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Let $H_1$ and $H_2$ be Hilbert spaces, then I would intuitively define the inner product on $H_1 \times H_2$ by $\langle (x_1,x_2),(y_1,y_2)\rangle = \langle x_1,y_1 \rangle + \langle x_2,y_2 \rangle$.

Now the inner product of the tensor space is given by $\langle (x_1 \otimes x_2),(y_1 \otimes y_2)\rangle = \langle x_1 y_1 \rangle \cdot \langle x_2,y_2 \rangle$ and I was wondering how this fits together? Could anybody comment on the issue why we have to define the inner products exactly in that way for the two different spaces?

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The way the cartesian and tensor product combine can be seen in the isomorphism of complex valued functions $$L^{2}(R\times R)\cong L^{2}(R)\otimes L^{2}(R)$$ The inner product on the left hand side is a double integral while that on the right is your product of inner products (integrals in one variable) . This identifies functions on $R\times R$ with limits of sums of functions of the form f(x)g(y) with respect to the norm generated by the appropriate inner product.

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First you treat Hilbert spaces as algebraic objects, vector spaces. Thus you may take an algebraic tensor product $H_1 \underline{\otimes} H_2$, however an algebraic tensor product of Hilbert spaces is not a Hilbert space. Therefore, we have to fix this obstacle, we define the inner product (as you done it in your question) on their algebraic tensor product and take the completion of $H_1 \underline{\otimes} H_2$ with respect to this inner product. This object equipped with the inner product which you have pointed out is a Hilbert space and we denote it as $H_1 \otimes H_2$ and call a (Hilbert space) tensor product of $H_1$ and $H_2$.

You may want to read this source, it will give you more explanation why this inner product is a natural one.

Comment to your first sentence: your intuition will bring you to the definition of a direct sum, rather than a tensor product.

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