Rotational invariance of many particle (quantum system) system? I am trying to prove pairwise coupled harmonically interacting (quantum) system of particles as rotationally invariant . $$H=\frac{1}{2}\sum_{j}p_{j}^2 +\frac{1}{2}k\sum_{i<k}(\vec{r_{i}}-\vec{r_{k}})^{2}$$ as rotationally invariant. I defined the (quantum mechanical) total angular momentum as $$\vec{L}=\vec{L_{1}}+\vec{L_{2}}+\vec{L_{3}}+......+\vec{L_{n}}$$ and expanded both LHS and RHS along the $$\vec{i},\vec{j},\vec{k}$$ components as follows.$$L_{x}=L_{x_{1}}+L_{x_{2}}+L_{x_{3}}+L_{x_{4}}+.....+L_{x_{n}}$$ similarly for y and z components $$L_{y}=L_{y_{1}}+L_{y_{2}}+L_{y_{3}}+L_{y_{4}}+.....+L_{y_{n}}$$
$$L_{z}=L_{z_{1}}+L_{z_{2}}+L_{z_{3}}+L_{z_{4}}+.....+L_{z_{n}}$$. Here $x_{1},x_{2}..x_{n}$ & $p_{x_{1}},p_{x_{2}}..p_{x_{n}} $ are the x components of position and momenta of all the n particles and similarly for $y$ and $z$. Also, $$L_{x_{1}}=y_{1}p_{z_{1}}-z_{1}p_{y_{1}}$$ and similarly for every particle
To prove the system to be rotationally invariant it should satisfy the following condition,$$[H,L_{x}]=0=[H,L_{y}]=[H,L_{z}]$$
Is it correct ? Am I making any mistake. Can I add quantum angular momentum vectors as mentioned above ?
 A: Yes, you define the total (orbital) angular momentum as you have written above. You can easily show that the total angular momentum still satisfies the angular momentum algebra 

[$L_i$,$L_j$] = i$\hbar$ $\varepsilon_{ijk} L_k$

because the position and momentum operators for different particles commute ($[x_i, p_{x_j}] = i\hbar \delta_{ij}$ and so on) and you can use the angular momentum algebra satisfied by angular momentum of individual particles. 
You are also correct that you need to show $[H,\vec{L}] = 0$ to show that your Hamiltonian $H$ is rotationally invariant. The reason is that $\vec{L}$ is the generator of the rotation operator (by angle $\phi$ about axis $\hat{n}$) $D(\phi,\hat{n}) = \exp(-i\phi \frac{\hat{n} \cdot \vec{L}}{\hbar})$ and [H,$\vec{L}$] =0 implies $D(\phi,\hat{n}) H D(\phi,\hat{n})^{\dagger} = H$, by BCH identity. 
To show $[H,\vec{L}]=0$ here is straightforward. First note that for individual particles, $[p_i,L_j] = i\hbar \varepsilon_{ijk} p_k$ (the indices here refer to x,y, and z). So $[p_i^2,L_j] =2i\hbar p_i p_k \varepsilon_{ijk} = 0$ and hence the kinetic energy part is rotationally invariant. Similarly, for individual particles, $[x_i, L_j] = i\hbar \varepsilon_{ijk} x_k$ (again, the indices are for x,y,and z, not the particle number) and you can use the same algebra to show the potential part is rotationally invariant as well.
