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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.

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  • $\begingroup$ 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
    Commented Sep 11, 2019 at 3:40

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

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

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

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