Prove/show that $\epsilon_{ijk}(\hat a_j \hat b_k-\hat b_k \hat a_j)=2\epsilon_{ijk}\hat a_j\hat b_k$

Question

I am simply trying to understand where the factor of $2$ comes from in the RHS of $$\epsilon_{ijk} \left[\hat a_j, \hat b_k \right]=\epsilon_{ijk}\left(\hat a_j \hat b_k-\hat b_k \hat a_j\right)=\epsilon_{ijk}\left(2\hat a_j\hat b_k\right)\tag{a}$$

Here, $\left[\hat a_j, \hat b_k \right]$ is the commutator of operators $\hat a_j,$ and $\hat b_k$ and the Einstein summation convention is being used. Here the operators must not be assumed to commute or anticommute.

I once asked my lecturer about why eqn. $(\mathrm{a})$ is true and the usual story is that the "Levi-Civita is anti-symmetric under exchange of $j$ and $k$, so the two terms are the same up to a minus sign which gives the extra factor of $2$". But I'm finding it hard to understand this statement.

So in trying to convince myself that $(\mathrm{a})$ is true, starting with the LHS of $(\mathrm{a})$, I write it out for $i$th component by summing over $j$ and $k$ which run from $1$ to $2$:

$$\epsilon_{ijk} \left[\hat a_j, \hat b_k \right]=\epsilon_{ijk}\left(\hat a_j \hat b_k-\hat b_k \hat a_j\right)\tag{1}$$ $$=\epsilon_{i12}\left(\hat a_1 \hat b_2-\hat b_2 \hat a_1\right)+\epsilon_{i21}\left(\hat a_2 \hat b_1-\hat b_1 \hat a_2\right)$$ $$=\epsilon_{i12}\left(\hat a_1 \hat b_2-\hat b_2 \hat a_1\right)-\epsilon_{i12}\left(\hat a_2 \hat b_1-\hat b_1 \hat a_2\right)$$ $$=\epsilon_{i12}\left((\hat a_1 \hat b_2-\hat a_2 \hat b_1)-(\hat b_2\hat a_1-\hat b_1 \hat a_2)\right)$$ $$\stackrel{\color{red}{?}}{=}\epsilon_{ijk}\left(\hat a_j \hat b_k-\hat b_k \hat a_j\right)\tag{2}$$ In going from the second to the third equality I used the fact that $\epsilon_{i21}=-\epsilon_{i12}$ due to the fact that the Levi-Civita is totally antisymmetric under interchange of $2$ indices. But the final expression, $(2)$, doesn't show where the factor of $2$ comes from in $(a)$. All the final expression in $(2)$ shows is that the starting equation, $(1)$ is true (assuming my reasoning is correct). However, I'm unsure about the final equality and indicated this with a question mark.

So instead, starting from the RHS of $(\mathrm{a})$ $$\epsilon_{ijk}\left(2\hat a_j\hat b_k\right)=2\left(\epsilon_{i12}\hat a_1\hat b_2+\epsilon_{i21}\hat a_2 \hat b_1 \right)\tag{3}$$ $$=2\left(\epsilon_{i12}\hat a_1\hat b_2-\epsilon_{i12}\hat a_2 \hat b_1 \right)=2\epsilon_{i12}\left(\hat a_1\hat b_2-\hat a_2 \hat b_1 \right)$$ $$=2\epsilon_{i12}\left(\hat a_1\hat b_2-\hat a_2 \hat b_1 \right)=2\epsilon_{ijk}\hat a_j\hat b_k \tag{4}$$

But once again, in eqn. $(4)$ I've proven absolutely nothing, assuming my manipulations are correct I have simply proven that the starting equation, $(3)$ is true.

So to summarize, can anyone please show me where this factor of $2$ comes from? Or put another way, show that $$\fbox{$\epsilon_{ijk}\left(\hat a_j \hat b_k-\hat b_k \hat a_j\right)=2\epsilon_{ijk}\hat a_j\hat b_k$}\tag{a}$$

The proof does not have to be rigorous, it need only apply to a particular case, (like I tried to calculate above for the case of $j$ and $k$ running from $1$ to $2$).

Answer 1

Starting with: $\varepsilon_{ijk}a_jb_ k-\varepsilon_{ijk}a_kb_j $

We take the second term: $-\varepsilon_{ijk}a_kb_j=+\varepsilon_{ikj}a_kb_j$. This works because $ijk\to ikj$ is an odd permutation, so it generates a negative sign. We’re summing over $j$ and $k$, so provided we stay consistent we can rename them. Thus, the second term becomes $+\varepsilon_{ijk}a_jb_k$, which is identical to the first term, hence the factor of 2.

EDIT

Starting with: $\varepsilon_{ijk}[a_j,b_k]=\varepsilon_{ijk}a_jb_k-\varepsilon_{ijk}b_ka_j$ (Compare this to the expression above; they are different)

Then, manipulating this:

$$\varepsilon_{ijk}a_jb_k-\varepsilon_{ijk}b_ka_j=\varepsilon_{ijk}a_jb_k+\varepsilon_{ikj}b_ka_j=\varepsilon_{ijk}a_jb_k+\varepsilon_{ijk}b_ja_k$$

(Note how this is different from the math before the edit. The idea is the same, but the operators are never interchanged)

Continuing forward by adding "$0$":

$$\varepsilon_{ijk}(a_jb_k+b_ja_k+a_jb_k-a_jb_k)=2\varepsilon_{ijk}a_jb_k+\varepsilon_{ijk}(b_ja_k-a_jb_k)$$

$$=2\varepsilon_{ijk}a_jb_k-\varepsilon_{ikj}b_ja_k-\varepsilon_{ijk}a_jb_k=2\varepsilon_{ijk}a_jb_k-\varepsilon_{ijk}b_ka_j-\varepsilon_{ijk}a_jb_k=2\epsilon_{ijk}a_jb_k-\varepsilon_{ijk}\{a_j,b_k\}$$

Where $\{a_k,b_j\}$ denotes an anticommutator. But this means:

$$\varepsilon_{ijk}[a_j,b_k]=2\varepsilon_{ijk}a_jb_k-\varepsilon_{ijk}\{a_j,b_k\}$$

Thus, the property you give in your question is only possible if $\{a_j,b_k\}=0$.

Answer 2

This didn't fit in the comment section:

Something is very wrong here. For the case $n=3$ and $i=1$ the only non-zero contribution to the contracted sum is

$$2\varepsilon_{1jk}\hat{a}_j \hat{b}_k=2(\varepsilon_{123}\hat{a}_2 \hat{b}_3+\varepsilon_{132}\hat{a}_3 \hat{b}_2)=2(\hat{a}_2 \hat{b}_3-\hat{a}_3 \hat{b}_2)$$

The same calculation for the expression $\varepsilon_{ijk}(\hat{a}_j \hat{b}_k-\hat{b}_k \hat{a}_j)$ ($n=3$ and $i=1$)

$$\varepsilon_{1jk}(\hat{a}_j \hat{b}_k-\hat{b}_k \hat{a}_j)=\hat{a}_2\hat{b}_3 - \hat{a}_3\hat{b}_2 + \hat{b}_2\hat{a}_3 - \hat{b}_3\hat{a}_2$$

Clearly, those two are not the same unless somehow

$$ \hat{a}_2\hat{b}_3 - \hat{a}_3\hat{b}_2 = \hat{b}_2\hat{a}_3 - \hat{b}_3\hat{a}_2 $$

One way you can make this work is to randomly add the anti symmetric condition $\hat{a}_j\hat{b}_k=-\hat{b}_k\hat{a}_j$

Answer 3

Not an answer but too long for a comment:

So the calculations I made in my question were nonsensical as I was trying to consider just the $i$th component only, but as mentioned by @ContraKinta $i,j,k$ run from $1$ to $3$ to give the $27$ ($6$ non-zero) components of the Levi-Civita.

What I was trying to do is to show that $$\varepsilon_{ijk}(\hat a_j \hat b_k-\hat b_k \hat a_j)=2\epsilon_{ijk}\hat a_j\hat b_k$$ by explicitly writing out each component:

$$\begin{align}\varepsilon_{ijk}\left(\hat{a}_j \hat{b}_k-\hat{b}_k \hat{a}_j\right)&=\varepsilon_{ijk}\hat{a}_j \hat{b}_k-\varepsilon_{ijk}\hat{b}_k \hat{a}_j\\&=\varepsilon_{123}\hat{a}_2\hat{b}_3 + \varepsilon_{132}\hat{a}_3\hat{b}_2 - \varepsilon_{123}\hat{b}_3\hat{a}_2 - \varepsilon_{132}\hat{b}_2\hat{a}_3 \\&+\varepsilon_{213}\hat{a}_1\hat{b}_3 + \varepsilon_{231}\hat{a}_3\hat{b}_1 - \varepsilon_{213}\hat{b}_3\hat{a}_1 -\varepsilon_{231}\hat{b}_1\hat{a}_3\\&+\varepsilon_{312}\hat{a}_1\hat{b}_2 + \varepsilon_{321}\hat{a}_2\hat{b}_1 - \varepsilon_{312}\hat{b}_2\hat{a}_1 - \varepsilon_{321}\hat{b}_1\hat{a}_2\\&=\varepsilon_{123}\hat{a}_2\hat{b}_3 - \varepsilon_{123}\hat{a}_3\hat{b}_2 - \varepsilon_{123}\hat{b}_3\hat{a}_2 + \varepsilon_{123}\hat{b}_2\hat{a}_3 \\&-\varepsilon_{123}\hat{a}_1\hat{b}_3 + \varepsilon_{123}\hat{a}_3\hat{b}_1 + \varepsilon_{123}\hat{b}_3\hat{a}_1 -\varepsilon_{123}\hat{b}_1\hat{a}_3\\&+\varepsilon_{123}\hat{a}_1\hat{b}_2 - \varepsilon_{123}\hat{a}_2\hat{b}_1 - \varepsilon_{123}\hat{b}_2\hat{a}_1 + \varepsilon_{123}\hat{b}_1\hat{a}_2\\&=\varepsilon_{123}\left(\hat a_2 \hat b_3 - \hat a_3\hat b_2\right)-\varepsilon_{123}\left(\hat b_2 \hat a_3 - \hat b_3\hat a_2\right)\\&+\varepsilon_{123}\left(\hat a_3 \hat b_1 - \hat a_1\hat b_3\right)-\varepsilon_{123}\left(\hat b_1 \hat a_3 - \hat b_3\hat a_1\right)\\&+\varepsilon_{123}\left(\hat a_1 \hat b_2 - \hat a_2\hat b_1\right)-\varepsilon_{123}\left(\hat b_2 \hat a_1 - \hat b_1\hat a_2\right)\end{align}$$ where after the third equality I have written all $12$ terms using only $\varepsilon_{123}$ in order to try to spot a pattern and write it in a compact way. But I don't see a way of doing this via this method to prove the statement in the title, so it seems the only way to show that $$\varepsilon_{ijk}(\hat a_j \hat b_k-\hat b_k \hat a_j)=2\epsilon_{ijk}\hat a_j\hat b_k$$ is to do it in an abstract way (without specifying the values the indices take) as is done in the answer by @moboDawn_φ .