# Tensor Product of Hilbert spaces

Let there be given a (monoidal) category ${\cal C}$, e.g., the category of finite dimensional vector spaces, the category of Hilbert spaces, etc.

In such a category ${\cal C}$, one typically has the isomorphism

$$\tag{1} {\cal H} \otimes {\cal K} \cong {\cal L}({\cal H}^{*}, {\cal K}), $$

where ${\cal H}^{*}$ is a dual object, and ${\cal L}$ is the pertinent space of morphisms ${\cal H}^{*}\to {\cal K}$.

Often textbooks don't provide the *actual definition* of a tensor product, which is nevertheless at least partly explained on Wikipedia, but instead cheat by using the isomorphism (1) as a *working definition* of a tensor product ${\cal H} \otimes {\cal K}$.

I don't think Wald ever defines a tensor product for infinite dimensional space in his GR text, so I presume your question is about the finite dimensional case where we simply write the tensor product as the vector space over pairs $u_iv_j$ where $u$ and $v$ are a basis. I will show the equivalence in that case.

If we have two finite dimensional Hilbert spaces $H_1$, $H_2$ we can take the orthonormal bases $u_i\in H_1$ , $v_j\in H_2$. Since everything is finite dimensional, everything is finite rank, so the the vector space is just the the space of linear maps from $H_1$ to $H_2$. Take a linear map $A$ and define $a_{ij}= \langle A(u_i),v_j\rangle = \langle u_i,A^{\dagger}(v_j)\rangle$. Using the orthonormality of the bases that means $a_{ij}$ is simply the matrix presentation of $A$, and the vector space is simply the appropriate vector space of matrices. Then we can interpret $Tr(A^\dagger B)$ as the usual matrix trace which gives $\sum_{ij} a_{ij}^*b_{ij}$.

This is equivalent to the usual notation whereby we write tensor products as elements $\sum_{ij}a_{ij}u_i\otimes v_j$. Again the vector space is the appropriately sized matrices. The inner product is defined to be $\langle a\otimes b, c\otimes d\rangle = \langle a,b\rangle\cdot\langle c,d\rangle$. This gives the same result as above after plugging in the basis.