Stern-Gerlach experiment

In the Stern Gerlach experiment, one can determine the value of $$j$$ (eigenvalue of $$J^2$$) by counting the number of discrete lines formed on the screen. For instance, if I count $$7$$ discrete lines on the screen then I can use the equation $$2j+1=7$$ to determine the value of $$j$$. Solving the equation gives me $$j=3$$. So the value of $$m$$ (eigenvalue of $$J_z$$) will be $$-3,-2,-1,0,+1,+2,+3$$. Clearly, all the lines are formed due to orbital angular momentum because the value of $$m$$ is an integer.

But what if some of the lines are formed due to spin angular momentum as well?

For example, let the value of $$l$$ (eigenvalue of $$L^2$$) be $$2$$. This provides me $$2 \times 2+1=5$$ discrete lines on the screen. The particles used in the experiment also have $$s$$ (eigenvalue of $$S^2$$) being equal to $$\frac{1}{2}$$. So the spin of the particles provides me with $$2\times\frac{1}{2}+1=2$$ discrete lines on the screen. In total, I will again have $$7$$ discrete lines ($$5$$ due to $$L$$ and $$2$$ due to $$S$$).

My question is, how will I know in the later case that the value of $$j$$ is both $$2$$ and $$\frac{1}{2}$$ but not $$j=3$$ (as calculated in the former case)?

• to be clear: the eigenvalues of $J^2$ are $j(j+1)$ not $j$. Aug 13, 2022 at 14:47

The lines are formed by the magnetic field coupling to the total angular momentum $$J$$, not separately to spin and orbital angular momentum. The way that angular momenta of components add gets quite complicated. The rule is that the $$z$$-components add ($$m_j = m_\ell + m_s$$). The rule for the total angular momentum is that $$j$$ can take take any value that you can form into a triangle with $$\ell$$ and $$s$$. In the particular example you give, $$j = 5/2$$ or $$3/2$$. If the atom is in a state of $$j=3/2$$ then you'll see 4 lines. If it's in any other state (mixed or pure $$j=5/2$$) then you'll see 6 lines.
I agree with Sean E. Lake. You need the total angular momentum $$J=L+S$$ for some atom of interest. Number of blobs is then $$2J+1$$. For ground state atoms, values of $$J$$ are tabulated as term symbols: https://en.wikipedia.org/wiki/Term_symbol
For example, $$\mathrm{Ag}$$ has $$J=1/2$$ $$\to$$ so 2 blobs. Fe has $$J=4$$ $$\to$$ that would give you 9 blobs. Ca has $$J=0$$ $$\to$$ only one central blob, i.e. no SG splitting of the beam.
By including spin, one does not add two lines but doubles the number of lines. Basically every $$m$$ state is split in two states with $$m+1/2$$ and $$m-1/2$$. Thus in your example with $$\ell=2$$ and $$s=1/2$$, one obtains $$10$$ states.
• How 10? For l=2, the values of $m_l$ are -2,-1,0,1,2 and for l=1/2, the values of $m_l$ are +1/2 and -1/2. Total number of states is 7 Aug 14, 2022 at 11:44
• for each $m_l$ values you get 2 states: spin up and spin down. Thus $2\times 5=10$. Aug 14, 2022 at 11:52