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Is $|\mathbf r\rangle$, the state of being exactly at $\mathbf r$, a direct sum or a tensor product of $|x\rangle$, $|y\rangle$ and $|z\rangle$. The same question for $|\mathbf p\rangle$. Now my attempt is the following:

If it were a direct sum, i.e. $|\mathbf r\rangle = |x\rangle +|y\rangle + |z\rangle$ then the position operator may be $\mathbf{\hat r} = \hat x \oplus \hat y \oplus \hat z$, that is a direct sum of the operators on $X = Y = Z = \mathbb R$, such that $$ \hat x \oplus \hat y \oplus \hat z:X\oplus Y \oplus Z \to X\oplus Y \oplus Z, $$ \begin{align} |x\rangle +|y\rangle + |z\rangle \mapsto &\ (\hat x \oplus \hat y \oplus \hat z)(|x\rangle +|y\rangle + |z\rangle)\\ &= \hat x|x\rangle + \hat y|y\rangle + \hat z|z\rangle \\ &= x|x\rangle + y|y\rangle + z|z\rangle \end{align} But if it were a tensor product, i.e. $|\mathbf r\rangle = |x\rangle \otimes |y\rangle \otimes |z\rangle$, then $\mathbf{\hat r} = \hat x \otimes \hat y \otimes \hat z$ $$ \hat x \otimes \hat y \otimes \hat z:X\otimes Y \otimes Z \to X\otimes Y \otimes Z, $$ \begin{align} |x\rangle \otimes |y\rangle \otimes |z\rangle \mapsto &\ (\hat x \otimes \hat y \otimes \hat z)(|x\rangle \otimes |y\rangle \otimes |z\rangle)\\ &= \hat x|x\rangle \otimes \hat y|y\rangle \otimes \hat z|z\rangle \\ &= x|x\rangle \otimes y|y\rangle \otimes z|z\rangle \\ &= xyz\ |x\rangle \otimes |y\rangle \otimes |z\rangle \end{align} which doesn't make a lot of sense in terms of eigenvalues and eigenvectors. So which one is it, or is it none of them ? and should $\mathbf{\hat r} |\mathbf r\rangle = \mathbf r|\mathbf r\rangle$ mean anything ? See this for more on the definitions of these maps.

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It’s a tensor product as the various kets you have live in distinct Hilbert spaces. In this space $\hat x$ really is $\hat x\otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}$, $\hat y$ is formally $\hat{\mathbb{I}}\otimes \hat y\otimes \hat{\mathbb{I}}$ etc. Indeed $\vert x\rangle $ is formally $\vert x\rangle \otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}$ and operator like $\hat x\otimes \hat y\otimes \hat z$ acting on $\vert \mathbf{r}\rangle$ would return $xyz\vert \mathbf{r}\rangle$.

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  • $\begingroup$ Could you tell me, in which textbook can I find this ? $\endgroup$
    – Physor
    Commented Nov 15, 2020 at 0:30
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    $\begingroup$ pretty much all the reasonable ones. I suppose the one by Cohen-Tannoudji is reasonably formal and should have this. It's really more a question of which ones not to look at. $\endgroup$ Commented Nov 15, 2020 at 0:35
  • $\begingroup$ I didn't understand your last sentence! $\endgroup$
    – Physor
    Commented Nov 15, 2020 at 0:36
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    $\begingroup$ If the one you're using is not clear on that stop using it and switch to an alternate source. $\endgroup$ Commented Nov 15, 2020 at 0:37
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    $\begingroup$ @ZeroTheHero "It's really more a question of which ones not to look at." words to live by... I wish there were a tactful formulation of this in the resource recommendation guidance... $\endgroup$ Commented Nov 15, 2020 at 1:35
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Just to flesh out @ZeroTheHero 's impeccable answer for you, with a hint of how to escape conceptual hash by small finite-dimensional matrices. I'll avoid addressing your unsound conjectures/probes, to protect your attention from expressions which are not even wrong, in favor of standard stuff.

The states are tensor product states, $$|\mathbf r\rangle = |x\rangle \otimes |y\rangle \otimes |z\rangle= |x\rangle |y\rangle |z\rangle,$$ whereas 3d vectors $\mathbf r= (x,y,z)^T$ are just that. They can be made into eigenvalues of 3d vectors of operators, $\hat {\mathbf r}= (\hat x,\hat y,\hat z)^T$, operators acting on the space of $ |\mathbf r\rangle$s, as the accepted answer details, $ \hat x |\mathbf r\rangle = x|\mathbf r\rangle$, etc. Whence your target 3d vector expression, $$\mathbf{\hat r} |\mathbf r\rangle = \mathbf r|\mathbf r\rangle,$$ quite meaningful indeed. Yet again, $|\mathbf r\rangle$ is not a vector, much unlike $\mathbf{ r}$. It yields the latter under action of the vector $\mathbf{\hat r} $.

You may further dot this 3-vector equation by a fixed 3-vector $\mathbf a,$ to reduce it to just one equation (scalar); or to itself, $$ \mathbf{ a}\cdot \mathbf{\hat r} |\mathbf r\rangle = \mathbf {a \cdot r}|\mathbf r\rangle ~;\\ \mathbf{ \hat r}\cdot \mathbf{\hat r} |\mathbf r\rangle = r^2|\mathbf r\rangle, $$ etc.

Your instructor must have taught you how to illustrate such Hilbert spaces by finite-dimensional vector spaces when you are groping for your bearings. Take x to only take 2 positions, so $|x\rangle$ is a 2-vector; y to only take 3 positions, so $|y\rangle$ is a 3-vector; and z to only take 4 positions, so $|z\rangle$ is a 4-vector.

Their direct product space $|\mathbf r\rangle$ then is 24d, (whereas their direct sum space would be 9d). All operators on this space are thus 24×24 matrices, trivial to visualize. So, if the two eigenvalues of $\hat x$ are $x_1$ and $x_2$, do you see the diagonal 24×24 $~~~\hat x$ matrix consisting of an upper 12×12 block with entries $x_1$ and a lower 12×12 block with entries $x_2$? Trying to visualize some of your proposed constructions, by contrast, this way, would be simply impossible/ inconceivable--not even wrong. This language should enable you to contrast the 3-vectors of the continuous case, also present and controlling here, to the 24d vector kets.

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  • $\begingroup$ What I understand is that $$ \hat {\mathbf r}= (\hat x,\hat y,\hat z)^T = (\hat x\otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}, \hat{\mathbb{I}}\otimes\hat y \otimes \hat{\mathbb{I}}, \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}\otimes \hat z)^T $$ but that defines a strange multiplication between vectors, i.e., $|\mathbf r\rangle = |x\rangle |y\rangle |z\rangle,$ with $\hat {\mathbf r}= (\hat x,\hat y,\hat z)^T$. And I'm sorry to say, that I didn't understand the dot product in the second part of your answer $\endgroup$
    – Physor
    Commented Nov 16, 2020 at 13:04
  • $\begingroup$ You got the first line correctly; the hatted operators are defined as such. That's why I invited you to check them out on the 24d rep. But $|\mathbf r\rangle$ is not a vector! Its eigenvalues under vectors of operators are vectors! I strove to get across that 3vectors have scalar products as normal, so the triplet of equations pushing you may reduce to just one eigenvalue equation as ${\mathbf a }\cdot \mathbf {r}$ is a scalar! Perhaps it would help you using explicit indices for the 3d space. $\endgroup$ Commented Nov 16, 2020 at 14:38

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