quantum mechanics - Adding Angular Momenta Operators in QM

Consider $j,m$ to be the angular momentum magnitude and $z$-projection eigenvalues corresponding to a total angular momentum operator $\hat{J}$, composed of angular momentum $\hat{J}_1$ and $\hat{J}_2$ with eigenvalues $j_1,m_1$ and $j_2,m_2$. We want to know what values $j$ and $m$ can take on in terms of $j_1,m_1,j_2,m_2$. It is commonly stated that

$$\hat{J}_z = \hat{J}_{1z} + \hat{J}_{2z},$$

from which one can immediately derive $m = m_1+m_2$.

What is the explanation for the simple addition of the $z$ operators? If there is some vectorial model explanation, then is it also true that $\hat{J}_x = \hat{J}_{1x} + \hat{J}_{2x}$, for example? Is there some other way to prove this?

Further, if we are looking at a vectorial model, why isn't it true that the magnitudes are the same, i.e. that $j = j_1+j_2$?

Answer

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.