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 $j = j_1+j_2$?

  • $\begingroup$ What does your textbook (Sakurai) say? $\endgroup$ Dec 27, 2019 at 22:03
  • $\begingroup$ Linked. $\endgroup$ Dec 28, 2019 at 2:52
  • 2
    $\begingroup$ I do not understand the last part of your question - even for ordinary vectors outside of quantum mechanics, unless they are parallel, the magnitude of their sum is not the sum of their magnitudes! $\endgroup$
    – ACuriousMind
    Dec 28, 2019 at 11:09
  • $\begingroup$ Related : Total spin of two spin-1/2 particles $\endgroup$
    – Frobenius
    Jan 19, 2020 at 19:56

3 Answers 3


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 $m_x \neq m_{x_1} + m_{x_2}$.

  • $\begingroup$ Thanks for your answer, @SuperCiocia. Another piece I've been a bit troubled by but forgot to add into my original question is this: if it's just vector addition, then why isn't it true that $j = j_1+j_2$ or that $j(j+1) = j_1(j_1+1) + j_2(j_2+1)$ or something like that? There must be more going on? And $\hat{J}^2$ commutes with $\hat{J}_{1,2}^2$. Thanks for your thoughts. $\endgroup$ Dec 28, 2019 at 2:42
  • 1
    $\begingroup$ Magnitudes don’t just sum. You have to take the magnitude of the new vector. The total angular momentum commutes with the angular momenta of the subsystems, yes. $\endgroup$
    – SuperCiocia
    Dec 28, 2019 at 3:26

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


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:


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:


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.

  • $\begingroup$ Thanks for your answer, Cryo. It was long, but that's exactly what I needed. Based on your explanation, it might be tempting to think that the magnitude of $J_{12}$ is equal to the length of $J_1+J_2$, i.e. that $j_{12} = j_1+j_2$. (This was another confusion I had but forgot to include originally in my question.) Do you have any thoughts on why this isn't true? Thanks again. $\endgroup$ Dec 28, 2019 at 2:43
  • 1
    $\begingroup$ @flevinBombastus. Glad it helps. Regarding the $j?j_1+j_2$, I would say that it is not true because it is not. In the description I gave above the j-values, as well as m-values are secondary. What matters is the irreducible sub-spaces of the Hilbert space. Based on this you can see that $j\neq j_1 + j_2$ in general. Lets say $j_1=3$ and $j_2=4$, then the representation theory of SO(3) readily suggests that the dimensionality of $\mathcal{H}_1$ is 7 i.e. ($2j_1+1$), and of $\mathcal{H}_2$ is 9. Multilinear algebra then stipulates that dimensionality of $\mathcal{H}_1\otimes\mathcal{H_2}$ is ... $\endgroup$
    – Cryo
    Dec 28, 2019 at 3:34
  • 1
    $\begingroup$ ... $7\times9=49$. Next consider $j_1+j_2=7$, the representation theory of SO(3) stipulates that dimensionality of Hilbert space with such $j$ would be $15$, but $15 \neq 49$, so we have a contradiction. The multilinear algebra part is not wrong (easy to check), so the problem is with $j\neq j_1 + j_2$ $\endgroup$
    – Cryo
    Dec 28, 2019 at 3:38
  • 2
    $\begingroup$ I would suggest looking up treatments of representations of SO(3) that do not invoke "physical reasoning" and classical mechanics as a justification for, what in reality is, a purely mathematical tranfsormation. This was an eye-opener for me. For example "Lie Groups, Lie Algebras, and Representations " by Brian Hall. Also Weinberg's "Lectures on Quantum Mechnics" (Ch 9) offers a similar approach (symmetries first, classical mechanics later). Another one is Ballentine "Quantum Mechanics: A Modern Development" (Ch3) $\endgroup$
    – Cryo
    Dec 28, 2019 at 3:47
  • $\begingroup$ I know I'm not supposed to just comment to thank, but this is really great, so—thanks so much, @Cryo! $\endgroup$ Dec 28, 2019 at 21:02

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.

  • $\begingroup$ I add that \mathcal J^2$ can be diagonalise by a change of basis and the relationship between the two basis is given by writing states of one as a linear combination of the other with the famous Clebsch-Gordon coefficients $\endgroup$
    – nox
    Jan 1, 2020 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.