Why the total angular momentum of two particles is a vector instead of a tensor in quantum mechanics?

Question

I am trying to compare the addition of angular momentum to the addition of different translations. If a particle could move along the $x$, $y$ and $z$ axes, the position operator would be a vector operator, which is $${\mathbf r}=X\otimes I_y \otimes I_z + I_x\otimes Y \otimes I_z +I_X\otimes I_y \otimes Z$$ where $X$ $Y$ $Z$ are scalar operator, but r is a vector operator.

Suppose that I have two particles: they have angular momentum, respectively, $J_1$ and $J_2$. The total angular momentum is $J=J_1 \otimes I_2+I_1 \otimes J_2$, where $I_1$ and $I_2$ are the identity operator in the Hilbert spaces of $J_1$ and $J_2$. If $J=J_1 \otimes I_2+I_1 \otimes J_2$ is right, and $J_1, J_2$ are vector operators, then J should be a tensor.

But in the textbook of quantum mechanics, it is often written as $J=J_1+J_2$, and J is a vector.I think $J=J_1+J_2$ is more than a notation, because we are actully using components expression like $J_x=J_1x+J_2x$, which could NOT be deduced from $\mathbf{J} = \mathbf{J}_1 \otimes I_2 + I_1 \otimes \mathbf{J}_2$.

My question is: What makes us treat J as a vector instead of a tensor?

Answer 1

This is a confusing situation because the words "scalar", "vector" and "tensor" have multiple meanings.

If we have vectors $\{u, v\}$ in some vector space $V$, then yes, $u \otimes v$ is a tensor, an element of $V \otimes V$. But $V \otimes V$ is also a vector space, which means that $u \otimes v$ is also a vector within its own space. In QM, when we consider composite systems (or a particle in 3D space) and take tensor products, we might treat the composite states as tensors for the purposes of doing calculations, but conceptually they're vectors in a larger space.

Now let's complicate things even more and talk about operators. A regular operator is a linear function $A: V \to V$; for example, the position operator $X$ in one dimension is a regular operator. When we take tensor products of our space and consider operators on the composite space, we don't think of them as tensors! I would not call $X \otimes I_y \otimes I_z$ a tensor; it's still a function from our Hilbert space (which is a tensor product of smaller spaces) to itself.

To actually get to what we usually call vector and tensor operators, we need to introduce components explicitly. The position vector operator is not as you have written, but

$${\mathbf r}= (X\otimes I_y \otimes I_z, I_x\otimes Y \otimes I_z, I_X\otimes I_y \otimes Z);$$

it's supposed to be a vector, after all! It has three components, each of which is a regular operator on the total Hilbert space $L^2(\mathbb{R}^3) = L^2(\mathbb{R})\otimes L^2(\mathbb{R})\otimes L^2(\mathbb{R})$. To put it another way, it's a linear function $\mathbf{r} \to V \times V \times V$. Notice that we take the Euclidean product, not the tensor product. A tensor operator would then be a sort of matrix of operators, and so on. (There are also some basis transformation properties which I'm ignoring.)

If the angular momenta are vector operators $\mathbf{J}_i$, then the sum is the vector operator

$$\mathbf{J} = \mathbf{J}_1 \otimes I_2 + I_1 \otimes \mathbf{J}_2$$

(where the tensor product distributes over components). Again, we took a tensor product, but we don't call the total operator a tensor. Whether it's a scalar, vector or tensor operator depends on whether it has components, not on whether it acts on a tensor product space. And writing $\mathbf{J} = \mathbf{J_1} + \mathbf{J_2}$ is simply an abuse of notation - we're ignoring the identity operators. We could also write $\mathbf{r} = (X, Y, Z)$ if we care about saving ink (or pixels).