# Regarding linear momentum operator as vector operator

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?

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.

• I think problem is in your transformation equation, this is how I will transform partial differential $\frac{\partial}{\partial{x'}}=\frac{\partial}{\partial{x}}\frac{\partial{x}}{\partial{x'}}+\frac{\partial}{\partial{y}}\frac{\partial{y}}{\partial{x'}}$ use this you will get $p_x'=cos{\theta}p_x+sin{\theta}p_y$ Commented Sep 11, 2022 at 14:52
• Thanks for the comment. We have got the transformation equation for $p_x$ But we actually have $p_x\hat x$ Then $\hat x$ will also get transform because $p_x$ is not a number it is an operator.
– Manu
Commented Sep 12, 2022 at 8:23

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 $$\vec u= \vec e_z$$.