ^{\nu \rho }}{M^{\mu \sigma }} - {\eta ^{\mu \rho }}{M^{\nu \sigma }} - {\eta ^{\mu \sigma }}{M^{\rho \nu }} + {\eta ^{\nu \sigma }}{M^{\rho \mu }},$$ $$i[{P^\mu },{M^{\rho \sigma }}] = {\eta ^{\mu \rho }}{P^\sigma } - {\eta ^{\mu \sigma }}{P^\rho }.$$

I try to derive the commutator between Pauli-Lubanski operator and a generator of Lorentz group, which is also given in our lecture $$i[{W^\alpha },{M^{\rho \sigma }}] = {\eta ^{\alpha \rho }}{W^\sigma } - {\eta ^{\alpha \sigma }}{W^\rho }.\tag{*}$$

But I only get $$\begin{align} i[{W^\alpha },{M^{\rho \sigma }}] =& i[\frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\mu \nu }},{M^{\rho \sigma }}] \\ =& \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }\left( {\delta _\nu ^\rho {M_\mu }^\sigma - \delta _\mu ^\rho {M_\nu }^\sigma - \delta _\mu ^\sigma {M^\rho }_\nu + \delta _\nu ^\sigma {M^\rho }_\mu } \right) \\ & + \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}\left( {\delta _\beta ^\rho {P^\sigma } - \delta _\beta ^\sigma {P^\rho }} \right)M_{\mu\nu}. \end{align}$$

Obviously I can contract out the delta's, but that does not get me any closer to the simpler result of (*). Can anyone point out what to do next?

  • 2
    $\begingroup$ Your index structure does not look right. For example in the last term the $\mu$ and $\nu$ indices are free but should be contracted. You also contract indices where both index are up. You need to use the relation $[AB,C] = A[B,C] + [B,C]A$ and the (anti-)symmetry properties of the indices. $\endgroup$
    – Heidar
    Commented Sep 28, 2012 at 15:57
  • $\begingroup$ @Heidar: Fixed. I did it correctly on scratch paper, but got it wrong when typesetting on a computer. $\endgroup$
    – Siyuan Ren
    Commented Sep 28, 2012 at 16:58
  • $\begingroup$ @Heidar: I tried to use anti-symmetry in the beginning to simplify the first four terms. I hoped they would cancel each other out, but they just got reduced into two different terms. Still nowhere near the end result. $\endgroup$
    – Siyuan Ren
    Commented Sep 28, 2012 at 17:00

5 Answers 5


Maybe, the answer will be non-useful for the author of the question, but it is important for the other people who want to derive this expression. [:)].

First method.

I made some hint: it's easy to show, that $$ \hat {W}^{\mu}\hat {P}_{\mu} = \frac{1}{2}\varepsilon^{\mu \alpha \beta \gamma}\hat {J}_{\alpha \beta}\hat {P}_{\gamma }\hat {P}_{\mu} = 0 $$ as the convolution of symmetrical $\hat {P}_{\gamma }\hat {P}_{\mu}$ and antisymmetrical $\varepsilon^{\mu \alpha \beta \gamma}$.

So the commutator $$ [\hat {J}_{\kappa \lambda}, \hat {W}^{\mu}\hat {P}_{\mu}] = 0. \qquad (.1) $$ But by the other hand $$ [ \hat {J}_{\kappa \lambda}, \hat {W}^{\mu}\hat {P}_{\mu}] = \hat {W}^{\mu}[ \hat {J}_{\kappa \lambda}, \hat {P}_{\mu}] + [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} = -\hat {W}^{\mu}i(g_{\mu \kappa }\hat {P}_{\lambda} - g_{\mu \lambda}\hat {P}_{\kappa }) + [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} . $$ So, by using $(.1)$ from the previous formula there follows the next: $$ [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}]\hat {P}_{\mu} = i\hat {W}^{\mu}(g_{\mu \kappa }\hat {P}_{\lambda} - g_{\mu \lambda}\hat {P}_{\kappa }) = i(\hat {W}_{\kappa}\hat {P}_{\lambda} - \hat {W}_{\lambda}\hat {P}_{\kappa}) = i\left(\hat {W}_{\kappa}\delta^{\mu}_{\lambda} - \hat {W}_{\lambda}\delta^{\mu}_{\kappa}\right)\hat {P}_{\mu}, $$ where $\delta^{0}_{0} = 1, \delta^{i}_{i} = -1$,

and, finally, $$ [\hat {J}_{\kappa \lambda }, \hat {W}^{\mu}] = i\left(\hat {W}_{\kappa}\delta^{\nu}_{\lambda} - \hat {W}_{\lambda}\delta^{\mu}_{\kappa}\right) \Rightarrow [\hat {J}_{\kappa \lambda }, \hat {W}_{\mu}] = i\left(\hat {W}_{\kappa}g_{\mu \lambda} - \hat {W}_{\lambda}g_{\mu \kappa}\right). $$

Second method.

It is similar to QMechanic method. Lets have general Poincare transformation matrix $U (\hat {\Lambda}, a^{\mu})$, where $a$ is translation 4-vector and $\hat {\Lambda}$ is a 3-rotations and Lorentz transformations matrix. Due to this transformation, $$ \hat {J}_{\mu \nu} \Rightarrow U (\hat {\Lambda}, a^{\mu}) J^{\mu \nu}U (\hat {\Lambda}, a^{\mu})^{+} = \Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}\hat {J}^{\rho \sigma} + a^{\mu}\Lambda^{\nu}_{\rho }\hat {P}^{\rho} - a^{\nu}\Lambda^{\mu}_{\rho}\hat {P}^{\rho }, $$ $$ \hat {P}^{\mu} \Rightarrow \Lambda^{\mu}_{\nu}\hat {P}^{\nu}. $$ By using these transformation rules you can get $$ U (\hat {\Lambda}, a) W_{\mu} U (\hat {\Lambda}, a)^{+} = \frac{1}{2} \varepsilon_{\mu \nu \rho \sigma}\left( \lambda^{\nu}_{\alpha}\Lambda^{\rho}_{\beta}\hat {J}^{\alpha \beta} + a^{\nu}\Lambda^{\rho}_{\alpha }\hat {P}^{\alpha} - a^{\rho}\Lambda^{\nu}_{\rho}\hat {P}^{\rho }\right) \Lambda^{\sigma}_{\delta}\hat {P}^{\delta } = $$ $$ =\frac{1}{2}\Lambda^{\alpha}_{\mu}\varepsilon_{\alpha \nu \rho \sigma }\hat {J}^{\nu \rho} \hat {P}^{\sigma} = \Lambda^{\alpha}_{\mu} \hat {W}_{\alpha}. $$ So $$ \frac{i}{2}[\hat {J}_{\mu \nu}, \hat {W}_{\sigma }] = \omega^{\mu \nu}g_{\sigma \mu}\hat {W}_{\nu} = \frac{1}{2}(g_{\sigma \mu}\hat {W}_{\nu} - g_{\sigma \nu}\hat {W}_{\mu})\omega^{\mu \nu}, $$ which leads to the targeted expression.

In general, this means that all of 4-operator $\hat {A}_{\gamma}$ commutes with $\hat {J}_{\alpha \beta}$ as $i(g_{\gamma \alpha}\hat {A}_{\beta} - g_{\gamma \beta}\hat {A}_{\alpha})$, because $\hat {J}_{\alpha \beta}$ represents the generators of the Lorentz group. Also, this means, that commutator of $\hat {J}_{\alpha \beta }$ with each 4-scalar will give zero.

  • 3
    $\begingroup$ If I understand correctly, in your first method, you rely on the claim that $\hat{A}_{\kappa\lambda}^\mu\hat{P}_\mu=0$ implies $\hat{A}_{\kappa\lambda}^\mu=0$ (where in your case $\hat{A}_{\kappa\lambda}^\mu=[\hat{J}_{\kappa\lambda},\hat{W}^\mu]-i(\hat{W}_{\kappa}\delta^\mu_\lambda - \hat{W}_{\lambda}\delta^\mu_\kappa)$). I don't think this is correct. $\endgroup$
    – Kubav
    Commented Oct 25, 2015 at 13:57

Here is a possible strategy:

  1. Consider an arbitrary (infinitesimal) real antisymmetric matrix $$\tag{1}\omega^{\mu\nu}~=~-\omega^{\nu\mu}.$$

  2. Define $$\tag{2}\omega^{\mu}{}_{\lambda}~:=~\omega^{\mu\nu}{}\eta_{\nu\lambda}.$$

  3. Define $$\tag{3}M~:=~ \frac{i}{2}M_{\mu\nu}\omega^{\nu\mu}.$$

  4. Then OP's equations can be reformulated as: $$\tag{4}W^{\alpha} ~:=~ \frac{1}{2}\varepsilon^{\alpha\beta\mu\nu}P_{\beta}M_{\mu \nu}, $$ $$\tag{5} [P_{\alpha}, M]~=~P_{\beta}\omega^{\beta}{}_{\alpha},$$ $$\tag{6} [M_{\mu\nu}, M]~=~M_{\mu\lambda}\omega^{\lambda}{}_{\nu}- (\mu \leftrightarrow \nu),$$ $$\tag{7} [W^{\alpha}, M]~=~-\omega^{\alpha}{}_{\beta}W^{\beta}. \qquad\qquad(*)$$

  5. In particular, note that OP's question(v2) can be reformulated as How to prove eq. (7)?

  6. Define the exponential matrix $$\tag{8} A^{\alpha}{}_{\beta}~:=~(e^{\omega})^{\alpha}{}_{\beta}~:=~ \delta^{\alpha}_{\beta} + \omega^{\alpha}{}_{\beta}+ \frac{1}{2}\omega^{\alpha}{}_{\gamma} \omega^{\gamma}{}_{\beta} + {\cal O}(\omega^3).$$

  7. Expand the following determinantal identity $$\tag{9} A^{\alpha}{}_{\alpha^{\prime}}A^{\beta}{}_{\beta^{\prime}}A^{\mu}{}_{\mu^{\prime}}A^{\nu}{}_{\nu^{\prime}}\varepsilon^{\alpha^{\prime}\beta^{\prime}\mu^{\prime}\nu^{\prime}} ~=~\det(A)\varepsilon^{\alpha\beta\mu\nu}$$ to first order in $\omega$ to deduce the following trace identity $$\tag{10} \omega^{\alpha}{}_{\alpha^{\prime}}\varepsilon^{\alpha^{\prime}\beta\mu\nu} +\omega^{\beta}{}_{\beta^{\prime}}\varepsilon^{\alpha\beta^{\prime}\mu\nu} +\omega^{\mu}{}_{\mu^{\prime}}\varepsilon^{\alpha\beta\mu^{\prime}\nu} +\omega^{\nu}{}_{\nu^{\prime}}\varepsilon^{\alpha\beta\mu\nu^{\prime}} ~=~{\rm tr}(\omega)\varepsilon^{\alpha\beta\mu\nu},$$ with the help of the determinant-trace identity $$\tag{11} \det(e^{\omega}) ~=~e^{{\rm tr}(\omega)}.$$ Eq. (10) may be viewed as an infinitesimal version of eq.(9), but because of linearity eq. (10) is actually valid for any$^1$ finite $4\times 4$ matrix.

  8. Use the antisymmetry (1) and the definition (2) to deduce that the trace vanishes $$\tag{12} {\rm tr}(\omega)~:=~\omega^{\mu}{}_{\mu}~=~0. $$

  9. Finally, use eqs. (4), (5), (6), (10) and (12) to prove the sought-for eq. (7): $$2[W^{\alpha}, M]~=~\varepsilon^{\alpha\beta\mu\nu}\left([P_{\beta},M] M_{\mu\nu}+P_{\beta} [M_{\mu\nu},M] \right) $$ $$~=~\varepsilon^{\alpha\beta\mu\nu} \left( P_{\lambda} \omega^{\lambda}{}_{\beta} M_{\mu\nu} - P_{\beta} M_{\nu\lambda} \omega^{\lambda}{}_{\mu} + P_{\beta} M_{\mu\lambda} \omega^{\lambda}{}_{\nu} \right)$$ $$~=~ \left( \omega^{\beta}{}_{\beta^{\prime}}\varepsilon^{\alpha\beta^{\prime}\mu\nu} +\omega^{\mu}{}_{\mu^{\prime}}\varepsilon^{\alpha\beta\mu^{\prime}\nu} +\omega^{\nu}{}_{\nu^{\prime}}\varepsilon^{\alpha\beta\mu\nu^{\prime}} \right) P_{\beta}M_{\mu \nu} $$ $$\tag{13}~=~ -\omega^{\alpha}{}_{\alpha^{\prime}}\varepsilon^{\alpha^{\prime}\beta\mu\nu} P_{\beta}M_{\mu \nu} ~=~ -2\omega^{\alpha}{}_{\beta}W^{\beta}.$$

$^1$ Alternatively, one may establish eq. (10) by purely combinatorical reasoning, without the aid of eq. (9).

  • $\begingroup$ (8)-(10) seems a bit unnatural. The only property of $\omega_{\mu\nu}$ is anti-symmetry, so (10) should be able to be derived from that only. $\endgroup$
    – Siyuan Ren
    Commented Sep 29, 2012 at 2:47
  • $\begingroup$ I got $[{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\mu \nu }},M] = {\varepsilon ^{\alpha \beta \mu \nu }}{P_\lambda }{M_{\mu \nu }}{\omega ^\lambda }_\beta + {\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\mu \lambda }}{\omega ^\lambda }_\nu - {\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\nu \lambda }}{\omega ^\lambda }_\mu $ by substituting in your suggestion. Pardon my stupidity, for I don't know how to proceed. Eq.(10) doesn't seem relevant. $\endgroup$
    – Siyuan Ren
    Commented Sep 30, 2012 at 9:16
  • $\begingroup$ I updated the answer. (Note that eq. numbers may have shifted.) $\endgroup$
    – Qmechanic
    Commented Oct 1, 2012 at 18:46

The first four terms of your calculation vanish. They can be brought into the form (using antisymmetry of the epsilon and of M)

$\epsilon^{\alpha\beta\nu}_\rho M_{\nu\sigma}-\epsilon^{\alpha\beta\nu}_\sigma M_{\nu\rho}$ If you stick in numbers for the indices you'll quickly see that this cancels. The other two terms should then give the wanted result.

The trick to use for the other two terms is (check out wikipedia for more)


  • $\begingroup$ The indices in $\epsilon^{\alpha\beta\nu}_\rho M_{\nu\sigma}-\epsilon^{\alpha\beta\nu}_\sigma M_{\nu\rho}$ is placed wrong. $\endgroup$
    – Siyuan Ren
    Commented Sep 29, 2012 at 1:40
  • $\begingroup$ Ups, indeed. In my sketch in a piece of paper i put all free indices downstairs. Ma bad :) Calculation goes through nonetheless. $\endgroup$ Commented Sep 29, 2012 at 8:28
  • $\begingroup$ I don't want to stick in numbers to check that blah blah vanishes. If I stick numbers to check this, I can simply stick numbers to check (*). $\endgroup$
    – Siyuan Ren
    Commented Sep 30, 2012 at 8:13

You can rewrite the Pauli-Lubanski vector as $W_{\mu}=(W_0,\mathbf{W})$ and prove the following identities first $$ P\cdot W=0,\quad W_0=\mathbf{P}\cdot\mathbf{J},\quad \mathbf{W}=P_0\mathbf{J}-\mathbf{P}\times\mathbf{K} $$ where $\mathbf{P}$, $\mathbf{J}$, and $\mathbf{K}$ are defined as (I am not sure whether we are using the same conventions) $$ \mathbf{P}=\left\{P^1,P^2,P^3\right\},\quad \mathbf{J}=\left\{M^{32},M^{13},M^{21}\right\},\quad \mathbf{K}=\left\{M^{01},M^{02},M^{03}\right\} $$ Then, the proofs for $[W^0,M^{\rho\sigma}]$ and $[W^i,M^{\rho\sigma}]$ will become much easier.


I found a more directly way to solve......

$$ \mathrm{R.H.S.}=g^{\mu \rho} W^{\sigma}-g^{\mu \sigma} J^{\rho}=-\frac{1}{2} \epsilon^{\rho \sigma \alpha \beta} \epsilon_{\theta \tau \alpha \beta} g^{\mu \theta} W^{\tau}=-\frac{1}{4} \epsilon^{\rho \sigma \alpha \beta} \epsilon_{\theta \tau \alpha \beta} g^{\mu \theta} \epsilon_{\phantom{\tau}\nu \iota \lambda}^{\tau} P^{\nu} J^{\iota \lambda}=-\frac{1}{2} \epsilon^{\rho \sigma \alpha \beta}\left(P^{\mu} J_{\alpha \beta}+P_{\alpha} J_{\beta}^{\phantom{\beta}\mu}-P_{\beta} J_{\alpha}^{\phantom{\alpha}\mu}\right) $$ $$ \mathrm{L.H.S}=\frac{1}{2} \epsilon_{\phantom{\mu}\kappa \iota \nu}^{\mu}\left[P^{\kappa}\left(g^{\nu \rho} J^{\iota \sigma}-g^{\iota \rho} J^{\nu \sigma}-g^{\sigma \iota} J^{\rho \nu}+g^{\sigma \nu} J^{\rho \iota}\right)+\left(g^{\kappa \rho} P^{\sigma}-g^{\kappa \sigma} P^{\rho}\right) J^{\iota \nu}\right] =\epsilon^{\mu \rho \kappa \iota} P_{\kappa} J_{\iota}^{\phantom{\iota}\sigma}-\epsilon^{\mu \sigma \kappa \iota} P_{\kappa} J_{\iota}^{\phantom{\iota}\rho}+\frac{1}{2}\left(\epsilon^{\mu \rho \iota \nu} P^{\sigma} J_{\iota \nu}-\epsilon^{\mu \sigma \iota \nu} P^{\rho} J_{\iota \nu}\right). $$ That is to prove: $$ \frac{1}{2} \epsilon^{\alpha \beta\{\rho \sigma} P^{\mu\}} J_{\alpha \beta}+\epsilon^{\alpha \beta\{\rho \sigma} P_{\alpha} J_{\beta}^{\phantom{\beta}\mu\}}=0. $$ where $f^{\{abc\}} = f^{abc}+f^{bca} +f^{cab}.$ obviously it was equivalent to: $$ \epsilon_{\theta \kappa \iota \tau} \epsilon^{\theta \rho \sigma \mu} \epsilon_{\alpha \beta}^{\phantom{\alpha\beta}\kappa \iota}\left(P^{\tau} J^{\alpha \beta}+P^{\alpha} J^{\beta \tau}+P^{\beta} J^{\tau \alpha}\right)=0. $$ in the same way: $$ \epsilon^{\theta \kappa \iota \tau} \epsilon_{\theta \rho \sigma \mu} \epsilon_{\phantom{\alpha\beta}\kappa \iota}^{\alpha \beta} \epsilon^{\mu \nu \lambda \omega} \epsilon_{\mu \tau \alpha \beta} P_{\nu} J_{\lambda \omega}=0. $$ sinse $\epsilon^{\theta \kappa \iota \tau} \epsilon_{\phantom{\alpha\beta}\kappa \iota}^{\alpha \beta} \epsilon_{\mu \tau \alpha \beta}=0$,the original proposition has been proved.


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