# How to get Levi-Cevita symbol in the derivation for angular momentum using Noether's theorem? (David Tong Ex Sheet 1 Q6)

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

• it's just the cross product. The expression is just a fancy way of writing $({\vec x} \times {\vec p})_{i}$ Commented Oct 29, 2020 at 0:53
• To be specific, $(\mathbf{a} \times \mathbf{b})^i = \epsilon_{ijk} a^j b^k$ Commented Oct 29, 2020 at 3:34
• @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 Commented Oct 29, 2020 at 14:44
• @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 Commented Oct 29, 2020 at 14:45
• @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 $\epsilon_{ijk}a^{j}b^{k}$ Commented Oct 29, 2020 at 15:24

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: $$L_i = \epsilon_{ijk}L^{jk}$$