# Why is the angular momentum added for two independent electron system? (no problem)

There is no problem now. But somebody may be confused by the same analysis when studying QM or Group theory. (actually my motivation for asking this question comes from the SU(5) Grand Unification theory where the hypercharge simply add up for the antisymmetric tensor representation of SU(5).)

When we consider two independent electron 1 and 2, we always have $$\tag1 J=J_1+J_2,$$ where $J_1$, $J_2$ are the angular momentum operators for the electron 1 and 2 respectively, and $J$ is the total angular momentum operator. But here now I am confused by Eq.(1). $J_1$ is an operator acting upon Hilbert space $\mathcal{H}_1$ of elector 1 and $J_2$ is an operator acting upon Hilbert space $\mathcal{H}_2$ of electron 2. Then what does it mean by $J_1+J_2$ when $J_1$, $J_2$ act upon different Hilbert space? You may think that $J_1$ actually is $$J_1\otimes I$$ and $J_2$ actually is $$I\otimes J_2,$$ where $I$ is the identity operator for the corresponding Hilbert space. Then $J$ actually is $$\tag2 J=(J_1\otimes I)+(I\otimes J_2)=J_1\otimes J_2. \text{ (note that this is incorrect)}$$ Then now comes another question. Suppose $J_1$ has eigenvalue $j_1$, $J_2$ has eigenvalue $j_2$, then how to prove that $j_1+j_2$ is the eigenvalue of $J\equiv J_1\otimes J_2$?

Now since Eq.(2) is incorrect, we may use $(J_1\otimes I)+(I\otimes J_2)$ rather than $J_1\otimes J_2$. Suppose $\psi_1$ is the eigenvector of $J_1$ with eigenvalue $j_1$, $\psi_2$ is the eigenvector of $J_2$ with eigenvalue $j_2$. Then we have $$\left((J_1\otimes I)+(I\otimes J_2)\right)(\psi_1\otimes\psi_2)=(j_1+j_2)(\psi_1\otimes\psi_2).$$ Q.E.D

• eqn (2) is not correct IMO. how did you get the tensor product in the end. – Bruce Lee Jan 7 '16 at 21:33

Your initial identification, $$J=(J_1\otimes I)+(I\otimes J_2),$$ is correct. However, it's incorrect to change this sum into a product: $$(J_1\otimes I)+(I\otimes J_2)\neq J_1\otimes J_2.$$ (If nothing else, you want $J_1+J_2$ to double if you double both $J_1$ and $J_2$, whereas $J_1\otimes J_2$ quadruples under that transformation.)
Then, if you want to translate eigenvectors $|\psi_1⟩$ of $J_1$ and $|\psi_2⟩$ of $J_2$ into an eigenvector of $J$, you can simply take their tensor product, $|\psi_1⟩\otimes|\psi_2⟩$, and this will obey $$(J_1\otimes I)|\psi_1⟩\otimes|\psi_2⟩=j_1|\psi_1⟩\otimes|\psi_2⟩$$ and $$(I\otimes J_2)|\psi_1⟩\otimes|\psi_2⟩=j_2|\psi_1⟩\otimes|\psi_2⟩,$$ giving an overall eigenvector of $J$ with eigenvalue $j_2$.
Note, however, that if these are total angular momentum eigenvectors for different particles, in general it is not the case that the eigenvalue of the total angular momentum of the system will be $j_1+j_2$. The reason for this is that the total angular momentum operator is really \begin{align} \mathbf J^2 &=(\mathbf J_1+\mathbf J_2)^2 =\mathbf J_1^2+2\mathbf J_1\cdot\mathbf J_2+\mathbf J_2^2 \\ & =\mathbf J_1^2+2J_{1,x} J_{2,x}+2J_{1,y} J_{2,y}+2J_{1,z} J_{2,z}+\mathbf J_2^2, \end{align} and the presence of all three $J_{1,i}$ makes it more difficult to find the new eigenstates. It's perfectly doable, of course, via the well-known procedure of addition of angular momenta in quantum mechanics, but you do need to do it carefully.