1
$\begingroup$

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)?

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

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – user231188
    Aug 14, 2022 at 11:44
  • $\begingroup$ for each $m_l$ values you get 2 states: spin up and spin down. Thus $2\times 5=10$. $\endgroup$ Aug 14, 2022 at 11:52

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.