Question with this heading has already been asked in this site but they have nothing to do with the particular thing I am asking.You can suggest me a better heading.Now to the question. Just because for a given angular momentum state $\ell$ there are $-\ell,...,\ell$ possible magnitudes at a particular direction (and they add to zero) doesn't mean the total angular momentum of the system is zero! There are still angular momentum components in other two directions (although can't be measured simultaneously) and they would not necessarily add up to zero. So what is happening here?

  • $\begingroup$ hint:i think one has to look up the antisymmetric nature of the filled up states and the clue lies there...these states are invariant under rotations. $\endgroup$
    – drvrm
    Mar 24, 2018 at 19:40
  • $\begingroup$ If I describe a state by saying that it has all the $\ell=3$ single-particle states filled, then I've given a full description of a uniquely defined state, and I've done it without even referring to a coordinate system. That guarantees that the state is spherically symmetric and has angular momentum zero. So if there is a mystery, it's not why the total angular momentum is zero, it's how this plays out when you write it in a particular $(\ell,m)$ basis. $\endgroup$
    – user4552
    Mar 25, 2018 at 14:51

2 Answers 2


The angular momentum of a quantum state $|\Psi\rangle$ is zero if and only if $J_z|\Psi\rangle = J_+|\Psi\rangle = J_- |\Psi\rangle = 0$, where $J_z$ and $J_{\pm}$ are associated with the total angular momentum.

In the example OP is interested in, $|\Psi\rangle$ is the many-electron state of the closed shell, which can be written as \begin{equation} |\Psi\rangle \propto\sum_{P}(-1)^P |l, \uparrow\rangle_{P(1)} \otimes |l, \downarrow\rangle_{P(2)}\otimes |l-1,\uparrow\rangle_{P(3)} \otimes \cdots\otimes |-l, \downarrow\rangle_{P(4l+2)}, \end{equation} where $P$ denotes permutation.

It is easy to see that $J_z|\Psi\rangle = 0$ because each term on the above has zero total angular momentum. To prove that $J_\pm|\Psi\rangle = 0$, let's consider an arbitrary term on the RHS of the above. Here, an attempt to raise or lower the orbital or spin angular momentum of any electron will result in two electrons having exactly the same set of quantum numbers. Summing over such a term over all possible permutations with the sign $(-1)^P$ will yield zero.

(In fact, one can prove that the total orbital angular momentum and the total spin angular momentum separately vanish.)


Something that is spherically symmetric has zero angular momentum. Note that:

$$\sum_{m=-1}^l{|Y_l^m(\theta, \phi)|^2}=\frac{2l+1}{4\pi}$$

The L.H.S represents the total probability distribution. It is the incoherent summation over the probability distributions of individual states in a closed shell (fixed $l$, all $m$ filled).

The R.H.S has no angular dependence, so a closed shell is spherically symmetric. It is a scalar--invariant under rotations. (Moreover, it represents the uniform distribution of $2l+1$ particles over $4\pi$ steradians.)


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