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The title pretty much says it all. For example, we sometimes think of the three momentum operators as components of a "vector operator" $\hat{ \vec{p}}=(\hat p_x,\hat p_y,\hat p_z)$. The terminology is often justified via commutation relations and can made rigorous using representation theory [1] [2].

But does there exist (or can we construct) a space in which the vector operator $\hat{\vec p}$ lives, as an actual vector (or generalized vector) such that the standard operators $\hat p_i$ are genuine components? This would be nice because then, for instance, we could derive the transformation laws for how the $\hat p_i$ change under a change of basis, rather than imposing them by hand.

A first thought would be something like End$(H)\times{}$End$(H)\times{}$End$(H)$. But this space has infinite dimension [assuming $H=L^2(\mathbb R^3)$], and we want a three-dimensional space. We also want the components to be operators, which a standard vector space can't accommodate. But maybe a module would work?

Qmechanic
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WillG
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    Related/possible duplicate (at least the final paragraph of my answer there would be my answer to this question, too): https://physics.stackexchange.com/q/525919/50583 – ACuriousMind Jul 12 '23 at 16:28
  • I'm not convinced this has anything to do with quantum mechanics per se. You're wondering whether functions from $A\to B^k$ can be seen as belonging to a $k$-dimensional space with components in $A\to B$. – benrg Jul 12 '23 at 20:04

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