# Pesking & Schroeder Eqn 3.18 (Lorentz algebra)

I am trying to show that (for 4x4 matrices) the representation given by equation 3.18 (Peskin and Schroeder, page 39):
$$(J^{\mu\nu})_{\alpha\beta} =i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})$$
implies the commutation relations in 3.17:
$$[J^{\mu\nu},J^{\rho\sigma}] =i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})$$
For some reason I cannot even get this to work for the $$(\mu,\nu,\rho,\sigma)=(0,1,1,2)$$ component:
$$[J^{01},J^{12}]_{\alpha\beta} =J^{01}_{\alpha\gamma}J^{12}_{\gamma\beta} -J^{12}_{\alpha\gamma}J^{01}_{\gamma\beta} =i(\delta^0_{\alpha}\delta^1_{\gamma}-\delta^0_{\gamma}\delta^1_{\alpha}) i(\delta^1_{\gamma}\delta^2_{\beta}-\delta^1_{\beta}\delta^2_{\gamma}) -i(\delta^1_{\alpha}\delta^2_{\gamma}-\delta^1_{\gamma}\delta^2_{\alpha}) i(\delta^0_{\gamma}\delta^1_{\beta}-\delta^0_{\beta}\delta^1_{\gamma})$$
Now, sums like $$\delta^1_{\gamma}\delta^2_{\gamma}$$ vanish, whereas $$\delta^1_{\gamma}\delta^1_{\gamma}$$ is 1, so we get:
$$[J^{01},J^{12}]_{\alpha\beta}= -\delta^0_{\alpha}\delta^2_{\beta} +\delta^0_{\beta}\delta^2_{\alpha} =i(J^{02})_{\alpha\beta}$$
On the other hand, the right hand side of 3.17 was supposed to give us:
$$i(g^{11}J^{02}-g^{01}J^{12}-g^{12}J^{01}+g^{02}J^{11})_{\alpha\beta} =i((-1)J^{02}-0-0+0)_{\alpha\beta}=-i(J^{02})_{\alpha\beta}$$
ugh...
I suspect I messed up with the metric at some point, but I don't see where.

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## Answers and Replies

dextercioby
Science Advisor
Homework Helper
There's something indeed dubious with your calculation, since there's no summation over \gamma anywhere (actually both \gamma's are downstairs, so they can't be summed over). So try to rewrite this again and sum correctly with the metric.

OK, changed this to
$$[J^{01},J^{12}]_{\alpha\beta} =(J^{01})_{\alpha\gamma}(J^{12})^{\gamma}_{\beta} -(J^{12})_{\alpha}^{\gamma}(J^{01})_{\gamma\beta} =(J^{01})_{\alpha\gamma}g^{\gamma\delta}(J^{12})_{\delta\beta} -(J^{12})_{\alpha\delta}g^{\delta\gamma}(J^{01})_{\\gamma\beta}$$
and this works. Thanks.

Fredrik
Staff Emeritus
Science Advisor
Gold Member
OK, changed this to
$$[J^{01},J^{12}]_{\alpha\beta} =(J^{01})_{\alpha\gamma}(J^{12})^{\gamma}_{\beta} -(J^{12})_{\alpha}^{\gamma}(J^{01})_{\gamma\beta} =(J^{01})_{\alpha\gamma}g^{\gamma\delta}(J^{12})_{\delta\beta} -(J^{12})_{\alpha\delta}g^{\delta\gamma}(J^{01})_{\\gamma\beta}$$
and this works. Thanks.
This time you're starting with an equality that's false. If you're denoting the component on row $\alpha$ column $\beta$ of the matrix $J^{\mu\nu}$ by $(J^{\mu\nu})_{\alpha\beta}$, you can't just change that to $(J^{\mu\nu})^\alpha_\beta$ in the middle of a calculation.

For some reason I cannot even get this to work for the $$(\mu,\nu,\rho,\sigma)=(0,1,1,2)$$ component:
What is
$$i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})$$when the only two indices with the same values are $\nu$ and $\rho$, and that value is 1?

This time you're starting with an equality that's false. If you're denoting the component on row $\alpha$ column $\beta$ of the matrix $J^{\mu\nu}$ by $(J^{\mu\nu})_{\alpha\beta}$, you can't just change that to $(J^{\mu\nu})^\alpha_\beta$ in the middle of a calculation.

i think dexter is implying that the J's are tensors, not matrices, and therefore the product of two Js is their contractions, and hence it is ok that one of the contracted indices is up and one is down (for otherwise, you cannot sum them)

What is
$$i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})$$when the only two indices with the same values are $\nu$ and $\rho$, and that value is 1?

the answer to this is at the last line of my first post:
$-iJ^{\mu\sigma}$
(but when $\mu$ or $\sigma$ are also 1, then you get 0 of course)

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Fredrik
Staff Emeritus
Science Advisor
Gold Member
I am trying to show that (for 4x4 matrices) the representation given by equation 3.18 (Peskin and Schroeder, page 39):
$$(J^{\mu\nu})_{\alpha\beta} =i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})$$
implies the commutation relations in 3.17:
$$[J^{\mu\nu},J^{\rho\sigma}] =i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})$$
I tried to do this calculation myself. I'm assuming that you meant that the first formula defines the 16 matrices $J^{\mu\nu}$ by, for each one of those matrices, specifying its component on row $\alpha$, column $\beta$, for all $\alpha,\beta$.

The result I'm getting is $$[J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}= i\delta^\nu_\gamma\delta^\rho_\gamma(J^{\mu\sigma})_{\alpha\beta}+\cdots$$ (Yes, I meant that there's a sum over all values of $\gamma$ even though they're both downstairs). Clearly each term of $\delta^\nu_\gamma\delta^\rho_\gamma$ is =0 when $\rho\neq\nu$. When $\rho=\nu$, one of the terms is 1 and the others are 0, so they add up to 1. So the result is never -1, and that means it can't be true that $\delta^\nu_\gamma\delta^\rho_\gamma=g^{\nu\rho}$ for all $\nu,\rho$.

I'm not ruling out the possibility that I made some silly mistake, but right now I think the result we're supposed to get looks wrong. Edit: The result isn't wrong, but maybe the definition of the matrices is.

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I tried to do this calculation myself. I'm assuming that you meant that the first formula defines the 16 matrices $J^{\mu\nu}$ by, for each one of those matrices, specifying its component on row $\alpha$, column $\beta$, for all $\alpha,\beta$.
right, this is what i meant
The result I'm getting is $$[J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}= i\delta^\nu_\gamma\delta^\rho_\gamma(J^{\mu\nu})_{\alpha\beta}+\cdots$$ (Yes, I meant that there's a sum over all values of $\gamma$ even though they're both downstairs). Clearly each term of $\delta^\nu_\gamma\delta^\rho_\gamma$ is =0 when $\rho\neq\nu$. When $\rho=\nu$, one of the terms is 1 and the others are 0, so they add up to 1. So the result is never -1, and that means it can't be true that $\delta^\nu_\gamma\delta^\rho_\gamma=g^{\nu\rho}$ for all $\nu,\rho$.

I'm not ruling out the possibility that I made some silly mistake, but right now I think the result we're supposed to get looks wrong.

yes, this is the same problem i was facing. i tried to demonstrate it for specific indices, in my first post.

i think it comes down to how we multiply one J by the other
Edit: if we multiply J's as matrices, then you are right and there is probably something wrong with formula 3.17 (the definition of the matrices)

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Fredrik
Staff Emeritus
Science Advisor
Gold Member
I think I got it. For each $\mu,\nu$, the formula $(J^{\mu\nu})_{\alpha\beta} =i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})$ defines 16 numbers, but those numbers aren't the $\alpha\beta$ components (row $\alpha$, column $\beta$) of the matrix $J^{\mu\nu}$. Instead $J^{\mu\nu}$ is defined as the matrix with $\alpha\beta$ component $(J^{\mu\nu})^\alpha{}_\beta =g^{\alpha\gamma}(J^{\mu\nu})_{\gamma\beta}$. Similarly, the $\alpha\beta$ component of the commutator is $[J^{\mu\nu},J^{\rho\sigma}]^\alpha{}_\beta$, not $[J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}$.

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