orbital angular momentum of the silver atom

In a silver atom, the first 46 electrons are all paired and according to David McIntyre in Quantum Mechanics,

The electrons in the closed shells can be represented by a spherically symmetric cloud with no orbital or spin angular momentum.

How do we show that the total orbital angular momentum for a spherically symmetric distribution is zero - using geometry?

furthermore, does this configuration of angular momentum - orbital angular momentum - change if the atoms are subject to an inhomogeneous magnetic field, like in the Stern Gerlach experiment? If not then why?

For a full shell, the addition of the expectation values of any angular momentum $L_i$ is zero, and similarly for the spin operators $\sigma_i$. This is not hard to see - for $l(l+1)$ as the expectation value of $L^2$ for a s,p,d,f subshell, the basis of that subshell is spanned by states indexed by integers between $-l$ and $l$, and since that l is also the actual value for the angular momentum in a given direction, full occupancy of these states implies vanishing total angular momentum. Even easier for spin - there are only up and down states, and equally many of them in every shell, so if they're all occupied, the net spin is zero.
• @ApprenticeHacker: There are $2l+1$ states in every subshell of constant $l$. If you evaluate an angular momentum operator upon one of them, the result is just the value between $-l$ and $l$ that labels the state. Now, given two electrons, two of these states are realized, and the total angular momentum is just the sum of their labels (because the label $m_l \in \{-l,\dots,l\}$ is defined as the eigenvalue of one of the $L_i$). Given a full subshell, the total angular momentum is just the sum of all labels, which is zero. – ACuriousMind Jan 28 '15 at 19:30
• This is not QM addition of probabilities, it is realizing that for two systems $H_1,H_2$ the total angular momentum upon the combined space $H_1 \otimes H_2$ is given by $L \otimes 1 + 1 \otimes L$, but I got the impression that I should not use the tensor space formalism to answer your question, so I tried to formulate it in natural language. – ACuriousMind Jan 28 '15 at 19:32
• Even though this answer have been accepted, be aware that there are a lot of handwaving arguments here. "If you evaluate an angular momentum operator upon one of them, the result is just the value between $-\ell$ and $\ell$ that labels the state." @ACuriusMind is probably talking about the projection of L. In the next sentence the total L and is projection is mixed. – yarchik Jan 19 '16 at 10:45