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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?

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    $\begingroup$ The position as you have written it is a scalar - it doesn't have components! $\endgroup$
    – Javier
    Commented Mar 3, 2022 at 14:28
  • $\begingroup$ @CosmasZachos But that's not what a vector operator usually means, is it? $\endgroup$
    – Javier
    Commented Mar 3, 2022 at 14:46
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    $\begingroup$ @Javier You are right, there is no way to project out vector components of that "vector", as the three terms are not trace-orthogonal. $\endgroup$ Commented Mar 3, 2022 at 15:01
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    $\begingroup$ As for the vector operator, I wasn't going to go there: it is the systematic mixing of components under rotations, and, indeed, any scheme to mix components should fail.... $\endgroup$ Commented Mar 3, 2022 at 15:03
  • $\begingroup$ @CosmasZachos Could you give a little bit of explanation of the relationship between vector split and trace orthogonal? $\endgroup$
    – Jack
    Commented Mar 3, 2022 at 15:34

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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).

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    $\begingroup$ The 500lb gorilla in the room is the fact that the coproducts J $_i$ obey the same mutual commutation relations with the other components as J $_{i~1}$ and J $_{i~2}$, so all three obey the same commutation relations (in different reps of the rotation algebra). This is the crucial feature of the vector component split. Each summand, 1 and 2, is traceless. Moreover, they are mutually trace orthogonal! $\endgroup$ Commented Mar 3, 2022 at 15:13
  • $\begingroup$ 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 $\mathbf{J} = \mathbf{J}_1 \otimes I_2 + I_1 \otimes \mathbf{J}_2$. $\endgroup$
    – Jack
    Commented Mar 3, 2022 at 15:38
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    $\begingroup$ @Jack writing $J_x = J_{1x} + J_{2x}$ is also an abuse of notation. The "correct" version would be $J_x = J_{1x} \otimes I_2 + I_1 \otimes J_{2x}$. $\endgroup$
    – Javier
    Commented Mar 3, 2022 at 16:13
  • $\begingroup$ @CosmasZachos I'm not sure I understand your comment. Why do the commutation relations or the trace matter here? I think OP's question is more about vector operators and tensor products, and not so much about the specific properties of the angular momentum operator. $\endgroup$
    – Javier
    Commented Mar 3, 2022 at 16:15
  • $\begingroup$ The trace is a quick handle on projecting components from a sum. You are right the commutation is not relevant to the OP's confusion, but it is the very reason this object was defined in the first place, and reminds him the sum rotates identically to its summands, so a vector. $\endgroup$ Commented Mar 3, 2022 at 16:23

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