The Kronecker Delta is commonly used to represent a diagonal matrix: $$ a_i \delta_{ij}=\left( \begin{array}{ccc} a_1 & 0 & 0\\ 0 & a_2 & 0\\ 0 & 0 & a_3 \end{array}\right) $$ for 3 dimensions (not Einstein Summation).
However, in Quantum Mechanics, or more specifically the book by Griffith, the Kronecker Delta was used to represent a"shifted" diagonal matrices, i.e.: $$ a_-=\sqrt{n'}\delta_{n,n'-1}= \left( \begin{array}{ccc} 0 & \sqrt{1} & 0 & 0 & 0 & ...\\ 0 & 0 & \sqrt{2} & 0 & 0 & ...\\ 0 & 0 & 0 & \sqrt{3} & 0 & ...\\ 0 & 0 & 0 & 0 & \sqrt{4} & ...\\ . & . & . & . & . & ... \end{array} \right) $$ where $n$ is the row index and $n'$ is the column index, and both $n$ and $n'$ begin from 0 (so the upper left entry is $n=0, n'=0$).
Now I was wondering if there are algebraic properties for the Kronecker Delta used in this way. If I multiplied two matrices in their Kronecker Delta form, is there a way to get the solution using simple manipulations of the Kronecker Deltas and their indices?
For example, computing something like: $$ (a_{-})^l(a_{+})^m $$ where $a_{+}$ is $(a_{-})^\dagger$ (conjugate transpose of the lowering operator), or $$ a_+=\sqrt{n}\delta_{n',n-1}=\sqrt{n'}\delta_{n,n'+1}= \left( \begin{array}{ccc} 0 & 0 & 0 & 0 & ...\\ \sqrt{1} & 0 & 0 & 0 & ...\\ 0 & \sqrt{2} & 0 & 0 & ...\\ 0 & 0 & \sqrt{3} & 0 & ...\\ 0 & 0 & 0 & \sqrt{4} & ...\\ . & . & . & . & ... \end{array} \right) $$
I worked out $(a_+)^m$ by taking successive products of $a_+$ and got the following by inspecting the pattern:
$$ (a_+)^m=\sqrt{\frac{(n'+m)!}{n'!}}\delta_{n,n'+m}=\sqrt{\frac{n!}{(n-m)!}}\delta_{n',n-m} $$ where if I had, for example: $$ (a_+)^3 = \sqrt{n}\delta_{n,n'+1}\sqrt{n'}\delta_{n',n''+1}\sqrt{n''}\delta_{n'',n'''+1}=\sqrt{n'''+3}\sqrt{n'''+2}\sqrt{n'''+1}\delta_{n,n'''+3} $$ I would systematically eliminate each delta by setting the delta indexes belonging to one delta equal to one another (and eliminating the "repeated" indices), until I am left with two indices, namely $n$ and $n'''$, and checking that it gives the correct matrix.
$n'=n''+1=n'''+2$
This is too tedious to do for the general case, and I haven't explored multiplication of non-identical matrices.
I was wondering if there are rules from, perhaps, Tensor Analysis that makes this easier.
For reference, $a_-$ is the lowering operator in the system of one electron in a Simple Harmonic Oscillator potential $V = \frac{1}{2}m\omega^2x^2$, and in terms of momentum $p$ and position $x$,
$$ a_-=\frac{1}{\sqrt{2m\hbar\omega}}(ip+m\omega x) $$ and $p = -i\hbar\frac{\partial}{\partial x}$ in position space. $a_-$ acts on a SHO energy eigenstate $|\psi_{n}>$ and lowers it: $$ a_-|\psi_{n}>=\sqrt{n}|\psi_{n-1}> $$