quantum mechanics - Rigorous mathematical definition of vector operator?

In standard quantum mechanics textbooks, the concept of operators is often introduced as linear maps that map a Hilbert space $H$ onto itself: $$ \hat{O}: H \rightarrow H \, . $$

However, directly after, we use the position operator $\hat{\vec{x}}$, which isn't of the said shape, but instead is like a triple of operators, e.g. $\hat{\vec{x}} = (\hat{x}, \hat{y}, \hat{z})$. What I thought now is that maybe, one could treat the position operator as a linear map $$ \hat{\vec{x}}:H \rightarrow H \otimes R^3 \, . $$

Is it wrong to say the that the position operator is that kind of map? Does it make sense?

Edit: I know that you can treat the position operator as three independent operators $\hat{x}$, $\hat{y}$, $\hat{z}$. I just want to know if my way of treating the position operator is an equivalent way to see things, or if it is wrong to treat it like that. If it is wrong, why is it wrong?

Answer

So there's a natural isomorphism

$$ \varphi: H^{\oplus 3} \to H \otimes \mathbb R^3\\ (a, b, c) \mapsto a \otimes e_0 + b \otimes e_1 + c \otimes e_2 $$

where $e_i$ is a basis for $\mathbb R^3$. The definition you're objecting to is

$$ \hat{\vec{x}}: H \to H^{\oplus 3}\\ v \mapsto (\hat x(v), ~ \hat y(v) , ~\hat z(v)) $$

You can convert it to your version by composing with the above isomorphism.

$$ \tilde x \equiv \varphi \circ \hat{\vec{x}}:~~ H \to H \otimes \mathbb R^3\\ v \mapsto \hat x(v) \otimes e_0 + ~ \hat y(v) \otimes e_1 + ~\hat z(v)\otimes e_2 $$

So your definition makes complete sense.

Comments

So this answers the question as stated. Judging by your comments on other answers I think you're looking for a nicer way to formalize the annoying statement 'blah operator transforms under blah group like a vector'. Whether this definition yields what you want should be the subject of another question.

Namely I think you're hoping somehow conjugation by a representation of the group will factor through your tensor product and have the group action on $\mathbb R^3$ just act on the second factor. Spent a couple of minutes trying to naively get this to work without much luck, will update if there's progress, or at least give a better explanation of why not.

Edit

I'm going to describe how this construct makes the "covariance under conjugation by some representation of some action on $\mathbb R^3$" condition more explicit, as OP had hoped.

For clarity let's start by giving a name to the operation of conjugating this "vector operator".

Let $G$ be some group that admits an action on $\mathbb R^3$. Let $U: G \to H$ be a unitary representation of $G$ on $H$. Define, for any $R \in G$

\begin{align} C_R: \mathrm{Mor}(H, H^{\oplus 3}) &\to \mathrm{Mor}(H, H^{\oplus 3})\\ f(\cdot) &\mapsto (U(R) f(\cdot)^i U(R)^\dagger)_{i \in (0, 1, 2)} \end{align}

where $\mathrm{Mor}(V, W)$ is collection of linear maps $V \to W$. This is just a name for component-wise conjugation of this tuple of operators.

Now the main claim

Proposition The statement

$$ C_R(\hat{\vec{x}}) = \left(\sum_{j}R_{ij}\hat{\vec{x}}(v)^j\right)_{i \in (0, 1, 2)} $$

is equivalent to

$$ \varphi \circ C_R(\hat{\vec{x}}) = (1 \otimes R)(\varphi \circ \hat{\vec{x}}) $$

Proof For brevity we give the $\implies$ direction, the other direction should be obvious from the same computation

\begin{align} \varphi\left(\left(\sum_{j}R_{ij}\hat{\vec{x}}(v)^j\right)_{i \in (0, 1, 2)}\right) &= \sum_i\left(\sum_{j} R_{ij} \hat{\vec{x}}(v)^j\right) \otimes e_i \\ &= \sum_{ij} \hat{\vec{x}}(v)^j \otimes R_{ij}e_i\\ &=\sum_{j} \left(\hat{\vec{x}}(v)^j \otimes \sum_i R_{ij}e_i\right)\\ &= (1 \otimes R)\left(\sum_{j} \hat{\vec{x}}^j(v) \otimes e_j \right)\\ &=(1 \otimes R)\left(\varphi(\hat{\vec{x}(v)}) \right) \end{align}