Tensor products and the relation between Hilbert space and the Real space

Question

My question concerns the tensor product of Hilber spaces and the connection to the real Cartesian space. I begin by presenting what my book says, and then what I have understood so far and what confuses me.

I am following Cohen-Tannoudji: Let us form the tensor product $$\mathcal{E}_{xyz}=\mathcal{E}_{x} \otimes \mathcal{E}_{y} \otimes \mathcal{E}_{z}.$$We note that the basis of $\mathcal{E}_{xyz}$ is built from the bases of $\mathcal{E}_x$, $\mathcal{E}_y$, and $\mathcal{E}_z$, such that $$|x, y, z\rangle=|x\rangle \otimes |y\rangle \otimes |z\rangle.$$ Furthermore, we note that the basis kets of $\mathcal{E}_{xyz}$ are simultaneously eigenvectors of $X$, $Y$, and $Z$ operators. Therefore, we have $$X\ |x, y, z\rangle=x\ |x, y, z\rangle\\Y\ |x, y, z\rangle=y\ |x, y, z\rangle\\Z\ |x, y, z\rangle=z\ |x, y, z\rangle.$$ Therefore, $\mathcal{E}_{xyz}$ coincides with $\mathcal{E}_{\bf{r}}$, the state space of a three-dimensional particle and $|x, y, z\rangle$ with $|\bf{r}\rangle$; thus, the $x$, $y$, and $z$ are precisely the Cartesian coordinates of $\bf{r}$.

My understanding and questions

Tensor products are applied to vector spaces and the result is another vector space. In the first line, we consider three 1D Hilbert vector spaces and build, via the tensor product, a 3D Hilbert space. Similarly, we can build the basis kets of this new Hilbert space from the basis kets of the 1D Hilbert spaces. Moreover, it makes sense that these new basis kets are simultaneous eigenvectors of $X$, $Y$, and $Z$, because the three Hilbert spaces are independent of one another so the three position operators should commute in $\mathcal{E}_{xyz}$ (that is, they should act independently). I am totally lost at the last sentence.

How do we go from Hilbert spaces with potentially infinite basis kets to the real Cartesian space with three axes $x$, $y$, and $z$? Also, am I right to think of all $\mathcal{E}$ spaces being Hilbert spaces? Also, is my reasoning for why $X$, $Y$, and $Z$ commute in $\mathcal{E}_{xyz}$ correct?

Answer 1

Yes, they commute because $X$, $Y$ and $Z$ act independently. For simplicity consider a 2d Hilbert space with states $|x,y\rangle$. Then when we write $X$ we really mean the operator $X \otimes \mathbb{1}$, and likewise $Y$ is shorthand for $\mathbb{1} \otimes Y$. The fact that $X$ and $Y$ commute then follows from the fact that $X$ and $Y$ both commute with the unit operator $\mathbb{1}$.

Cohen-Tannoudji means something very simple. He has probably discussed the Hilbert space $\mathcal{E}_{\mathbf{r}}$ before, which is just the space of states labeled by a three-vector $\mathbf{r}$ instead of a single coordinate $x$ --- it's the Hilbert space of a single particle living in $\mathbb{R}^3$. We all know that $\mathbb{R}^3 = \mathbb{R} \oplus \mathbb{R} \oplus \mathbb{R}$. But it's not a priori clear what the relation between $\mathcal{E}_\mathbf{r}$ and the Hilbert space $\mathcal{E}_x$ of a 1d particle is. Mathematically speaking, the discussion you summarize is just the statement that $$ \mathcal{E}_{\mathbf{r}} = \mathcal{E}_x \otimes \mathcal{E}_y \otimes \mathcal{E}_z$$ with the identification $|\mathbf{r}\rangle = |x,y,z\rangle$ if $\mathbf{r} = (x,y,z)^t$ - it's an isomorphism between two Hilbert spaces.

Of course there's also a bijection from $\mathbb{R}^3$ to $\mathcal{E}_\mathbf{r}$: just send $\mathbf{r} \mapsto |\mathbf{r}\rangle$. But this is not an isomorphism of Hilbert spaces: $\mathbb{R}^3$ has dimension 3 and $\mathcal{E}_\mathbf{r}$ is infinite-dimensional. The above map isn't even linear: $$| \mathbf{r}_1 +\mathbf{r}_2 \rangle \neq | \mathbf{r}_1 \rangle + | \mathbf{r}_2 \rangle.$$ Both spaces also have very different inner products.