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Outer product on operators?

by dipole
Tags: operators, outer, product
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dipole
#1
Nov28-12, 10:46 PM
P: 436
In my QM textbook, there's an equation written as:

[itex] \vec{J} = \vec{L}\otimes\vec{1} + \vec{S}\otimes\vec{1} [/itex]

referring to angular momentum operators (where [itex]\vec{1} [/itex] is the identity operator). I don't really understand what the outer product (which I'm assuming is what the symbol [itex]\otimes[/itex] 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. :\
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Fightfish
#2
Nov29-12, 01:40 AM
P: 623
[tex]\otimes[/tex] 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
#3
Nov29-12, 04:22 AM
P: 123
First of all, I think that the formula should be J = L[itex]\otimes[/itex]1 + 1[itex]\otimes[/itex]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[itex]\otimes[/itex]HB (the vectors of that new space are defined in this way:say ΨΑ[itex]\in[/itex]HA and ΨΒ[itex]\in[/itex]HΒ, then the vectors Ψ=ΨA[itex]\otimes[/itex]ΨB for all ΨA and ΨB are the vectors of H. ΨA[itex]\otimes[/itex]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[itex]\otimes[/itex]B , where this new operator is defined by:
O Ψ [itex]\equiv[/itex](A[itex]\otimes[/itex]B) (ΨA[itex]\otimes[/itex]ΨB) = (AΨA)[itex]\otimes([/itex]BΨB).
When the operators are represented by matrices, then the matrix A[itex]\otimes[/itex]B is defined as:
[A[itex]\otimes[/itex]B]aa',bb' = Aaa'Bbb'


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