Product of Two Levi-Civita Symbols
📂Mathematical Physics Product of Two Levi-Civita Symbols Theorem The ϵ i j k \epsilon_{ijk} ϵ ijk Levi-Civita symbol .
ϵ i j k = { + 1 if ϵ 123 , ϵ 231 , ϵ 312 − 1 if ϵ 132 , ϵ 213 , ϵ 321 0 if i = j or j = k or k = i
\epsilon_{ijk} = \begin{cases} +1 & \text{if} \ \epsilon_{123}, \epsilon_{231}, \epsilon_{312}
\\ -1 & \text{if} \ \epsilon_{132}, \epsilon_{213}, \epsilon_{321}
\\ 0 & \text{if} \ i=j \ \text{or} \ j=k \ \text{or} \ k=i \end{cases}
ϵ ijk = ⎩ ⎨ ⎧ + 1 − 1 0 if ϵ 123 , ϵ 231 , ϵ 312 if ϵ 132 , ϵ 213 , ϵ 321 if i = j or j = k or k = i
The δ i j \delta_{ij} δ ij Kronecker delta .
δ i j : = { 1 , i = j 0 , i ≠ j
\delta_{ij} := \begin{cases} 1,&i=j
\\ 0, & i\ne j \end{cases}
δ ij := { 1 , 0 , i = j i = j
Between the product of two Levi-Civita symbols and the Kronecker delta, the following relationships hold:
(a) When one index is the same: ϵ i j k ϵ i l m = δ j l δ k m − δ j m δ k l \epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} ϵ ijk ϵ i l m = δ j l δ km − δ jm δ k l
(b) When two indices are the same: ϵ i j k ϵ i j m = 2 δ k m \epsilon_{ijk}\epsilon_{ijm}=2\delta_{km} ϵ ijk ϵ ijm = 2 δ km
(c) When all three indices are the same: ϵ i j k ϵ i j k = 6 \epsilon_{ijk}\epsilon_{ijk}=6 ϵ ijk ϵ ijk = 6
Explanation Note that the summation symbol ∑ \sum ∑ Einstein notation . This applies to the formulas above as well. Memorizing (a) can be very useful as it’s frequently used. A simple way to remember it is as follows.
Proof (a) Let e i \mathbf{e}_{i} e i ( i = 1 , 2 , 3 ) (i=1,2,3) ( i = 1 , 2 , 3 ) standard unit vectors in 3-dimensional space.
e 1 = ( 1 , 0 , 0 ) , e 2 = ( 0 , 1 , 0 ) , e 3 = ( 0 , 0 , 1 )
\mathbf{e}_{1} = (1, 0, 0),\quad \mathbf{e}_{2} = (0, 1, 0),\quad \mathbf{e}_{3} = (0, 0, 1)
e 1 = ( 1 , 0 , 0 ) , e 2 = ( 0 , 1 , 0 ) , e 3 = ( 0 , 0 , 1 )
Let P i j k P_{ijk} P ijk 3 × 3 3 \times 3 3 × 3 matrix whose 1st, 2nd, and 3rd rows are e i \mathbf{e}_{i} e i e j \mathbf{e}_{j} e j e k \mathbf{e}_{k} e k
P i j k = [ — e i — — e j — — e k — ]
P_{ijk} = \begin{bmatrix}
\text{— } \mathbf{e}_{i} \text{ —} \\
\text{— } \mathbf{e}_{j} \text{ —} \\
\text{— } \mathbf{e}_{k} \text{ —}
\end{bmatrix}
P ijk = — e i — — e j — — e k —
Then, by the properties of determinants, it’s easy to see that det P i j k = ϵ i j k \det P_{ijk} = \epsilon_{ijk} det P ijk = ϵ ijk P 123 P_{123} P 123 identity matrix , hence its determinant is 1 1 1
det P 123 = det P 231 = det P 312 = 1
\det P_{123} = \det P_{231} = \det P_{312} = 1
det P 123 = det P 231 = det P 312 = 1
When different rows are swapped an odd number of times, the sign of the determinant changes, hence,
det P 132 = det P 213 = det P 321 = − 1
\det P_{132} = \det P_{213} = \det P_{321} = -1
det P 132 = det P 213 = det P 321 = − 1
The determinant of a matrix with two or more identical rows is 0 0 0 0 0 0 det P i j k = ϵ i j k \det P_{ijk} = \epsilon_{ijk} det P ijk = ϵ ijk determinants .
ϵ i j k ϵ i l m = det [ — e i — — e j — — e k — ] det [ — e i — — e l — — e m — ] = det [ — e i — — e j — — e k — ] det [ ∣ ∣ ∣ e i e l e m ∣ ∣ ∣ ] ( ∵ det A = det A T ) = det ( [ — e i — — e j — — e k — ] [ ∣ ∣ ∣ e i e l e m ∣ ∣ ∣ ] ) ( ∵ ( det A ) ( det B ) = det ( A B ) ) = det [ e i ⋅ e i e i ⋅ e l e i ⋅ e m e j ⋅ e i e j ⋅ e l e j ⋅ e m e k ⋅ e i e k ⋅ e l e k ⋅ e m ]
\begin{align*}
\epsilon_{ijk}\epsilon_{ilm}
&= \det \begin{bmatrix}
\text{— } \mathbf{e}_{i} \text{ —} \\
\text{— } \mathbf{e}_{j} \text{ —} \\
\text{— } \mathbf{e}_{k} \text{ —}
\end{bmatrix} \det \begin{bmatrix}
\text{— } \mathbf{e}_{i} \text{ —} \\
\text{— } \mathbf{e}_{l} \text{ —} \\
\text{— } \mathbf{e}_{m} \text{ —}
\end{bmatrix} \\
&= \det \begin{bmatrix}
\text{— } \mathbf{e}_{i} \text{ —} \\
\text{— } \mathbf{e}_{j} \text{ —} \\
\text{— } \mathbf{e}_{k} \text{ —}
\end{bmatrix} \det \begin{bmatrix}
\vert & \vert & \vert \\
\mathbf{e}_{i} & \mathbf{e}_{l} & \mathbf{e}_{m} \\
\vert & \vert & \vert
\end{bmatrix} & (\because \det A = \det A^{T}) \\
&= \det \left( \begin{bmatrix}
\text{— } \mathbf{e}_{i} \text{ —} \\
\text{— } \mathbf{e}_{j} \text{ —} \\
\text{— } \mathbf{e}_{k} \text{ —}
\end{bmatrix} \begin{bmatrix}
\vert & \vert & \vert \\
\mathbf{e}_{i} & \mathbf{e}_{l} & \mathbf{e}_{m} \\
\vert & \vert & \vert
\end{bmatrix} \right) & \Big(\because (\det A) (\det B) = \det (AB) \Big) \\
&= \det \begin{bmatrix}
\mathbf{e}_{i} \cdot \mathbf{e}_{i} & \mathbf{e}_{i} \cdot \mathbf{e}_{l} & \mathbf{e}_{i} \cdot \mathbf{e}_{m} \\
\mathbf{e}_{j} \cdot \mathbf{e}_{i} & \mathbf{e}_{j} \cdot \mathbf{e}_{l} & \mathbf{e}_{j} \cdot \mathbf{e}_{m} \\
\mathbf{e}_{k} \cdot \mathbf{e}_{i} & \mathbf{e}_{k} \cdot \mathbf{e}_{l} & \mathbf{e}_{k} \cdot \mathbf{e}_{m}
\end{bmatrix}
\end{align*}
ϵ ijk ϵ i l m = det — e i — — e j — — e k — det — e i — — e l — — e m — = det — e i — — e j — — e k — det ∣ e i ∣ ∣ e l ∣ ∣ e m ∣ = det — e i — — e j — — e k — ∣ e i ∣ ∣ e l ∣ ∣ e m ∣ = det e i ⋅ e i e j ⋅ e i e k ⋅ e i e i ⋅ e l e j ⋅ e l e k ⋅ e l e i ⋅ e m e j ⋅ e m e k ⋅ e m ( ∵ det A = det A T ) ( ∵ ( det A ) ( det B ) = det ( A B ) )
Since e i \mathbf{e}_{i} e i e i ⋅ e j = δ i j \mathbf{e}_{i} \cdot \mathbf{e}_{j} = \delta_{ij} e i ⋅ e j = δ ij
ϵ i j k ϵ i l m = det [ δ i i δ i l δ i m δ j i δ j l δ j m δ k i δ k l δ k m ]
\epsilon_{ijk}\epsilon_{ilm} = \det \begin{bmatrix}
\delta_{ii} & \delta_{il} & \delta_{im} \\
\delta_{ji} & \delta_{jl} & \delta_{jm} \\
\delta_{ki} & \delta_{kl} & \delta_{km}
\end{bmatrix}
ϵ ijk ϵ i l m = det δ ii δ ji δ ki δ i l δ j l δ k l δ im δ jm δ km
Note that we are only considering cases where i i i j , k , l , m j, k, l, m j , k , l , m j , k , l , m j, k, l, m j , k , l , m i i i ϵ i j k ϵ i l m = 0 \epsilon_{ijk}\epsilon_{ilm} = 0 ϵ ijk ϵ i l m = 0
ϵ i j k ϵ i l m = det [ 1 0 0 0 δ j l δ j m 0 δ k l δ k m ] = δ j l δ k m − δ j m δ k l
\epsilon_{ijk}\epsilon_{ilm} = \det \begin{bmatrix}
1 & 0 & 0 \\
0 & \delta_{jl} & \delta_{jm} \\
0 & \delta_{kl} & \delta_{km}
\end{bmatrix} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}
ϵ ijk ϵ i l m = det 1 0 0 0 δ j l δ k l 0 δ jm δ km = δ j l δ km − δ jm δ k l
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(b) This is the case where l = j l=j l = j (a) . Thus, it can be expressed as follows.
ϵ i j k ϵ i j m = δ j j δ k m − δ j m δ k j
\epsilon_{ijk}\epsilon_{ijm} = \delta_{jj}\delta_{km} - \delta_{jm}\delta_{kj}
ϵ ijk ϵ ijm = δ jj δ km − δ jm δ kj
Here, δ j j = 3 \delta_{jj}=3 δ jj = 3 δ j m δ k j = δ m k \delta_{jm}\delta_{kj}=\delta_{mk} δ jm δ kj = δ mk
ϵ i j k ϵ i j m = δ j j δ k m − δ j m δ k j = 3 δ k m − δ m k = 2 δ k m
\epsilon_{ijk}\epsilon_{ijm} = \delta_{jj}\delta_{km} - \delta_{jm}\delta_{kj} = 3\delta_{km} - \delta_{mk} = 2\delta_{km}
ϵ ijk ϵ ijm = δ jj δ km − δ jm δ kj = 3 δ km − δ mk = 2 δ km
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(c) This is the case where m = k m=k m = k (b) , hence,
ϵ i j k ϵ i j k = ∑ k = 1 3 2 δ k k = 2 δ 11 + 2 δ 22 + 2 δ 33 = 2 + 2 + 2 = 6
\epsilon_{ijk}\epsilon_{ijk} = \sum_{k=1}^{3}2\delta_{kk} = 2\delta_{11} + 2\delta_{22} + 2\delta_{33} = 2 + 2 + 2 = 6
ϵ ijk ϵ ijk = k = 1 ∑ 3 2 δ kk = 2 δ 11 + 2 δ 22 + 2 δ 33 = 2 + 2 + 2 = 6
Alternatively, by explicitly writing out all non-zero terms, the following can be obtained.
ϵ i j k ϵ i j k = ∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 1 ϵ i j k ϵ i j k = ϵ 123 ϵ 123 + ϵ 231 ϵ 231 + ϵ 312 ϵ 312 + ϵ 132 ϵ 132 + ϵ 213 ϵ 213 + ϵ 321 ϵ 321 = 6
\begin{align*}
\epsilon_{ijk}\epsilon_{ijk} &=\sum \limits _{i=1} ^{3}\sum \limits _{j=1} ^{3}\sum \limits _{k=1} ^{1} \epsilon_{ijk}\epsilon_{ijk}
\\ &=\epsilon_{123}\epsilon_{123}+\epsilon_{231}\epsilon_{231}+\epsilon_{312}\epsilon_{312}+\epsilon_{132}\epsilon_{132}+\epsilon_{213}\epsilon_{213}+\epsilon_{321}\epsilon_{321}
\\ &=6
\end{align*}
ϵ ijk ϵ ijk = i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 1 ϵ ijk ϵ ijk = ϵ 123 ϵ 123 + ϵ 231 ϵ 231 + ϵ 312 ϵ 312 + ϵ 132 ϵ 132 + ϵ 213 ϵ 213 + ϵ 321 ϵ 321 = 6
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