# Outer product on operators?

1. Nov 28, 2012

### 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. :\

2. Nov 29, 2012

### 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.

3. Nov 29, 2012

### cosmic dust

First of all, I think that the formula should be J = L$\otimes$1 + 1$\otimes$S . About it's meaning, when you have two operators (say A and B) which operate on two, in general different, Hilbert spaces (say HA and HB), then you can create a new Hilbert space by the direct product of the two of them, H = HA$\otimes$HB (the vectors of that new space are defined in this way:say ΨΑ$\in$HA and ΨΒ$\in$HΒ, then the vectors Ψ=ΨA$\otimes$ΨB for all ΨA and ΨB are the vectors of H. ΨA$\otimes$ΨB is a new item that has two independent parts, ΨA and ΨB , pretty much like when you have two reals a and b, you can create a new item (a,b) which represents a point in a plane) . The operators on this new Hilbert space are then created by the direct product of the operators that operate in the two initial spaces, i.e. O = A$\otimes$B , where this new operator is defined by:
O Ψ $\equiv$(A$\otimes$B) (ΨA$\otimes$ΨB) = (AΨA)$\otimes($BΨB).
When the operators are represented by matrices, then the matrix A$\otimes$B is defined as:
[A$\otimes$B]aa',bb' = Aaa'Bbb'