Pascal's Triangle for non-commutative?

Pythagorean

I was curious if there was a non-commutative version of Pascal's triangle for operators (such as those used in brah-ket notation)

The important note is that (a + b)^2 = a^2 + b^2 + ab +ba
where ba != ab

Pythagorean

I'm guess if it's an even power such as (x + y)^4

you can always assume the cross-terms are split between their orderings, so from pascal's triangle, which gives the coefficients 1 4 6 4 1, the middle coefficient corresponds to 3x^2y^2 + 3y^2x^2.

Not quite sure how to handle odd degrees without doing pages of algebra.

I'm posting in this forum because tensors have the non-commutative property and I happen to be applying brah-ket notation so I thought it fit. Apologies if not.

ObsessiveMathsFreak

This is interesting.

Going by (a+b)^3 = a^3 + a^2*b+aba + ab^2 + ba^2 + bab + b^2*a + b^3

It would seem that the only possible row in the "triangle" for general noncommuntative numbers would be "1 1 1 1 1 1 1 1". The exact form may depend on just how uncommutative the number are.

Pythagorean

wow, that looks like it may be difficult to generalize with something like Pascal's triangle. Maybe not the coefficients themselves, but the degree of each term. You'd have to have two degree functions for each variable (i.e. one for <x| and one for |x>)

HallsofIvy

The point of Pascal's triangle is that the i,j entry simply counts the number of ways you can order i "x"s and j-i "y"s. If your multiplication is not commutative, those do not add. All terms are distinct. That's why ObsessiveMathFreak says you just get "1 1 1 ...".

Hurkyl

Just to make sure it's said... those terms do not appear in the expansion of (x+y)^4. (except possibly for special cases)