# Is this eigenstate a tensor product or a direct sum?

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.

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 $$xyz\vert \mathbf{r}\rangle$$.

• Could you tell me, in which textbook can I find this ? Commented Nov 15, 2020 at 0:30
• 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. Commented Nov 15, 2020 at 0:35
• I didn't understand your last sentence! Commented Nov 15, 2020 at 0:36
• If the one you're using is not clear on that stop using it and switch to an alternate source. Commented Nov 15, 2020 at 0:37
• @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... Commented Nov 15, 2020 at 1:35

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.

• 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 Commented Nov 16, 2020 at 13:04
• 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. Commented Nov 16, 2020 at 14:38