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Let $H_1, H_2$ be Hilbert spaces and consider their Hilbert space tensor product $$H_1 \hat{\otimes} H_2$$ which is the completion of the algebraic tensor product $H_1 \otimes H_2$ with respect to the unique inner product on $H_1 \otimes H_2$ satisfying $$\langle x \otimes y, x' \otimes y'\rangle = \langle x , x' \rangle \langle y, y'\rangle$$

If $E_1$ is an orthonormal basis for $H_1$ and $E_2$ is an orthonormal basis for $H_2$, I proved that $$E_1 \otimes E_2:= \{x \otimes y: x \in E_1, y\in E_2\}$$ is an orthonormal basis for $H_1 \hat{\otimes} H_2$. From this, I want to deduce that $$\dim(H_1 \hat{\otimes} H_2 ) = \dim (H_1) \dim (H_2)$$ (product of cardinal numbers). I see that it suffices to check that the map $$E_1 \times E_2 \to E_1 \otimes E_2: (x,y) \mapsto x \otimes y$$ is injective, but I can't see why this holds: $$x \otimes y = x' \otimes y' \implies x= x', y = y'$$ must not be true for general pure tensors, but maybe because we have the orthogonality we can say something more?

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3 Answers 3

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Hint: If $X,Y$ are vector spaces and $\{x_1,\dots,x_n\}\subset X$ is linearly independent, then $\sum_{i=1}^nx_i\otimes y_i=0$ implies that $y_i=0$ for all $i$.

In your case: suppose that $x\neq x'$ or $y\neq y'$. WLOG suppose that $x\neq x'$. If $x\otimes y=x'\otimes y'$, then $x\otimes y + x'\otimes(-y')=0$. Since $\{x,x'\}$ are linearly independent, this yields $y=-y'=0$, so $y=y'=0$, a contradiction, since $y,y'$ are unit vectors.

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  • $\begingroup$ I can add details for the proof of the hint if you need them $\endgroup$ Sep 20, 2020 at 12:13
  • $\begingroup$ No it's fine. The hint is proven in the book I'm reading :) $\endgroup$
    – user745578
    Sep 20, 2020 at 12:14
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Assume $(x,y)\ne (x',y')$ and consider a bilinear map $f:H_1\times H_2\to\Bbb C$ defined on the bases as $$f(e_1,e_2)=\left\{\matrix{1&\text{ if }e_1=x\text{ and }e_2= y\\ 0&\text{ otherwise.}}\right.$$ This factors through $H_1\otimes H_2$ thus distinguishing $x\otimes y$ and $x'\otimes y'$.

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  • $\begingroup$ are you using the universal property ?, other wise how do you that the map out of $H_1 \otimes H_2$ is well defined, If you can't tell yet when two elements in $H_1 \otimes H_2$ are equal ? $\endgroup$
    – Physor
    Sep 20, 2020 at 12:53
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The way you should approach this ultimately depends on your definition of the (algebraic) tensor product.

One approach is as follows: it suffices to note that $x \otimes y = x' \otimes y'$ only if there exists a scalar $\alpha$ for which $\alpha x = x'$ and $y = \alpha y'$. If $x',x$ are part of the same orthonormal basis, then they are either equal or fail to be multiples of each other. So, your map is indeed injective.

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  • $\begingroup$ How do you know there is such a scalar? (New to tensor products, sorry if the question is basic) $\endgroup$
    – user745578
    Sep 20, 2020 at 12:10
  • $\begingroup$ @user745578 What is your definition of a tensor product? Is this list of properties the definition that you're used to? $\endgroup$ Sep 20, 2020 at 12:10
  • $\begingroup$ I constructed it as a quotient space of a free vector space imposing the tensor relations, but I'm also familiar with the universal property that turns bilinear maps into linear maps. $\endgroup$
    – user745578
    Sep 20, 2020 at 12:11
  • $\begingroup$ Yes, these properties I know. $\endgroup$
    – user745578
    Sep 20, 2020 at 12:12
  • $\begingroup$ @user745578 If you're familiar with the universal property, then I think Berci's answer is the more reasonable way to go $\endgroup$ Sep 20, 2020 at 12:13

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