Sum with multiple Kronecker deltas

Question

I have a problem dealing with the following expression: $$\sum_{i,j,m}\delta_{ij}\delta_{mj}\delta_{jm}x_ip_j$$ which I know it should yield the following result: $$ 3\mathbf{x}\cdot\mathbf{p} $$

I was thinking in terms of matrices and thought that because we have that $$\sum_{j,m}\delta_{mj}\delta_{jm} =\mathbf{I}\mathbf{I} = \mathbf{I}^2 = \mathbf{I}, \sum_{i,j}\delta_{ij}x_ip_j = \mathbf{x} \cdot \mathbf{p}$$ then the output would simply be: $\mathbf{x \cdot p}$, nevertheless it's wrong. since there's a missing $3$ factor out of everything. Hence, I'd like to see how to get the correct result, which I mentioned above. Thanks in advance

Answer 1

You have $δ_{ij} = 1$ iff $i=j$, else it is $0$. Hence your summand $δ_{ij}δ_{mj}δ_{jm} = 1$ iff $i=j=m$, so $$\sum_{i,j,m} δ_{ij}δ_{mj}δ_{jm} x_i p_j = \sum_{i} \sum_j\sum_{m} δ_{ij}δ_{mj}δ_{jm} x_i p_j = \sum_i \sum_j δ_{ij}δ_{jj}δ_{jj} x_i p_j = \sum_i δ_{ii}x_ip_i = \mathbf{x}\cdot \mathbf{p}.$$

Answer 2

You should simply have asked this on the P.SE rather than M.SE.

You typed the original question wrong. It is not $$\delta_{ij}\delta_{mj}\delta_{jm}x_ip_j$$ but rather it is $$\delta_{ij}\delta_{mn}\delta_{nm}x_ip_j$$ with one more summation term. Only then will you get the $3\vec x\cdot\vec p$.