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I have seen that one can make an operator with $$ L^i=\epsilon^{ijk}x_{j}\partial_{k} $$ But what if I want to make instead items that are sums, instead of differences. For instance $$\mathcal{L}^z=x\partial_y +y\partial_x$$ Is there an object like $|\epsilon_{ijk}|$ that has only 1s (no -1s)? Thanks, |
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The short answer is No. The difference between the right $\epsilon_{abc}$ that changes the sign under transpositions – the antisymmetric one – and the wrong $\epsilon_{abc}$ that doesn't – yours – is that the right one is invariant under rotations while yours is not. The right epsilon tensor is a privileged invariant tensor; yours is just a random tensor, a random collection of components. One may demonstrate it in this way. The $\epsilon$ symbol is a set of coefficients that may be used to construct a single quantity – scalar $Q$ – out of three vectors $S,T,U$: $$ Q = \sum_{a,b,c=1}^3 \epsilon_{abc} S_a T_b U_c $$ If you think about it, $Q=(S\times T)\cdot U$ which is the inner product of a vector and a cross product of two other vectors. Its value is given purely geometrically; it doesn't depend on the coordinates. This $Q$ may also be interpreted as the determinant of a matrix with vectors $S,T,U$ as its rows (or columns). If you replaced the mixed-sign $\epsilon$ tensor by the "purely positive ones", the determinant would change to the sum over permutations without the sign and it wouldn't be rotationally invariant. Consequently, all objects constructed out of your version of $\epsilon$ would depend on the choice of the coordinates. For example, consider $S=e_x$, $T=e_y$, $U=e_z$, unit vectors in the direction of the axes. You get $Q=1$. But you may also rotate $S,T,U$ by 90 degrees around the $z$-axis. Then you will have $S=e_y$, $T=-e_x$, $U=e_z$. The $Q$ constructed out of these vectors will still give $Q=1$ but yours would change to $Q=-1$ so it isn't rotationally symmetric. All tensors that are symmetric under rotations are polynomials in $\epsilon_{abc}$ and $\delta_{ab}$. |
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