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  • $\begingroup$ This is pretty close to what I came up with. The /: and Notation is a neat trick and will definitely clean up the code some. I am curious though why del /: del[i_, j_]^_ := del[i, j]; is necessary? $\endgroup$ Commented Feb 7, 2013 at 14:31
  • $\begingroup$ When you use the commutator rule two or more times in succession it generates products of Kronecker deltas, so terms such as del[i,j]^2 (etc) can arise. The del /: del[i_, j_]^_ := del[i, j] definition simplifies these. $\endgroup$ Commented Feb 7, 2013 at 17:17
  • $\begingroup$ Ah I was reading that as del /: del[i_, j_] ^= del[i, j]. One additional question, is there any benefit to the above over something like del[i_, j_]^_ := del[i, j] without the del /:? $\endgroup$ Commented Feb 7, 2013 at 17:37
  • $\begingroup$ The effect of del /: is to make the definition associate itself with del, which is what you would want. If you omit del /: then it defaults to trying to associate itself with the Head of del[i_, j_]^_ which is Power, which is not what you you would want. Try a_^b_ := f[a,b] to see what happens. $\endgroup$ Commented Feb 8, 2013 at 0:19