Product of two Kronecker delta symbols as a determinant

  • #1
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Homework Statement:

The books says that "it is fairly easy" to verify the following product of two kronecker delta symbols : ##\epsilon_{ijk} \epsilon_{pqr} = \begin{vmatrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \\ \end{vmatrix}##

Relevant Equations:

We know the definition of the krokecker alternating symbol. Here all variables can only take values between 1 and 3. If any of the indices repeat, the value of the symbol is 0. If the indices alternate, the symbol takes on a minus sign.
I don't have a clue as to how to go about proving (or verifying) the equation above. It would be very hard to take individual values of i,j and k and p,q and r for each side and evaluate ##3^6## times! More than that, I'd like a proof more than a verification.

Any help would be welcome.
 

Answers and Replies

  • #2
Infrared
Science Advisor
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Both sides vanish if any two of ##i,j,k## or any two of ##p,q,r## are equal (the LHS is zero since a LC symbol with repeated index is zero, and the RHS is zero because two columns or two rows will be equal).

So we're in the case that ##i,j,k## and ##p,q,r## are pairwise distinct. Since both sides are anti-symmetric under changing two indices from either set, it's enough to check that your identity is true for just one particular choice of indices, say ##i=1, j=2, k=3## and ##p=1,q=2, r=3.## You should think about why this is sufficient if it's not obvious at first.

You can probably also expand the determinant, but I find this way of thinking to be simpler.
 
  • #3
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You should think about why this is sufficient if it's not obvious at first.
I have got your answer mostly, except your quote above. It is not obvious to me as to why should the values of ##i, j \,\text{and}\, k## and those of ##p,q\, \text{and}\, r## be those as you said : 1, 2 and 3. Why not 2, 3 and 1, respectively, for example?
 
  • #4
Infrared
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It follows from the antisymmetry of both sides. Changing ##(i,j,k)=(1,2,3)\to (2,1,3)\to (2,3,1)## changes both ##\varepsilon_{ijk}\varepsilon_{pqr}## and your determinant by a factor of ##-1## twice, and hence leaves them unchanged.
 

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