What is the Meaning of [J1,J2] = 0 in the Addition of Angular Momenta?

[tommy01]

hi everybody.

$$\textbf{J}_1$$ and $$\textbf{J}_2$$ are angular momentum (vector-)operators.
In many textbooks $$\left[\textbf{J}_1,\textbf{J}_2\right] = 0$$ is stated to be a condition to show that $$\textbf{J}=\textbf{J}_1+\textbf{J}_2$$ is also an angular momentum (vector-)operator. But what is meant with $$\left[\textbf{J}_1,\textbf{J}_2\right] = 0$$. When i show that $$\textbf{J}$$ is an angular momentum operator (i.e. $$\left[J_x,J_y\right]=iJ_z$$ ...) i always need the condition $$\left[(\textbf{J}_1)_x,(\textbf{J}_2)_x\right]$$ and the like. So the components of $$\textbf{J}_1$$ and $$\textbf{J}_2$$ should mutually commute. Is this the meaning of $$\left[\textbf{J}_1,\textbf{J}_2\right] = 0$$? For me it looks like $$(\textbf{J}_1)_x(\textbf{J}_2)_x+(\textbf{J}_1)_y(\textbf{J}_2)_y+(\textbf{J}_1)_z(\textbf{J}_2)_z-(\textbf{J}_2)_x(\textbf{J}_1)_x-(\textbf{J}_2)_y(\textbf{J}_1)_y-(\textbf{J}_2)_z(\textbf{J}_1)_z=0$$ and this does not imply the conditions i need (as far as i see).
I know Operators acting on different spaces commute and this fact is often used but i want to know how to treat the situation above only with the formal condition $$\left[\textbf{J}_1,\textbf{J}_2\right] = 0$$.

thanks and greetings tommy.

 

[strangerep]

[...] Is this the meaning of $$\left[\textbf{J}_1,\textbf{J}_2\right] = 0$$? For me it looks like $$(\textbf{J}_1)_x(\textbf{J}_2)_x+(\textbf{J}_1)_y(\textbf{J}_2)_y+(\textbf{J}_1)_z(\textbf{J}_2)_z-(\textbf{J}_2)_x(\textbf{J}_1)_x-(\textbf{J}_2)_y(\textbf{J}_1)_y-(\textbf{J}_2)_z(\textbf{J}_1)_z=0$$

Huh? You have two separate copies of the so(3) Lie algebra, so that commutation relation
just says that every generator from copy #1 commutes with every generator from copy #2.

Or did I misunderstand the question?

 

[tommy01]

Hi. Thank you for your answer. As i mentioned i know that every component of $$\textbf{J}_1$$ has to commute with every component of $$\textbf{J}_2$$ to show that $$\textbf{J}_1+\textbf{J}_2$$ is an angular momentum operator (another generator of the group) and because they act on different subspaces of the system. But the explicit form of $$\left[\textbf{J}_1,\textbf{J}_2\right]=0$$ (which is the only given condition) is $$(\textbf{J}_1)_x(\textbf{J}_2)_x+(\textbf{J}_1)_y( \textbf{J}_2)_y+(\textbf{J}_1)_z(\textbf{J}_2)_z-(\textbf{J}_2)_x(\textbf{J}_1)_x-(\textbf{J}_2)_y(\textbf{J}_1)_y-(\textbf{J}_2)_z(\textbf{J}_1)_z=0$$. So is $$\left[\textbf{J}_1,\textbf{J}_2\right]=0$$ just an abbreviation for $$\left[(\textbf{J}_1)_i,(\textbf{J}_2)_j\right]=0 \forall i,j$$ or does the explicit form of the commutator imply this?
greetings.

 

[jensa]

So is $$\left[\textbf{J}_1,\textbf{J}_2\right]=0$$ just an abbreviation for $$\left[(\textbf{J}_1)_i,(\textbf{J}_2)_j\right]=0 \forall i,j$$ or does the explicit form of the commutator imply this?

Yes, the abbrevation you wrote is correct (there is no scalar product involved in the commutator, statement is about each element of the vectors $$\mb{J}_1,\mb{J}_2$$). If it helps you can write the elements of what is basically $$\mathfrak{su}(2)\otimes\mathfrak{su}(2)$$ algebra as $$(J_1)_i=J_i\otimes 1,$$ and $$(J_2)_i=1\otimes J_i$$. Then it becomes more obvious that the two copies of the underlying algebra commute.

Hope this helps

 

[tommy01]

many thanks!