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I have seen the tensor product written as

$$ \left( \begin{array}{c} a \\ b \\ \end{array} \right) \otimes \left( \begin{array}{c} c \\ d \\ \end{array} \right) = \left( \begin{array}{c} ac \\ ad \\ bc \\ bd \\ \end{array} \right)$$ However I have also seen it written as $$ \left( \begin{array}{c} a \\ b \\ \end{array} \right) \otimes \left( \begin{array}{c} c \\ d \\ \end{array} \right) = \left( \begin{array}{c} a \\ b \\ \end{array} \right) \left( \begin{array}{c} c & d \\ \end{array} \right) = \left( \begin{array}{c} ac & ad\\ bc & bd \\\end{array} \right) $$ Which I have seen in the context of outer products. Why are there two ways to do a tensor product?

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  • $\begingroup$ The results are the same unless you put more structure on these things than vector space. $\endgroup$ Oct 3, 2016 at 11:30
  • $\begingroup$ I have seen the use of the first way when describing a two particle system in quantum mechanics, but I don't understand why there should be two ways to do it. If I did it in index notation, it seems the more natural way is the second way. $\endgroup$
    – Matt0410
    Oct 3, 2016 at 11:33
  • $\begingroup$ Related: math.stackexchange.com/q/1757901/538722 $\endgroup$ Oct 27, 2020 at 19:22

1 Answer 1

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Both are different realizations of the tensor product.

Consider $V=\mathbb{R}^ 2$. Your first "equality" arises from the isomorphism

$$V \otimes V \to \mathbb{R}^4$$ $$e_1 \otimes e_1 \mapsto e_1;$$ $$e_1 \otimes e_2 \mapsto e_2;$$ $$e_2 \otimes e_1 \mapsto e_3;$$ $$e_2 \otimes e_2 \mapsto e_4.$$

The second "equality" arises from the isomorphism

$V \otimes V \simeq V \otimes V^* \simeq Hom(V,V)$, where the first isomorphism comes from the isomorphism of the dual with the space arising from the standard inner product of $\mathbb{R}^2$ (which is essentially just transposition), and the second one comes from $(w,v^*) \mapsto v^*(\cdot) w.$ Note that it becomes a $2 \times 2$ matrix in the end, which is exactly (not exactly, but represents canonically) an element of $Hom(\mathbb{R}^2, \mathbb{R}^2)$.

This is mathematically speaking. I don't know why one is more appropriate than the other in your context. What I can say is that the second way is very useful, because it allows us to translate an endomorphism in terms of something structurally and algebraically rich such as the tensor product. The first one seems to be simply a down to earth immediate way to realize the tensor product as an array.

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