E-permutation and Kronecker delta identity

[hotvette]

Homework Statement

\text{Show that } \epsilon_{ijk} \epsilon_{mjk} = 2\delta_{im}

Homework Equations

<br /> \begin{equation*}<br /> \epsilon_{ijk} \epsilon_{mnp} =<br /> \left| \! \begin{array}{ccc} <br /> \delta_{im} &amp; \delta_{in} &amp; \delta_{ip} <br /> \\ \delta_{jm} &amp; \delta_{jn} &amp; \delta_{jp}<br /> \\ \delta_{km} &amp; \delta_{kn} &amp; \delta_{kp} \end{array} \! \right| <br /> \end{equation*}

The Attempt at a Solution

<br /> \begin{align*}<br /> &amp;\text{Evaluating the first term of the determinant, I get:}<br /> \\<br /> &amp;\delta_{im}<br /> \left| \! \begin{array}{cc} <br /> \delta_{jj} &amp; \delta_{jk}<br /> \\ \delta_{kj} &amp; \delta_{kk} \end{array} \! \right| <br /> = \delta_{im}(\delta_{jj}\delta_{kk}-\delta_{jk}\delta_{kj}) = \delta_{im}(9-3) = 6\delta_{im}<br /> \\<br /> &amp; \text{But I&#039;m not sure that&#039;s correct. If I just look at } \delta_{ii} \text{ , I get} \\<br /> &amp; \delta_{ii} = \delta_{11} + \delta_{22} + \delta_{33} = 3<br /> \text{ which means }<br /> \delta_{jj}\delta_{kk} = 3(3) = 9<br /> \end{align*}
Am I doing something wrong?

 

[Fightfish]

Why do you think you're doing something wrong though? While I prefer sneakier (or rather more lazy) approaches, this seems to work out fine. The other two terms in the determinant should add up nicely with the first term you got to give you the correct answer.

 

[hotvette]

Hmm, I guess I picked the easiest term to evaluate If I now look at the 2nd term of the determinant:
\begin{equation*}
- \delta_{ij}(\delta_{jm}\delta_{kk}-\delta_{jk}\delta_{km})= - \delta_{ij}\delta_{jm}\delta_{kk} + \delta_{ij}\delta_{jk}\delta_{km}
\end{equation*}
I get for the first part:
\begin{equation*}
- \delta_{ij}\delta_{jm}\delta_{kk} =-3\delta_{ij}\delta_{jm}
\end{equation*}
And I'm not sure what to do next. I can see that the expression is zero unless j=i, which means:
\begin{equation*}
-3\delta_{ij}\delta_{jm} = -3\delta_{jj}\delta_{im} = -9 \delta_{im}
\end{equation*}
For the 2nd part of the 2nd term:
\begin{equation*}
\delta_{ij}\delta_{jk}\delta_{km}
\end{equation*}
Using the same logic, the sum will be zero unless i=j=k. Thus:
\begin{equation*}
\delta_{ij}\delta_{jk}\delta_{km} = \delta_{jj}\delta_{kk}\delta_{im} = 9\delta_{im}
\end{equation*}
Something tells me this isn't right.

 

[Fightfish]

And I'm not sure what to do next. I can see that the expression is zero unless j=i, which means:
\begin{equation*}
-3\delta_{ij}\delta_{jm} = -3\delta_{jj}\delta_{im} = -9 \delta_{im}
\end{equation*}

Nope, this isn't right. $\delta_{ij}\delta_{jm} = \delta_{im}$. You can't replace $\delta_{ij}$ with $\delta_{jj}$ because "the expression is zero unless j=i" - that statement is what $\delta_{ij}$ itself means!

Using the same logic, the sum will be zero unless i=j=k. Thus:
\begin{equation*}
\delta_{ij}\delta_{jk}\delta_{km} = \delta_{jj}\delta_{kk}\delta_{im} = 9\delta_{im}
\end{equation*}

Similarly, $\delta_{ij}\delta_{jk}\delta_{km} = \delta_{im}$.

As a general rule of thumb, the "index replacement rule" $\delta_{ij} A_{j} = A_{i}$ can be applied to products of kronecker deltas as well.

 

[PeroK]

Am I doing something wrong?

What about a more intuitive approach? If $i \ne m$ then at least one of $\epsilon_{ijk}, \epsilon_{mjk}$ must be $0$

And if $i = m$ you have $\epsilon_{ijk}\epsilon_{ijk}$

For any $i$ there are only two choices for $j, k$ for which $\epsilon_{ijk} \ne 0$ $\dots$

 

[hotvette]

Thanks for the replies. Let me ponder them.

 

[hotvette]

I think I understand now. One way to think of \delta_{ij} is that it is either numeric or an operator. For example: \begin{align*}&\delta_{ii}=3 \\&\delta_{ik}\delta_{jk}=\delta_{ij}\\&\delta_{ik}\delta_{jk}\delta_{nn}=3\delta_{ij}\end{align*}

 

[Fightfish]

I think I understand now. One way to think of \delta_{ij} is that it is either numeric or an operator. For example: \begin{align*}&\delta_{ii}=3 \\&\delta_{ik}\delta_{jk}=\delta_{ij}\\&\delta_{ik}\delta_{jk}\delta_{nn}=3\delta_{ij}\end{align*}

Well, the kronecker delta is a tensor, and when you have repeated indices, you are actually performing a contraction that reduces the rank of the tensor. Another way of seeing it is that the kronecker delta can be represented as an identity matrix, so that $\delta_{ik}\delta_{kj}$ is equivalent to a matrix multiplication of two identity matrices and $\delta_{ii}$ is simply the trace of the identity matrix.

 

[hotvette]

I think I understand. Thanks!