Regarding linear momentum operator as vector operator

Question

The vector operator $\hat V$ are defined as the vectors which satisfies the commutator,
$$[\hat L_i,\hat V_j]=i\hbar\epsilon_{ijk}\hat V_k.$$
$\hat L$ is the angular momentum operator.

Thus, if the coordinates are rotated by an angle $\theta$ in anticlockwise direction, then
$\begin{pmatrix}\hat p_x' \\ \hat p_y'\\ \hat p_z'\end{pmatrix}=\begin{pmatrix} \cos\theta &\sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}\hat p_x \\ \hat p_y\\ \hat p_z\end{pmatrix}\tag{1}$

But I am not able to prove $(1)$ from the definition of momemtum operator and apply transformation.

After the passive transformation, $\hat p'(x',y',z')=\hat p(x(x',y',z'),y(x',y',z'),z(x',y',z') \\ =\hat p_x (x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat x(x',y',z')+ \hat p_y' (x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat y'(x',y',z') + \hat p_z'(x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat z'(x',y',z') \tag{2}$
As $\begin{pmatrix}\hat x' \\ \hat y'\\ \hat z'\end{pmatrix}=\begin{pmatrix} \cos\theta &\sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\hat x\\ \hat y\\ \hat z\end{pmatrix}$

Inverting the above matrix we get $x(x',y',z'),y(x',y',z'),z(x',y',z')$ and so on.
$\begin{pmatrix}\hat x \\ \hat y\\ \hat z\end{pmatrix}=\begin{pmatrix} \cos\theta &-\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\hat x'\\ \hat y'\\ \hat z'\end{pmatrix}$

So, $(2)$ becomes,
$\boxed{\hat p'_x(x',y',z')=\Big(\hat p_x(x(x',y',z'),y(x',y',z'),z(x',y',z'))\cos\theta -\hat p_y(x(x',y',z'),y(x',y',z'),z(x',y',z'))\sin\theta\Big)}$
$\boxed{\hat p'_y(x',y',z')=\Big(\hat p_x(x(x',y',z'),y(x',y',z'),z(x',y',z'))\sin\theta +\hat p_y(x(x',y',z'),y(x',y',z'),z(x',y',z'))\cos\theta\Big)}$

$\boxed{\hat p'_z(x',y',z')= \Big(p_z(x(x',y',z'),y(x',y',z'),z(x',y',z'))\Big)}$
We can see that the form of $(1)$ and the boxed equations differ by the sign of $\sin\theta$ are same.

Also by transformation equation,
$\hat p_x(x(x',y',z'),y(x',y',z'),z(x',y',z')) = -i\hbar\frac{\partial}{\partial (x'\cos\theta-y'\sin\theta)}$

After doing the complete transformation we can replace the prime from the coordinates without loss of generality.

But from the above analysis I am not able to get $(1)$.

Can somebody help me in proving why $(1)$ holds true?

Addendum.
If we consider the definition that $[L_i,\hat V_j]=i\hbar\epsilon_{ijk}\hat V_k$.
Then using the definition of Heisenberg operator $\hat V_H=e^{\frac{i}{\hbar}L_z}\hat Ve^{-\frac{i}{\hbar}L_z}$ and Baker Campbell Hausdroff formula we can get $(1)$.

But my question is that how to prove $(1)$ say for momentum operator (as we know the form of it) using the transformation equation of the coordinates itself.

Answer 1

One way to do this is to solve Heisenberg’s equations of motion. Writing $\vec V(\phi)$ a vector rotated by angle $\phi$ about an axis $\vec u$, you have by definition: $$ \vec V(\phi)=e^{i\hbar u \cdot \vec L} \vec Ve^{-i\hbar u \cdot \vec L} $$ so taking the derivative with respect to $\phi$ you get Heisenberg’s equations: $$ \frac{d}{d\phi}\vec V= \frac{1}{i\hbar}[\vec u \cdot \vec L, \vec V] \\ = \vec u \times \vec V $$ (second line using the commutation relations)

You can solve this linear equation by calculating the exponential of the operator: $f: \vec x \to \vec u \times \vec x$. This is most conveniently done in a direct orthonormal basis with $\vec u$ as the last vector where the corresponding matrix is: $$ M_f= \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ e^{\phi M_f}= \begin{pmatrix} \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ You can calculate the exponential by diagonalizing the $2\times2$ upper left block for example. You get what you want by taking $\vec V=\vec L$ and $\vec u= \vec e_z$.

Hope this helps