# Dimension of Hilbert space tensor product

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?

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.

• I can add details for the proof of the hint if you need them Sep 20, 2020 at 12:13
• No it's fine. The hint is proven in the book I'm reading :)
– user745578
Sep 20, 2020 at 12:14

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 $$x'\otimes y'$$.

• 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 ? Sep 20, 2020 at 12:53

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.

• How do you know there is such a scalar? (New to tensor products, sorry if the question is basic)
– user745578
Sep 20, 2020 at 12:10
• @user745578 What is your definition of a tensor product? Is this list of properties the definition that you're used to? Sep 20, 2020 at 12:10
• 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.
– user745578
Sep 20, 2020 at 12:11
• Yes, these properties I know.
– user745578
Sep 20, 2020 at 12:12
• @user745578 If you're familiar with the universal property, then I think Berci's answer is the more reasonable way to go Sep 20, 2020 at 12:13