The definition of addition on the tensor product of Hilbert spaces

Question

Let $H_1$ and $H_2$ be finite-dimensional Hilbert spaces with inner products $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ respectively. Construct the tensor product of $H_1$ and $H_2$ as $H_1\otimes H_2$, then we can define an inner product by setting $\langle u_1\otimes u_2,v_1\otimes v_2\rangle:=\langle u_1,v_1\rangle_1\langle u_2,v_2\rangle_2$ for $u_1,v_1\in H_1,u_2,v_2\in H_2$. This is how all textbooks define it.

My question is: Can we write the sum $u_1\otimes u_2+x_1\otimes x_2$ as one pure tensor product and consider something like $\langle u_1\otimes u_2+x_1\otimes x_2,v_1⊗v_2\rangle$?

Answer 1

It sounds like you're asking different questions. The answer to the second question is that the inner product is first defined on pure tensors (those which are the tensor product of two vectors) and then extended by linearity, so

$$\langle u_1 \otimes u_2 + x_1 \otimes x_2, v_1 \otimes v_2 \rangle = \langle u_1 \otimes u_2, v_1 \otimes v_2 \rangle + \langle x_1 \otimes x_2, v_1 \otimes v_2 \rangle.$$

The answer to the first question is that a sum of pure tensors can be written as a pure tensor iff at least one of $H_1$ and $H_2$ are one-dimensional. One direction is straightforward. In the other, if $\dim H_1$ and $\dim H_2$ are both greater than $1$, then any element of $H_1 \otimes H_2$ defines a linear operator $H_2^{\ast} \to H_1$ via tensor contraction: that is, tensor contraction gives a natural map

$$(H_1 \otimes H_2) \otimes H_2^{\ast} \cong H_1 \otimes (H_2 \otimes H_2^{\ast}) \to H_1.$$

The pure tensors describe maps of rank at most $1$ (exercise), and if $\dim H_1$ and $\dim H_2$ are both greater than $1$ then it is easy to write down linear maps $H_2^{\ast} \to H_1$ of rank greater than $1$.

In quantum mechanics, this observation leads to the important phenomenon of quantum entanglement.