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