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.

  • $\begingroup$ 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 $\endgroup$
    – nox
    Sep 11, 2019 at 3:40

1 Answer 1


You may find the following satisfying. Let $R$ be the rotation matrix for position vectors. The answer follows from a simple substitution $\mathbf{x}\rightarrow R\mathbf{x}$.

$$ \langle \mathbf{\hat{X}} \rangle_{\Psi’} = \int \overline{\Psi}(R^{-1}\mathbf{x})\cdot\mathbf{x}\cdot\Psi(R^{-1}\mathbf{x})\text{ d}\mathbf{x} $$ $$ = \int \overline{\Psi}(\mathbf{x})\cdot(R\mathbf{x})\cdot\Psi(\mathbf{x})\text{ d}\mathbf{x} $$ $$ =\langle R\mathbf{\hat{X}} \rangle_{\Psi}$$

  • $\begingroup$ 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. $\endgroup$ Sep 11, 2019 at 8:22
  • $\begingroup$ I finally did not edit the question. $\endgroup$ Sep 11, 2019 at 9:01
  • $\begingroup$ Yes sorry the prime shouldn’t be there $\endgroup$ Sep 11, 2019 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.