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I am trying to check that the "1" component of the Pauli-Lubanski vector for a massless particle with $P^{\mu} = (E, 0, 0, E)^{\mu}$ is $E(-J_1 + K_2)$, but I keep getting $E(-J_1 - K_2)$.

Starting from: $W_{\mu} = \frac{1}{2}\epsilon_{\mu \alpha \beta \gamma}P^{\alpha}M^{\beta \gamma}$,

$M_{0 i} = K_i$,

and $\frac{1}{2}\epsilon_{ijk}M_{jk} = J_i$.

I get,

$W_1 = \frac{1}{2}\epsilon_{1 \alpha \beta \gamma}P^{\alpha}M^{\beta \gamma}$

$= \frac{1}{2}\epsilon_{1 0\beta \gamma}EM^{\beta \gamma} + \frac{1}{2}\epsilon_{1 3 \beta \gamma}E M^{\beta \gamma}$

$= \frac{E}{2}(- \epsilon_{0 1 \beta \gamma}M^{\beta \gamma} + 2 \epsilon_{1 3 0 2 }M^{02}) $

$= \frac{E}{2}(- \epsilon_{0 1 i j}M^{ij} + 2\epsilon_{1 3 0 2 }M^{02})$

$= \frac{E}{2}(- \epsilon_{0 1 i j}M_{ij} - 2\epsilon_{1 3 0 2 }M_{02})$

$= E(-J_1 - \epsilon_{1 3 0 2 }M_{02})$

$= E(-J_1 - \epsilon_{1 3 0 2 }K_2$

$= E(-J_1 + \epsilon^{1 3 0 2 }K_2$

$= E(-J_1 - K_2)$.

But the answer is supposed to be $W_1 = E(-J_1 + K_2$.

I would appreciate it if someone could point out what I'm doing wrong please.

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1 Answer 1

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Check how you define the Levi-Civita symbols, there are a lot of conventions. In your derivation, you use the following relations: $$\varepsilon^{1302}=-1, \qquad \varepsilon_{01ij} = \varepsilon_{1ij}$$ Now, the first condition is equivalent to $\varepsilon^{0123}=+1$, so you are using the convention $\varepsilon_{0123}=-1$. On the other hand, the second condition is only consistent if you define $\varepsilon_{123} =-1$. Even if you could define $\varepsilon_{123}$ this way, I've never seen it, so I will assume it is not your intention.

This means that either $\varepsilon^{1302}=-1$ is wrong or $\varepsilon_{01ij} = \varepsilon_{1ij}$ is wrong. In particular, if you choose the convention $\varepsilon_{0123}=+1$, which is also very common, you would obtain $\varepsilon^{1302}=+1$ solving your sign problem.

EDIT: If you want to define $\varepsilon^{0123}=+1$ and $\varepsilon_{123} =+1$, then you need to modify the second relation to be $\varepsilon_{01ij} = -\varepsilon_{1ij}$, and you would obtain the result $$W_1 = E(J_1 - K_2)$$ which has the correct relative sign. Probably other conventions can account for the last $-$ sign.

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  • $\begingroup$ Thank you. The convention I am following is $\epsilon^{0123} = 1$, so thank you for uncovering at least one of my errors. However, if I then correct the error in the $\epsilon_{01ij}$ term, is the final result not unchanged? $\epsilon_{01ij}M^{ij} = -\epsilon^{01ij}M^{ij}= 2 J^1 = -2 J_1$..? And it certainly doesn't affect the $K_2$ term, which is off by a sign. $\endgroup$
    – Gleeson
    Feb 15, 2023 at 16:13
  • $\begingroup$ Then I don't see any other error in your derivation. But as I said, there are a lot of conventions involved here. For example, if I remember correctly, in Ryder's QFT book, the PL pseudovector is defined with a - sign (and I think he also uses the convention $\varepsilon^{0123}=+1$). So check that you are following all the conventions correctly. $\endgroup$
    – Gaussian97
    Feb 15, 2023 at 16:17
  • $\begingroup$ Once one has chosen on the 4d convention for the epsilon (whether epsilon(0123)= 1 when the indices are up, or down), then how does one infer the inherited implications for the 3d symbol when the symbols are up or down? I am comfortable in the regular 3d case where up/down didn't matter, but it seems to happen often that in a 4d calculation we need to "restrict" to 3d and it is not clear to me how this plays out. Is there a straightforward way? For example, if $\epsilon^{0123} = 1$, then is $\epsilon^{0ijk} = \epsilon^{i1jk} = \epsilon^{ij2k} = \epsilon^{ijk3} = \epsilon^{ijk}$? $\endgroup$
    – Gleeson
    Feb 15, 2023 at 17:07
  • $\begingroup$ I think the most used convention for the 3D symbol is precisely to ignore whether the indices are up or down, simply define cyclic as +1. Of course, you can also define $\varepsilon_{123}=+1$ and change the sign every time you raise an index (though I haven't seen this convention very often). At the end, conventions are completely arbitrary. Once you have defined how the 3D and 4D symbols are defined, finding a relation between them should be trivial. If you want to use $\varepsilon_{0123}=-1$ and the usual 3D symbol, then you need the relation $\varepsilon_{0ijk}=-\varepsilon_{ijk}$. $\endgroup$
    – Gaussian97
    Feb 15, 2023 at 17:23

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