# Addition of angular momenta

1. Jan 11, 2009

### 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.

2. Jan 11, 2009

### strangerep

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?

3. Jan 12, 2009

### 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.

4. Jan 12, 2009

### jensa

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

5. Jan 12, 2009

### tommy01

many thanks!

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