# Computation involving Pauli-Lubanski vector

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.

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.
• 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. Feb 15, 2023 at 16:13
• 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. Feb 15, 2023 at 16:17
• 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 $\epsilon^{0ijk} = \epsilon^{i1jk} = \epsilon^{ij2k} = \epsilon^{ijk3} = \epsilon^{ijk}$? Feb 15, 2023 at 17:07
• 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 $\varepsilon_{0ijk}=-\varepsilon_{ijk}$. Feb 15, 2023 at 17:23