At page 60 of Weinberg book "Quantum Theory of Fields" (vol 1) it is proved that $J^{\mu\nu}$ transform as a tensor, but I can't understand the step from eq (2.4.7) to (2.4.8). It seems to me that he assumes the following relation:

$(\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} = \omega_{\rho\sigma}\Lambda_\mu^{\,\,\rho}\Lambda_\nu^{\,\,\sigma}J^{\mu\nu}$

where $(\Lambda^{-1})^{\rho}_{\,\,\mu} = \Lambda_\mu^{\,\,\rho}$

but I can't find a way to prove that relation, cause in my calculation I get

$(\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} = \omega_{\rho\sigma}(\Lambda^T)_{\phantom.\mu}^{\rho}\Lambda_\nu^{\phantom.\sigma}J^{\mu\nu}$

Note the different position of the indexes and the transpose in the first $\Lambda$.

In the Wu-Ki Tung book he proves that $J_{\mu\nu}$ transform as a tensor using the following theorem (page 185, 10.2-14):

$\Omega\Lambda(\omega)\Omega^{-1} = \Lambda(\omega^\prime)\,\,,$ where $\omega^{\prime\mu\nu}=\Omega^{\mu}_{\,\,\lambda}\Omega^{\nu}_{\,\,\sigma}\omega^{\lambda\sigma}$

$\Omega$ is a generic Lorentz transformation and $\Lambda(\omega) = e^{-i/2\omega^{\mu\nu}J_{\mu\nu}}$, but again there's no explicit proof of that theorem.

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    $\begingroup$ Unfortunately the reply to the old post elaborate the steps for Weinberg book, but it still doesn't give explanation of the step that I can't understand $\endgroup$
    – Andrea
    Jan 3, 2022 at 11:08
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    $\begingroup$ This is the step there that I still can't prove: $\frac{1}{2}(\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} + .... = \frac{1}{2}\Lambda_\mu^{\phantom\mu\rho}\omega_{\rho\sigma} (\Lambda^{-1})^\sigma_{\phantom\sigma\nu} J^{\mu\nu} + ... $ $\endgroup$
    – Andrea
    Jan 3, 2022 at 11:15
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    $\begingroup$ @G.Blaickner I can't figure out why it uses $\Lambda_{\mu}^{\,\,\rho}$ which is $\Lambda^{-1}$, instead of $\Lambda_{\,\,\mu}^{\rho}$ which is $\Lambda$ $\endgroup$
    – Andrea
    Jan 3, 2022 at 12:10
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    $\begingroup$ My understanding is that $\Lambda$, $\omega$, and $\Lambda^{-1}$ are three matrixes so it's the product of three matrixes. In coordinate $\Lambda$ is $\Lambda^{\rho}_{\,\,\mu}$, not $\Lambda^{\,\,\rho}_{\mu}$. I am fine with $(\Lambda^{-1})^{\rho}_{\,\,\mu} = \Lambda_\mu^{\,\,\rho}$ $\endgroup$
    – Andrea
    Jan 3, 2022 at 12:24
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    $\begingroup$ no, ${\Lambda_\mu}^\rho \neq {(\Lambda^T)^\rho}_\mu$. The right identity is ${\Lambda_\mu}^\rho = {(\Lambda^T)_\rho}^\mu$. $\endgroup$ Jan 7, 2022 at 16:00

3 Answers 3


Is there a typo in the specific relation you're wondering about?

By definition $$ \Lambda=({\Lambda^\rho}_\mu)\qquad\text{and}\qquad\Lambda^{-1}=({\Lambda_\nu}^\sigma). $$ So by definition $$(\Lambda\omega\Lambda^{-1})_{\mu\nu}={\Lambda^\rho}_\mu\omega_{\rho\sigma}{\Lambda_\nu}^\sigma=\omega_{\rho\sigma}{\Lambda^\rho}_\mu{\Lambda_\nu}^\sigma $$ and should the relation be $$ (\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu}={\Lambda^\rho}_\mu\omega_{\rho\sigma}{\Lambda_\nu}^\sigma J^{\mu\nu}=\omega_{\rho\sigma}{\Lambda^\rho}_\mu{\Lambda_\nu}^\sigma J^{\mu\nu}\,? $$ Specifically, note the indices in ${\Lambda^\rho}_\mu$, which is where the typo seems to be? And this is all just the definitions of the tensors in use?

  • $\begingroup$ My calculation is the same as yours and that's the reason I was asking here. it's not a typo in the book cause actually the correct transformation is the one in the book (cause it's the same on other book) $\endgroup$
    – Andrea
    Jan 6, 2022 at 20:45
  • $\begingroup$ Being the same in two books doesn't necessarily mean it's correct. I think the relation you believe is assumed is definitely incorrect. So either it wasn't used or there is a typo/mistake in the books. $\endgroup$
    – g.s
    Jan 6, 2022 at 23:18

Weinberg is correct. Let us interpret the action of the Lorentz group actively after having chosen a Minkowskian reference frame. In that case the Lorentz transformation is a tensor ${\Lambda^a}_b$ of type $(1,1)$, it being a liner transformation of vectors. We can therefore use the standard procedure of lowering and raising indices by means of the Minkowski metric $\eta$.

First of all notice that the coefficients $\omega$ are introduced in the book with the same type of indices as the ones of the Lorentz matrices in the fundamental representation: $${\omega^\mu}_\nu\:.$$

That is the way one has to interpret the matrix in the parenthesis of $(\Lambda \omega \Lambda^{-1})_{\mu\nu}$. So,

first one computes ${(\Lambda \omega \Lambda^{-1})^\mu}_\nu$,

next he/she lowers the index $\mu$ using the metric.

We also take advantage of the identity $${(\Lambda^{-1})^a}_b= {\Lambda_b}^a$$ (see below why it is correct from two viewpoints).

$${(\Lambda \omega \Lambda^{-1})^\mu}_\nu = {\Lambda^\mu}_\rho {\omega^\rho}_\sigma {(\Lambda^{-1})^\sigma}_\nu= {\Lambda^\mu}_\rho {\omega^\rho}_\sigma {\Lambda_\nu}^\sigma= {\Lambda^\mu}^\rho {\omega_\rho}_\sigma {\Lambda_\nu}^\sigma\:.$$ Hence $$(\Lambda \omega \Lambda^{-1})_{\mu\nu} = {\Lambda_\mu}^\rho {\omega_\rho}_\sigma {\Lambda_\nu}^\sigma = \omega_{\rho\sigma} {\Lambda_\mu}^\rho {\Lambda_\nu}^\sigma\:.$$ That is the wanted identity.


The identity $${(\Lambda^{-1})^a}_b= {\Lambda_b}^a \tag{1}$$ has a correct meaning in this formalism. Indeed, it is equivalent to another true identity as all the following passages are reversible. From (1), $${(\Lambda^{-1T})^b}_a= {\Lambda_a}^b\tag{2}$$ and thus $$\eta_{cb}{(\Lambda^{-1T})^b}_a\eta^{ad}= \eta_{cb}{\Lambda_a}^b\eta^{ad} = {\Lambda^d}_c$$ In matrix form $$\eta (\Lambda^{-1T}) \eta = \Lambda$$ which is true because it immediately arises from the known identities for the elements of the Lorentz group $$\Lambda^{-1} = \eta \Lambda^T \eta\:, \quad \eta= \eta^T = \eta^{-1}$$ where $$\eta := diag(-1,1,1,1)\:.$$

Finally, (2) is also coherent with the fact that ${\Lambda_a}^b$ is the standard notation for the (active) action of the Lorentz group on covariant vectors (the fundamental representation ${\Lambda^a}_b$ acts on contravariant vectors). Let us prove it. If $$G \ni g \mapsto L_g \in GL(V)$$ is a representation of a group, referred to a basis $e_1,\ldots, e_n$ of the vector space $V$ of dimension $n$, the dual action acting on the dual space $V^*$ is just $$G \ni g \mapsto L'_g := (L_g)^{-1T} \in GL(V^*)$$ referred to the dual basis $e^{*1}, \ldots, e^{*n}$. This is nothing but (2) when $G$ is the Lorentz group.

  • $\begingroup$ My understanding is that you get the identity raising/lowering index of the Lorentz matrix. I thought index lowering/raising could be done only on tensor, while $\Lambda$ is a matrix, not a tensor. So in that formalism I can also safely lower/raise the index of Lorentz transformation matrix? $\endgroup$
    – Andrea
    Jan 7, 2022 at 15:46
  • $\begingroup$ Yes it works: just interpret actively the action of the Lorentz group. In that case it describes a linear map from vectors to (different!) vectors. A linear map between tensors is a tensor as well. $\endgroup$ Jan 7, 2022 at 15:53
  • $\begingroup$ It seems to me that lowering/raising of Lorentz matrix can lead to wrong results. As an example, we know that in matrix notation $\Lambda^T\eta\Lambda = \eta$ then multiplying both side by $\Lambda^{-1}\eta$ you get $\Lambda^T = \eta\Lambda^{-1}\eta$. Using that index notation it can be written as $(\Lambda^T)^{\mu}_{\phantom.\nu} = \eta_{\nu\alpha}(\Lambda^{-1})^{\alpha}_{\phantom.\beta}\eta^{\beta\mu} = \eta_{\nu\alpha}\Lambda_{\beta}^{\phantom.\alpha}\eta^{\beta\mu}=\Lambda^{\mu}_{\phantom.\nu}$ then $\Lambda^T=\Lambda$ which obviously is wrong. Is there an error in my calculus? $\endgroup$
    – Andrea
    Jan 9, 2022 at 17:17
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    $\begingroup$ Your first identity is incorrect because in the right hand side $\nu$ should be the rightmost index and $\mu$ the leftmost one. I agree with you that this approach is awkward and confused and that the identity in Weinberg's book should be established without referring to that lowering and raising procedure with Lorentz matrices. However the final identity is correct: the selfadjoint generators of the unitary rep of Lorentz group must transform as tensors. $\endgroup$ Jan 9, 2022 at 20:31

I believe a cleaner way to prove it is the following.

It is not difficult to verify that the 4-vectors matrix representation of the generators of the Lorentz algebra can be written as follow (see Wu-Ki Tung "Group Theory" page 185 or Maggiore "A modern introduction to Quantum Field Theory" page 19):

$(J^{\mu\nu})^\sigma_{\phantom.\alpha} = i(\eta^{\mu\sigma}\delta^\nu_\alpha - \eta^{\nu\sigma}\delta^\mu_\alpha)$

where $\mu\nu$ identifies the generator and $\sigma\alpha$ the raw,column matrix element of that generator.

Remind also the formula for the inverse of the Lorentz trasformation matrix:

$(\Lambda^{-1})^\rho_{\phantom.\sigma} = \Lambda^\chi_{\phantom.\phi}\eta_{\sigma\chi}\eta^{\rho\phi}$

Using the two equations above is not difficult to get the following two equations:

$(\Lambda^{-1})^\rho_{\phantom.\sigma}(J^{\mu\nu})^\sigma_{\phantom.\alpha}\Lambda^\alpha_{\phantom.\beta} = i(\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\beta} - \Lambda^\mu_{\phantom.\beta}\Lambda^\nu_{\phantom.\phi})\eta^{\rho\phi}$

$(J^{\phi\sigma})^\rho_{\phantom.\beta}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma} = i(\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\beta} - \Lambda^\mu_{\phantom.\beta}\Lambda^\nu_{\phantom.\phi})\eta^{\rho\phi}$

Since the RHS are equal then also the LHS are equal, and you get:

$(\Lambda^{-1})^\rho_{\phantom.\sigma}(J^{\mu\nu})^\sigma_{\phantom.\alpha}\Lambda^\alpha_{\phantom.\beta} = (J^{\phi\sigma})^\rho_{\phantom.\beta}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma}$

Note that in the LHS you have the product of the three matrices $(\Lambda^{-1}J^{\mu\nu}\Lambda)$, while in the RHS you have a linear combination of the matrices $J^{\phi\sigma}$, and on both side $(\rho,\beta)$ is the (raw,column) that identify the element in the resultant matrix: $(M^{\mu\nu})^\rho_{\phantom.\beta}$.

So as a matrix equation It can be written as:

$\Lambda^{-1}J^{\mu\nu}\Lambda = J^{\phi\sigma}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma}$

Which proves that the generators $J^{\mu\nu}$ transform as a tensor under similarity Lorentz transformation.

The proof is much more easier when the generators of the Lorentz algebra are represented by differential operators instead of matrices (for the definition of $J^{\mu\nu}$ with differential operator see Schwartz "Quantum Filed Theory" page 161, or Zee "Group Theory" page 434, or Maggiore "A modern introduction to Quantum Field Theory" page 30):

$ J^{\mu\nu} = i(x^\mu\partial^\nu - x^\nu\partial^\mu) $

$ x^\mu \xrightarrow[\text{}]{\text{Lorentz}} \Lambda^\mu_{\phantom.\rho} x^{\rho}\,\,,\,\, \partial^\mu \xrightarrow[\text{}]{\text{Lorentz}} \Lambda^\mu_{\phantom.\rho} \partial^{\rho}$

$ J^{\mu\nu} \xrightarrow[\text{}]{\text{Lorentz}} i(\Lambda^\mu_{\phantom.\rho} x^{\rho}\Lambda^\nu_{\phantom.\sigma}\partial^\sigma - \Lambda^\nu_{\phantom.\rho} x^{\rho}\Lambda^\mu_{\phantom.\sigma}\partial^\sigma) = i(\Lambda^\mu_{\phantom.\rho}\Lambda^\nu_{\phantom.\sigma}x^\rho\partial^\sigma - \Lambda^\nu_{\phantom.\sigma} \Lambda^\mu_{\phantom.\rho}x^\sigma\partial^\rho) = \Lambda^\mu_{\phantom.\rho}\Lambda^\nu_{\phantom.\sigma} J^{\rho\sigma} $


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