Meaning of Kronecker Product in Partially Expanded Operators

Question

I am studying operators in quantum mechanics and have reached confusion in the meaning of the Kronecker product of such operators. I am fairly lost so please excuse any errors in the following text.

Suppose I have an operator $\mathscr{O}:A\otimes B\to A\otimes B$ which can be expanded as a matrix acting in $B$ with entries being operators acting in $A$. Let $\mathscr{O}_M$ be this matrix. If $\otimes_K$ denotes the Kronecker product and I define $\mathscr{U}_M:=\mathscr{O}_M\otimes_K \mathscr{O}_M$, what intuition can I apply to the transformation $\mathscr{U}$ described by $\mathscr{U}_M$?

For example, $A=\mathbb{R}$ and $B=\mathbb{R}^3$, then $\mathscr{O}_M$ is simply a $3\times 3$ real matrix and $\mathscr{U}_M$ is a $9\times 9$ real matrix, which I believe just describes $\mathscr{U}=\mathscr{O}\otimes\mathscr{O}$. I, then, roughly understand what the transformation $\mathscr{U}$ is and am happy.

I am confused, however if, for example, $A,B=\mathbb{C}^2$. In this case $\mathscr{O}_M$ is a $2\times 2$ matrix with entries being operators in $\mathbb{C}^2$. Thus, $\mathscr{U}_M$ is a $4\times 4$ matrix with entries being operators in $\mathbb{C}^2$. Can I understand the operator $\mathscr{U}$ in terms of $\mathscr{O}$, without going through the matrix description? Clearly $\mathscr{U}\neq \mathscr{O}\otimes\mathscr{O}$ since $\dim(A\otimes B)=4$ so $\dim(A\otimes B\otimes A\otimes B)=16$ while $\mathscr{U}$ only appears to represent an $8$ dimensional transformation. It appears that $\mathscr{U}$ is somehow a map from $A\otimes B^{\otimes 2}$ to itself, but I do not see how to construct this map out of $\mathscr{O}$ without going through the matrix description.

Thank you!

Answer 1

You seem to be lighting a bonfire of symbols you don't fully appreciate, and getting lost in virulently abstract notation. Since this is a physics site, I will ignore/undo your substantially nonstandard/perverse conventions, and give you the WP or physics-text ones.

Consider your vector space $A\otimes B$ and take it to mean a-dim vectors V in A and b-dim vectors W in B in the basis where each component of V is multiplied by an entire vector W. In all, such vectors $V\otimes W$ are ab-dimensional.

They map onto each other through ab×ab matrices $\mathscr{O}:A\otimes B\to A\otimes B$. These matrices are the Kronecker product of a×a matrices $M_A$ with b×b matrices $M_B$, where each entry of $M_A$ is multiplied by an entire matrix $M_B$, $$ M_{A\otimes B}= M_A\otimes M_B . $$

So you should convince yourself all your block matrix ab×ab operations amount to separate a and b operations, as well, $$ M_{A\otimes B} (V\otimes W) = (M_A V)\otimes (M_B W), $$ and develop comfort with the mixed-product property when composing such maps, in WP.

Here is a trivial example in these conventions, for a=2 and b=3, $$ \sigma_1 \otimes \operatorname{diag} (1,0,-1) = \begin{pmatrix} 0&0&0&1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&-1\\ 1&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&-1&0&0&0 \end{pmatrix}\neq \operatorname{diag} (1,0,-1)\otimes \sigma_1 ~. $$

Admittedly, you'd rarely see Kronecker products of rotation generators like these: you'd see coproducts, instead; or else plain Kronecker products of entire rotation matrices; but such issues outrange your question. After you understand coproducts, you'll be ready to compose rotation-group representations the nice way.