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Working through David Tong's sheet here https://www.damtp.cam.ac.uk/user/tong/qft/oh1.pdf and can't follow how to get the Levi-Cevita symbol out the front? Its equation 15. I was looking at trying to use an identity with the Kronecker Deltas in the energy momentum tensor but I am really stuck. Below is the conserved charge, where does the $\epsilon_{ijk}$ come from?

$$Q_i = \epsilon_{ijk} \int d^3x (x^jT^{0k} - x^kT^{0j})$$

I can follow up to the conserved current and get the correct value for charge but I am missing the Levi-Cevita

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    $\begingroup$ it's just the cross product. The expression is just a fancy way of writing $({\vec x} \times {\vec p})_{i}$ $\endgroup$
    – Zo the Relativist
    Commented Oct 29, 2020 at 0:53
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    $\begingroup$ To be specific, $(\mathbf{a} \times \mathbf{b})^i = \epsilon_{ijk} a^j b^k$ $\endgroup$
    – Nihar Karve
    Commented Oct 29, 2020 at 3:34
  • $\begingroup$ @JerrySchirmer thanks, I feel silly asking this but where exactly does the cross product come into this? I've tried expanding by substituting in what I know $T^{0k}$ is but its getting kinda messy $\endgroup$
    – milkcookie
    Commented Oct 29, 2020 at 14:44
  • $\begingroup$ @NiharKarve thanks, where are you getting the cross product in the equation? I'm a bit stuck still sorry, i have the conserved current $j^\mu$ but struggling to use the Levi-Civita still in the charge $\endgroup$
    – milkcookie
    Commented Oct 29, 2020 at 14:45
  • $\begingroup$ @milkybean: $T^{0k}$ is the same thing as the three-momentum ${\vec p}^{k}$. Then, remember that determinant rule for calculating the cross product? Well, the signs in the terms in the determiant alternate in exactly the same way that $\epsilon_{ijk}$ do, right? So, the $x$ component of ${\vec a} \times {\vec b}$ is $a_{y}b_{z} - a_{z}b_{y}$, for example. Put it all together, and you can work out that the $i$ component of ${\vec a} \times {\vec b}$ is $\epsilon_{ijk}a^{j}b^{k}$ $\endgroup$
    – Zo the Relativist
    Commented Oct 29, 2020 at 15:24

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The conserved quantity that arises due to spatial rotational invariance is $$ L^{jk} = Q^{jk} = \int \mathrm{d}^3 x\ (x^j T^{0k} - x^k T^{0j}) $$

Unlike the classical angular momentum pseudovector $\mathbf{x \times p}$, this is expressed as a matrix:

$$ L^{ij} = \begin{pmatrix} 0 & L_{xy} & L_{xz} \\ -L_{xy} & 0 & L_{yz} \\ -L_{xz} & -L_{yz} & 0 \end{pmatrix} $$

whose components may look familiar - they are similar to the magnetic field components in the electromagnetic field tensor $F^{jk}$. Again, similar to the electromagnetic case, to convert $L^{jk}$ back into the classical vector form, you have to use the Levi-Civita symbol: $$ L_i = \epsilon_{ijk}L^{jk} $$

However, as Luboš Motl remarks here, it is usually more efficient to leave it tensor form.

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