# Behavior of total kinetic energy operator under local rotations

Consider the space of square integrable complex functions $$f(x_1,\ldots,x_n)$$ on $$\mathbb{R}^{3n}$$ taking $$n$$ three-dimensional vectors $$x_i$$ as arguments. Assign to each vector $$x_i$$ a mass $$m_i$$ and define the total mass $$m_o = \sum_{j=1} m_j$$ and the center of mass $$x_o = \frac{\sum_{j=1}^n m_j x_j}{m_o}\,\,\,.$$

We define also a total kinetic energy operator $$t_o=\sum_{j=1}^n t_j$$ where $$t_j = \frac{1}{m_j}\nabla^2_j$$ where $$\nabla^2_j f$$ is the Laplacian of a function $$f$$ with respect to the argument $$x_j$$.

We can define an operator $$O$$ affecting a collective rotation $$x_i \to x_i' = R x_i$$ of all the vectors $$x_i$$, where $$R \in O(3)$$ is a proper or improper rotation, i.e. $$Of(x_1,\ldots,x_n) = f(Rx_1,\ldots,Rx_n)$$ and one can notably show that $$O$$ commutes with $$t_o$$.

However we can also imagine, given a operation $$R \in O(3)$$, an operator $$O'$$ affecting a "local" rotation $$x_i \to R\left(x_i - x_o\right) + x_o\,\,\,.$$

My first question is: would $$O'$$ also commute with $$t_o$$?

Finally, we can also imagine an operator $$O''$$ affecting the coordinate transformation: $$x_i \to R'(x_o) (x_i - x_o) + x_o$$ where $$R' : \mathbb{R}^3 \to O(3)$$ is some map taking a center of mass $$x_o$$ and return a proper or improper rotation $$R'(x_o)$$. Does $$O''$$ commute with $$t_o$$ for any $$R'$$? I would not think so, especially since we have no conditions on the map $$R'$$.

I was able to verify via "brute force" that $$O'$$ does in fact commute with $$T$$. I also found a counter-example demonstrating that there are operators $$O''$$ which do not commute with $$T$$. I provide here a sketch of the calculations.
In general for an operator $$O$$ affecting a coordinate transformation $$y_i \to y_i'(\ldots,y_j,\ldots)$$, i.e. an operator $$O$$ acting on a function $$f(\ldots,y_j,\ldots)$$ so that $$Of(\ldots,y_i,\ldots)=f\left(\ldots,y_i'\left(\ldots,y_j,\ldots\right),\ldots\right)$$ then we have $$\partial_i O = \left(\partial_i y'_j\right) O \partial_j \tag{1}$$ where $$\partial_i \equiv \frac{\partial}{\partial y_i}$$ is an operator acting on functions in the obvious way. In our case we have an operator $$O'$$ affecting a coordinate transformation $$x_{\alpha i} \to x_{\alpha i}'\left(\ldots,x_{\beta j},\ldots\right)=\sum_{j=1}^3 R_{ij}\left(x_{\alpha j} - x^o_j\right) - x^o_i$$ where $$x_{\alpha i}$$ denotes component $$i$$ of particle $$\alpha$$, $$R_{ij}$$ is a $$3 \times 3$$ orthogonal matrix, and $$x^o_i = \frac{\sum_{\alpha=1}^n m_\alpha x_{\alpha i}}{\sum_{\alpha=1}^n m_\alpha}$$ is component $$i$$ of the center of mass. Defining $$m\equiv\sum_{\alpha=1}^n m_\alpha$$, we have then from Eq. (1) $$[\partial_{\alpha i},O'] = O' \left(R_{ji} - \delta_{ij}\right)\left(\delta_{\alpha \beta} - m_\alpha / m\right)\partial_{\beta j}\equiv O' M^{\alpha i}_{\beta j} \partial_{\beta j}$$ with implicit summing over repeating indices. From $$[A^2,B] = A[A,B] + [A,B]A$$ we obtain $$[\partial_{\alpha i}^2,O'] = O'\left(2 M^{\alpha i}_{\beta j} \partial_{\alpha i}\partial_{\beta j} + M^{\alpha i}_{\gamma k}M^{\alpha i}_{\beta j}\partial_{\gamma k}\partial_{\beta j}\right)$$ By carefully plugging expanding out all the terms and applying the equations $$\sum_{\alpha=1}^n m_\alpha = m$$ and $$\sum_{k=1}^3 R_{ik} R_{jk} = \delta_{ij}$$ one can show that indeed $$[T,O']=[\sum_{\alpha=1}^n m_\alpha^{-1} \sum_{i=1}^3 \partial^2_{\alpha i},O'] = 0$$ A perhaps cleaner approach requires first defining the coordinate system $$x^o_i,\Delta x_{2i},\ldots,\Delta x_{ni}$$ where the $$x^o_i$$ for $$i=1,2,3$$ are the same center of mass coordinates while the $$\Delta x_{\alpha i} \equiv x_{\alpha i} - x^o_i$$ for $$\alpha=2,\ldots,n$$ are the relative coordinates. One can show that $$T = \frac{1}{m}\sum_{i=1}^3 \tilde{\partial}^2_{oi} - \frac{1}{m}\sum_{\alpha,\beta=2}^n \sum_{i=1}^3 \tilde{\partial}_{\alpha i} \tilde{\partial}_{\beta i} + \sum_{\alpha=2}^n \frac{1}{m_\alpha} \sum_{i=1}^3 \tilde{\partial}_{\alpha i}^2$$ where the $$\tilde{\partial}$$ denotes a partial derivative with respect to the new coordinate system. Specifically we have $$\tilde{\partial}_{oi} \equiv \frac{\partial}{\partial x^o_i}$$ and $$\tilde{\partial}_{\alpha i} \equiv \frac{\partial}{\partial \Delta x_{\alpha i}}$$. It is more straightforward to show that $$O'$$ commutes with $$T$$ in these coordinates.
If one generalizes the operator $$O'$$ to an operator $$O''$$ affecting "local rotations" corresponding to a coordinate transformation $$x_{\alpha i} \to x'_{\alpha i} = R'\left( x^o \right)_{ij} \left( x_{\alpha j} - x^o_j \right) - x^o_i$$ where the rotation matrix $$R'_{ij}$$ is a function of the center of mass coordinates $$x^o_i$$, then in general we have $$[T,O''] \neq 0$$. I derived a rather complicated expression for the commutator which I am not sure is fully simplified. Instead I will give an example of a function $$f$$ of $$n=2$$ particles and a local rotation function $$R'_{ij}\left(x^o\right)$$ where the $$[T,O'']f \neq 0$$. Denoting the three center of mass coordinates $$X$$, $$Y$$, and $$Z$$, and the sole triplet of relative coordinates $$\mathbf{r} = \left(x,y,z\right)$$, we define a function $$f$$ so that $$f(X,Y,Z;x,y,z) = f(Z;\mathbf{r}) = e^{iZ} \left(R^y_Z \left(Z\right) \mathbf{\hat{z}}\right) \cdot \mathbf{r}$$ where $$R^a_\phi$$ denotes a rotation by an angle $$\phi$$ about the $$a$$ axis. We define the local rotation so that $$R'\left(Z\right) = R^y_Z R^z_{-\pi/2} R^y_{-Z}$$ If we set the two masses to be equal then the kinetic energy operator $$T$$ has the form $$\frac{1}{m} \left( \sum_{i = X,Y,Z} \partial_i^2 + \sum_{i = x, y, z} \partial_i^2 \right)$$ The second derivatives with respect to the relative coordinates will annihilate $$f$$, and we also have $$O''f = e^{iZ} \left( R^y_Z \mathbf{\hat{z}} \right) \cdot \left( R'(Z) \mathbf{r} \right) = e^{iZ} \left( R^y_Z R^z_{\pi/2} R^y_{-Z} R^y_Z \mathbf{\hat{z}} \right) \cdot \mathbf{r} = f$$ so that $$[\sum_{i = x, y, z} \partial_i^2, O'']f = 0$$ So if we can show $$[\sum_{i=X,Y,Z}\partial_i^2,O'']f = [\partial_Z^2,O'']f \neq 0$$ then it follows that $$[T,O'']f \neq 0$$ and thus $$[T,O''] \neq 0$$. To this end we note that $$\partial_Z^2 f = 2\left( i g - f\right)$$ where $$g(Z;\mathbf{r}) = e^{iZ} \left( R^y_Z \mathbf{\hat{x}} \right) \cdot \mathbf{R}$$ i.e. $$g$$ is obtained from $$f$$ by a substitution $$\mathbf{\hat{z}} \to \mathbf{\hat{x}}$$. We already noted that $$O'' f = f$$, while for $$g$$ we have $$O''g = e^{iZ} \left( R^y_Z R^z_{\pi / 2} R^y_{-Z} R^y_Z \mathbf{\hat{x}} \right) \cdot \mathbf{r} = e^{iZ} \left( R^y_Z \mathbf{\hat{y}} \right) \cdot \mathbf{r} \equiv h$$ so while $$\partial_Z^2 O'' f = \partial_Z^2 f = 2( i g - f )$$ we on the other hand have $$O'' \partial_Z^2 f = 2 O'' \left( i g - f \right) = 2 \left( i h - f \right)$$ so that $$[ T, O'' ]f = \frac{2i}{m} \left( g - h \right) \neq 0$$ Note that we have the commutator fails to vanish even for a local rotation function $$R'(Z)$$ that remains for all $$Z$$ in the same conjugacy class $$\cong \pi / 2$$.