- #1
Eye_in_the_Sky
- 331
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on the "Tensor Product"
In response to some remarks made in the thread "How do particles become entangled?", as well as a number of private messages I have received, I feel there is some need to post some information on the notion of a "tensor product".
Below, a rather intuitive look at the idea of the "tensor product" is taken. For simplicity, the vector spaces involved are assumed to be finite-dimensional. The infinite-dimensional case can be accommodated with only some minor amendments to the presentation.
(Note: The usual symbol for the tensor product is an "x" with a "circle" around it, but below I will use the symbol "x".)
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Let U and V be finite-dimensional vector spaces over C with bases {ui} and {vj}, respectively. For each ui and vj , define an object "ui x vj", and construe the full collection of these objects to be a basis for a new vector space W. That is,
W ≡ {∑ij αij(ui x vj) | αijЄC} ,
where, by definition,
if ∑ij αij(ui x vj) = 0 , then αij=0 for all i,j .
The above then makes W a vector space over C such that
Dim(W) = Dim(U)∙Dim(V) .
However ... had we chosen a different set of basis vectors for U or V, then the vector space W thereby obtained would be formally distinct from the one obtained above. There would be no way to 'link' the bases for each of the two W's.
Let us now introduce some additional 'structure' on the operation "x", such that all W's obtained by the above construction will be formally identical no matter which bases are chosen for U and V. Specifically, we extend the definition of "x" to be bilinear, thus allowing any vector of U to be placed in the left "slot", and any vector of V to be placed in the right "slot". We do this as follows:
For any u,u'ЄU , v,v'ЄV , and αЄC , let
(u + u') x v = (u x v) + (u' x v) ,
u x (v + v') = (u x v) + (u x v') ,
α(u x v) = (αu) x v = u x (αv) .
Now all W's are one and the same.
The next thing we need is an inner product <∙|∙> on W. Let <∙|∙>1 and <∙|∙>2 be the inner products on U and V, respectively. Then, for any u x v and u' x v' Є W , define
<u x v|u' x v'> ≡ <u|u'>1∙<v|v'>2 .
Finally, extend <∙|∙> to the whole of W by "antilinearity" in the first slot and "linearity" in the second slot.
It now follows that <∙|∙> is an inner product on W.
Moreover, if {ui} and {vj} are orthonormal bases of U and V respectively, then {ui x vj} is an orthonormal basis of W.
In response to some remarks made in the thread "How do particles become entangled?", as well as a number of private messages I have received, I feel there is some need to post some information on the notion of a "tensor product".
Below, a rather intuitive look at the idea of the "tensor product" is taken. For simplicity, the vector spaces involved are assumed to be finite-dimensional. The infinite-dimensional case can be accommodated with only some minor amendments to the presentation.
(Note: The usual symbol for the tensor product is an "x" with a "circle" around it, but below I will use the symbol "x".)
----------------------------------------
Let U and V be finite-dimensional vector spaces over C with bases {ui} and {vj}, respectively. For each ui and vj , define an object "ui x vj", and construe the full collection of these objects to be a basis for a new vector space W. That is,
W ≡ {∑ij αij(ui x vj) | αijЄC} ,
where, by definition,
if ∑ij αij(ui x vj) = 0 , then αij=0 for all i,j .
The above then makes W a vector space over C such that
Dim(W) = Dim(U)∙Dim(V) .
However ... had we chosen a different set of basis vectors for U or V, then the vector space W thereby obtained would be formally distinct from the one obtained above. There would be no way to 'link' the bases for each of the two W's.
Let us now introduce some additional 'structure' on the operation "x", such that all W's obtained by the above construction will be formally identical no matter which bases are chosen for U and V. Specifically, we extend the definition of "x" to be bilinear, thus allowing any vector of U to be placed in the left "slot", and any vector of V to be placed in the right "slot". We do this as follows:
For any u,u'ЄU , v,v'ЄV , and αЄC , let
(u + u') x v = (u x v) + (u' x v) ,
u x (v + v') = (u x v) + (u x v') ,
α(u x v) = (αu) x v = u x (αv) .
Now all W's are one and the same.
The next thing we need is an inner product <∙|∙> on W. Let <∙|∙>1 and <∙|∙>2 be the inner products on U and V, respectively. Then, for any u x v and u' x v' Є W , define
<u x v|u' x v'> ≡ <u|u'>1∙<v|v'>2 .
Finally, extend <∙|∙> to the whole of W by "antilinearity" in the first slot and "linearity" in the second slot.
It now follows that <∙|∙> is an inner product on W.
Moreover, if {ui} and {vj} are orthonormal bases of U and V respectively, then {ui x vj} is an orthonormal basis of W.
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