# What is the difference between a tensor product and an outer product?

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?

• The results are the same unless you put more structure on these things than vector space. Oct 3, 2016 at 11:30
• 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. Oct 3, 2016 at 11:33
• Oct 27, 2020 at 19:22

Consider $V=\mathbb{R}^ 2$. Your first "equality" arises from the isomorphism
$$V \otimes V \to \mathbb{R}^4$$ $$e_1 \otimes e_1 \mapsto e_1;$$ $$e_1 \otimes e_2 \mapsto e_2;$$ $$e_2 \otimes e_1 \mapsto e_3;$$ $$e_2 \otimes e_2 \mapsto e_4.$$
$V \otimes V \simeq V \otimes V^* \simeq Hom(V,V)$, 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 $Hom(\mathbb{R}^2, \mathbb{R}^2)$.