- #1
S. Moger
- 53
- 2
Homework Statement
[itex]M= \begin{pmatrix}
2 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 2\\
\end{pmatrix}[/itex]
Compute [itex]\frac{1}{6}\epsilon_{ijk}\epsilon_{lmn} M_{il} M_{jm} M_{kn}[/itex] .
The Attempt at a Solution
I computed the result which is 4, by realizing that there are 36 non-zero levi-civita containing components to sum. Within this group, for each fixed {i,l} there are 4 possible sets of {in},{km}. As the order of the M_{ab}'s inside the product doesn't matter they can be rearranged to form sums of 2 each at a time. The positive contribution to the sum arises from the (now) three occurences of the product of the diagonal elements [itex]M_{ii} M_{jj} M_{kk} = 2 \cdot 2 \cdot 2 = 8[/itex] (not meant to be read as sums). The negative contributions arise from products of specific non-diagonal components multipied by a specific diagonal component. Because of interchangeability inside the product (or alternatively, in this case, symmetry) all get bundled in pairs of two. Each non-zero contribution (i.e. non-{1,3} permutation) then equals [itex]2 \cdot (-1 \cdot -1 \cdot 2) = 4[/itex], and there are three of them. So the final result turns out to be
[itex]2 (3 \cdot 8 - 3 \cdot 4) = 24[/itex], which divided by 6 leaves us with 4.
However, it feels like I'm missing something here (I'm not using the symmetry in any kind of crucial way). Is this the way you would solve this problem? Would it be possible to "see" the result by simply looking at the expression (i.e. interpreting it before doing the math)? Or for example by expressing the epsilons in deltas?( Edit: I now see it's the determinant. But still, it isn't super apparent from just doing the math mindlessly (which is my type of thing until I feel I grasp the basics.)
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