# Adding Angular Momenta Operators in QM

I am gonna give a much shorter answer than @Cyro.

- Yes it is also true for the other components of angular momentum.
- It is simply due to vector addition. The total angular momentum of the system (were it classical) would be $\mathbf{J} = \sum_i\mathbf{J}_i$. For quantum, it's the same thing, but you just quantise the observable into an operator.
- The "catch" is that while $J_{\mathrm{tot}}$ and $J_z$ commute, the $x,y,z$ components of the angular momentum do not commute among themselves. The choice of the $z$ axis is conventional in this context. So you only know the value of the total angular momentum $\sqrt{j(j+1)}$
*and*the value of its $z$ projection $m_z = m_{z_1} + m_{z_2}$.

In other words, $m_z = m_{z_1} + m_{z_2}$ but $m_x \neq m_{x_1} + m_{x_2}$.

I think this is a great question. It also puzzled me for a while.

The key here is irreducible representations of the rotation group. You start with one quantum particle, the state of this quantum partile is $|\psi\rangle_1$ which is a vector in some Hilbert space $\mathcal{H}_1$. You also have a set of operators $\exp\left(iJ_{1x}\theta_x\right),\, \exp\left(iJ_{1y}\theta_y\right),\, \exp\left(iJ_{1z}\theta_z\right)$ that change this state to appear as it would appear to some other observer rotated by angles $-\theta_{x,y,z}$ arround the corresponding axis.

More generally you have an operator

$R_1\left(\boldsymbol{\theta}\right)=\exp\left(i\left[J_{1x}\theta_x+J_{1y}\theta_y+J_{1z}\theta_z\right]\right)$

That changes your state into one that would be observed by another, 'rotated', observer.

What you are after is description of the system that would be 'independent' of observer position. Whilst observers may not agree on all aspects of the state, they may agree on some of its aspect, more specifically they will agree on whether state is in a specific irreducible representation of $R_1\left(\boldsymbol{\theta}\right)$. More generally all observers can agree on the decomposition of $|\psi\rangle_1$ into sub-spaces of $\mathcal{H}_1$ that are mapped into themselves by all $R_1\left(\boldsymbol{\theta}\right)$. The simplest form of this is spherical symmetry, i.e. all observers will agree if the state is spherically symmetric. However, there are other forms of this, and that's the irreducible representations. More specifically, that's the irreducible representations of the SO(3) Lie Group, with elements $R_1\left(\boldsymbol{\theta}\right)$. If you look at the representation theory of this group, you will find that for a given representation it is sufficient, and much easier to find irreducible representations of the Lie algebra (rather than the actual group), i.e. the irreducible representations of $\mathbf{J}_1=\left(J_{1x},\,J_{1y},\,J_{1z}\right)$.

Now consider two such particles. The full state of the system is now $|\psi_1\psi_2\rangle=|\psi\rangle_1\otimes|\psi\rangle_2$ that is a vector in the tensor product space of the two underlying Hilbert spaces, $\mathcal{H}_1\otimes\mathcal{H}_2$. The rotations of this state are now represented by:

$R_{12}\left(\boldsymbol{\theta}\right)=R_{1}\left(\boldsymbol{\theta}\right)R_{2}\left(\boldsymbol{\theta}\right)$

And you are still seeking to find irreducible sub-spaces of this new representation of SO(3) group. Assuming that $\left[\mathbf{J}_1, \mathbf{J}_2\right]=0$ we have:

$R_{12}\left(\boldsymbol{\theta}\right)=\exp(i\mathbf{J}_1.\boldsymbol{\theta})\exp(i\mathbf{J}_2.\boldsymbol{\theta})=\exp(i\left(\mathbf{J}_1+\mathbf{J}_2\right).\boldsymbol{\theta})=\exp(i\mathbf{J}_{12}.\boldsymbol{\theta})$

i.e. the Lie algebra of this new representation is simply $\mathbf{J}_{12}=\mathbf{J}_1+\mathbf{J}_2$. Therefore the irreducible subspaces you need to find are the irreducible sub-spaces of $\left(\mathbf{J}_1+\mathbf{J}_2\right)$. These will be the subspaces of $\mathcal{H}_1\otimes\mathcal{H}_2$ that all observers will agree on. These will also turn out to be subspaces with specific angular momentum numbers ($j$), but that's peculiarities of the SO(3) representation theory (see https://en.wikipedia.org/wiki/Casimir_element).

Sorry if my explanation is a bit muddled, but I hope it conveys the general idea. The reason you add up angular momentum operators is that you multiply the rotation operators, and the reason for that, is that you combine different Hilbert spaces through tensor products.

*The point of this explanation is that it does not need classical mechanics, or even notion of angular momentum operator.* The reasoning here can be conducted entirely in terms of observers and seeking to find unique ways to represent states of the system. The connection to classical angular momentum comes much later, you find quantity that is conserved as a result of isotropy of space ($j$), and in classical mechanics this quantity is angular momentum, so you link the two.

For the "addition of angular momentum" the idea is that you're studying a system with two sources of angular momentum. For example, if your system consists of two particles, each of which having their own spin or orbital angular momentum then the total angular momentum will be a combination of them both; or if you just have one particle with both spin and orbital angular momentum, describing both of these angular momenta requires the same machinery.

Let $|\psi_{1}\rangle \in \mathscr{H}_{1}$ and $|\psi_{2}\rangle \in \mathscr{H}_{2}$ be the quantum states associated to each piece of angular momentum and $\mathscr{H}_{1,2}$ be the associated Hilbert spaces. Then the angular momentum of the system as a whole is described by the *tensor product* space $|\Psi\rangle = |\psi_{1}\rangle \otimes |\psi_{2}\rangle$.

On this space the angular momentum operators of subspace $1$ and $2$ acts as follows: $\mathcal{J}_{1i} = J_{1i}\otimes I_{2}$ and $\mathcal{J}_{2i} = I_{1}\otimes J_{2i}$ where $J_{ai}$ are the angular momentum operators on $\mathscr{H}_{a}$ and $I_{a}$ is the identity on $\mathscr{H}_{a}$. We can define operators of the total angular momentum as the sum of these: $$\mathcal{J}_{i} := J_{1i}\otimes I_{2}+ I_{1}\otimes J_{2i}.$$

Now we get to your question. $J_{az}$ are special because we use their eigenstates as a basis in $\mathscr{H}_{a}$. In other words, in the bases $\{|j_{1}, m_{1}\rangle\}$ and $\{|j_{2}, m_{2}\rangle\}$, the operators $J_{az}$ are *diagonal*. Now let's see how $\mathcal{J}_{z}$ acts on the tensor product state $|j_{1}, m_{1}\rangle \otimes |j_{2}, m_{2}\rangle$:
$$\mathcal{J}_{z}|j_{1}, m_{1}\rangle \otimes |j_{2}, m_{2}\rangle = \left(J_{1z}\otimes I_{2}+ I_{1}\otimes J_{2z} \right)|j_{1}, m_{1}\rangle\otimes |j_{2}, m_{2}\rangle \\ \qquad \qquad \quad = (m_{1} + m_{2})|j_{1}, m_{1}\rangle \otimes |j_{2}, m_{2}\rangle $$
so we see that $\mathcal{J}_{z}$ is diagonal in this basis. Hence people loosely write $\mathcal{J}_{z} = ``J_{1z} + J_{2z}''$ but it is just a shortcut for
$$\mathcal{J}_{i} := J_{1i}\otimes I_{2}+ I_{1}\otimes J_{2i}.$$

This will not work in the same way for $\mathcal{J}_{x, y}$ because the individual operators $J_{ax, y}$ are not diagonal. Moreover, $\mathcal{J}^{2}$ *will not* be diagonal, and its form is more complicated, $\mathcal{J}^{2} \mathbf{\boldsymbol \neq }J_{1}^{2} \otimes I_{2} + I_{1} \otimes J_{2}^{2}$ which I leave as an exercise to show.