# Why is the position operator a vector operator?

Suppose we don't know (yet) about the angular momentum operator, and just associate a (unitary) rotation operator $$\hat{R}$$ to every element $$R$$ of the 3D rotation group.

My question is : why does the position $$\hat{\boldsymbol{X}}$$ operator transforms like a regular vector when the system is rotated : $$\langle \hat{X}_i \rangle_{\Psi'} = R_{i,j} \langle \hat{X}_j \rangle_{\Psi} \quad \text{with} \quad |\Psi'\rangle=\hat{R}|\Psi\rangle$$ $$R_{i,j}$$ being the rotation matrix.

• In fact this is a good question because operators should transform as $T x T^{-1}$ if the wavefunction transforms under T. Here the operation is actually defined s as to rotate all vectors to its right
– nox
Sep 11, 2019 at 3:40

You may find the following satisfying. Let $$R$$ be the rotation matrix for position vectors. The answer follows from a simple substitution $$\mathbf{x}\rightarrow R\mathbf{x}$$.
$$\langle \mathbf{\hat{X}} \rangle_{\Psi’} = \int \overline{\Psi}(R^{-1}\mathbf{x})\cdot\mathbf{x}\cdot\Psi(R^{-1}\mathbf{x})\text{ d}\mathbf{x}$$ $$= \int \overline{\Psi}(\mathbf{x})\cdot(R\mathbf{x})\cdot\Psi(\mathbf{x})\text{ d}\mathbf{x}$$ $$=\langle R\mathbf{\hat{X}} \rangle_{\Psi}$$
• Did you mean $\Psi$ rather than $\Psi'$ in the two integrals ? If it is the case, maybe I wasn't clear, but I don't want to assume that $\Psi'(\mathbf{x})=\Psi(R^-1\mathbf{x})$ in the first place (without the phase). In fact, I want to prove that the phase $\alpha$ can be dropped, and I just realized I need something stronger : that $\hat{R}\hat{\mathbf{X}}\hat{R}^{-1}=R\hat{\mathbf{X}}$, so I edited my question. Sep 11, 2019 at 8:22