# What qualifies a set of operators as a "vector operator" in QM?

In quantum mechanics, there are many vector operators like position, momentum, all the types of angular momentum, etc. In Binney's QM book, he often references vector operators and scalar operators. My question is what makes operators like $$J_x$$, $$J_y$$, $$J_z$$ form a "vector operator" and some other set of operators like $$J_x$$, $$J_y$$, $$P_z$$ not form a vector operator? I ask because Binney often references properties that only apply to vector operators and not scalar operators so there is definitely some difference between them.

• The observable is a vector or scalar. Apr 6 '19 at 1:55

Let $$R$$ be a rotation, $$\widehat{R}$$ the unitary operator representing the rotation and $$\mathcal{R}_{ij}$$ be the rotation matrix asoaciated to $$R$$.

The set $$K=(K_{1},K_{2},K_{3})$$ form a vector operator if the components of $$K$$ as ransform under a rotation as

$$K_{i}'=\widehat{R}K_{i}\widehat{R}^{-1}=\mathcal{R}_{ij}^{-1}K_{j}.$$

Do note that normal algebraic vectors $$V=(V_{1},V_{2},V_{3})$$ transform as

$$V_{i}'=\mathcal{R}_{ij}V_{j}.$$

EDITED to correct some typos.

• Although your answer was accepted by the questioner, I can't understand it. Maybe in the first equation there is some apex lacking? (A $K'$ at LHS?). What do you mean by "transform"? Transform when? under what action? Your second equation too looks strange, under two respects: 1) the same $V$ on both sides? 2) why "normal" algebraic vectors should transform in the opposite way to quantum ones? Apr 6 '19 at 14:08
• Indeed, the LHS of the first equation should be a K′, and in the second equation it should be a V′. When I say the operator "transform" it means transform under a unitary transformation, in this case the unitary transformation that represent a rotation. (2) See section 11 of chapter 11 of Albert Messiah "quantum mechanics". The reason is basically that probabilities have to remain the same under rotation in some sense. Apr 6 '19 at 14:18

My question is what makes operators like Jx, Jy, Jz form a "vector operator" and some other set of operators like Jx, Jy, Pz not form a vector operator?

$$\mathbf{L} = \mathbf{x}\times\mathbf{p}$$ is the angular momentum operator. $$\mathbf{A} = (A_1, A_2, A_3)$$ is a vector operator if and only if

$$[L_i, A_j] = L_iA_j - A_jL_i = i\epsilon_{ijk}A_k$$

It follows that $$\mathbf{L}$$ is a vector operator since

$$[L_i, L_j] = i\epsilon_{ijk}L_k$$

• Why does this statement qualify as a reasonable condition to call it a vector? Apr 6 '19 at 4:59
• It comes from analyzing infinitesimal rotations of the operator. Apr 6 '19 at 14:19