Graduate Difference Between Outer and Tensor

[devd]

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?

 

[fresh_42]

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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