Transformation of position operator under rotations

[Anj123]

In the momentum representation, the position operator acts on the wavefunction as

1) $X_i = i\frac{\partial}{\partial p_i}$

Now we want under rotations $U(R)$ the position operator to transform as

$U(R)^{-1}\mathbf{X}U(R) = R\mathbf{X}$

How does one show that the position operator as represented in 1) indeed transforms like this?

 

[vanhees71]

In non-relativistic QT you can separate the angular-momentum operator into orbital and spin angular momentum. The spin part commutes. So you are left with the orbital part. Then from the Heisenberg commutator relations you get
$$[x_k,L_l]=[x_k,\epsilon_{lab} x_a p_b]=x_a \epsilon_{lab} [x_k,p_b]=x_a \epsilon_{lab} \mathrm{i} \delta_{kb}=\mathrm{i} \epsilon_{lak} x_a=-\mathrm{i} \epsilon_{lka} x_a,$$
i.e., the infinitesimal rotation is correctly represented on the position operators. Exponentiation yields that indeed $\vec{x}$ are vector operators under rotation.

As you see you don't need a cumbersome explicit representation but you can deal with the abstract operator algebra only!