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A Difference Between Outer and Tensor

  1. Feb 26, 2017 #1
    Say, we have two Hilbert spaces ##U## and ##V## and their duals ##U^*, V^*##.
    Then, we say, ##u\otimes v~ \epsilon~ U\otimes V##, where ##'\otimes'## is defined as the tensor product of the two spaces, ##U\times V \rightarrow U\otimes V##.

    In Dirac's Bra-Ket notation, this is written as, ##u\otimes v:=\left|u\right>\otimes \left| v\right>:=\left|u\right>\left| v\right>##


    What is the outer product, then?

    In Dirac notation, it is written as ##\left|u\right> \left< v\right|##. Which makes it clear that the outer product is a linear functional acting on some ##v'\epsilon ~V## and giving some ##u'\epsilon~ U##, such that ##\left| u'\right>=\left<v|v'\right>\left|v\right>##.
    So, would it be correct to say that (##\otimes_O## is the outer product)
    $$u\otimes _Ov:=u\otimes v^* $$
    for, ##u,~ v,~v^*~\epsilon ~U, V, V^*##, respectively?

    Also, would it be correct to say that ##u\otimes v## is a type ##(2,0)## tensor while, ##u\otimes _O v## is a type ##(1,1)## tensor?
     
  2. jcsd
  3. Feb 26, 2017 #2

    fresh_42

    Staff: Mentor

    The term "outer product" is context sensitive. There are three multiplications which are occasionally called outer:
    1. Graßmann algebra: https://en.wikipedia.org/wiki/Exterior_algebra
    2. Tensor product: https://en.wikipedia.org/wiki/Outer_product
    3. Cross product: https://en.wikipedia.org/wiki/Cross_product
    Personally I think that only the first one deserves this attribute, which makes it a tensor product with the additional requirement that ##u \wedge u = 0##. The usage you spoke about looks like you refer to the second one.
     
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