How Does the Outer Product Operate on Quantum Mechanical Operators?

[dipole]

In my QM textbook, there's an equation written as:

\vec{J} = \vec{L}\otimes\vec{1} + \vec{S}\otimes\vec{1}

referring to angular momentum operators (where \vec{1} is the identity operator). I don't really understand what the outer product (which I'm assuming is what the symbol \otimes means here) means when dealing with operators (which can be represented as matrices).

What happens when you outerproduct one operator with another? Unfortunately there is no explanation in the text, I guess it's assumed this is obvious or that the reader knows about this kind of math. :\

 

[Fightfish]

\otimes is not outer product. It is a tensor product.
Could you provide the context?
I am guessing that this means that you act the angular momentum operator only on the first particle but leave the second particle untouched.

 

[cosmic dust]

First of all, I think that the formula should be J = L\otimes1 + 1\otimesS . About it's meaning, when you have two operators (say A and B) which operate on two, in general different, Hilbert spaces (say H_A and H_B), then you can create a new Hilbert space by the direct product of the two of them, H = H_A\otimesH_B (the vectors of that new space are defined in this way:say Ψ_Α\inH_A and Ψ_Β\inH_Β, then the vectors Ψ=Ψ_A\otimesΨ_B for all Ψ_A and Ψ_B are the vectors of H. Ψ_A\otimesΨ_B is a new item that has two independent parts, Ψ_A and Ψ_B , pretty much like when you have two reals a and b, you can create a new item (a,b) which represents a point in a plane) . The operators on this new Hilbert space are then created by the direct product of the operators that operate in the two initial spaces, i.e. O = A\otimesB , where this new operator is defined by:
O Ψ \equiv(A\otimesB) (Ψ_A\otimesΨ_B) = (AΨ_A)\otimes(BΨ_B).
When the operators are represented by matrices, then the matrix A\otimesB is defined as:
[A\otimesB]_aa',bb' = A_aa'B_bb'